UP Board Solutions Class 12 Maths Chapter 5 Continuity and Differentiability

Exercise 5.1

Question 1. सिद्ध कीजिए कि फलन \( f(x) = 5x - 3 \), \( x = 0 \), \( x = -3 \) तथा \( x = 5 \) पर संतत है। हल- यहाँ, \( f(x) = 5x - 3 \)

(i) \( x = 0 \) पर, \( \lim_{x \to 0} f(x) = \lim_{x \to 0} (5x - 3) = 5 \times 0 - 3 = -3 \)
तथा \( f(0) = -3 \)
\( \therefore \lim_{x \to 0} f(x) = -3 = f(0) \)
अतः \( x = 0 \) पर \( f \) संतत है।
(ii) \( x = -3 \) पर, \( \lim_{x \to -3} f(x) = \lim_{x \to -3} (5x - 3) = 5 \times (-3) - 3 \)
\( = -15 - 3 = -18 \)
तथा \( f(-3) = -18 \)
अतः \( x = -3 \) पर \( f \) संतत है।
(iii) उपरोक्त की भाँति
स्वयं हल कीजिए। नोट-चूँकि दिया गया फलन \( f(x) = 5x - 3 \) बहुपद है । \( \therefore \) यह \( \forall x \in R \) के लिए संतत है।
Answer: The function \( f(x) = 5x - 3 \) is continuous at \( x=0, x=-3 \) and \( x=5 \).
In simple words: A polynomial function is always continuous. Since \( f(x) = 5x - 3 \) is a polynomial, it is continuous for all real values of \( x \), including \( 0, -3 \), and \( 5 \).

🎯 Exam Tip: When proving continuity for polynomial functions, it's often sufficient to state that all polynomial functions are continuous over their domain, but explicitly checking limits and function values at specific points demonstrates a thorough understanding.

 

Question 2. \( x = 3 \) पर फलन \( f(x) = 2x^2 - 1 \) के सातत्य की जाँच कीजिए। हल- यहाँ, \( f(x) = 2x - 1 \)
\( x = 3 \) पर, \( \lim_{x \to 3} f(x) = \lim_{x \to 3} (2x^2 - 1) = 2 \times 9 - 1 = 18 - 1 = 17 \)
तथा \( f(3) = 17 \)
\( \lim_{x \to 3} = 17 = f(3) \)
अतः \( x = 3 \) पर \( f \) संतत है।
Answer: The function \( f(x) = 2x^2 - 1 \) is continuous at \( x = 3 \).
In simple words: To check continuity at a point, we compare the function's value at that point with the limit of the function as \( x \) approaches that point. Since both are equal (17), the function is continuous at \( x=3 \).

🎯 Exam Tip: For polynomial functions, evaluating the limit directly by substituting the value of \( x \) is a common and correct approach to establish continuity at a point.

 

Question 3. निम्नलिखित फलनों के सांतत्य की जाँच कीजिए
(a) \( f(x) = x - 5 \)
(b) \( f(x) = \frac{1}{x - 5} \)
(c) \( f(x) = \frac{x^2 - 25}{x + 5}, x \neq 5 \)
(d) \( f(x) = |x - 5| \)
हल-
(a) \( f(x) = x - 5 \)
\( \because (x - 5) \) एक बहुपद है।
अतः प्रत्येक बिन्दु \( x \in R \) पर \( f \) संतत है।
(b) \( f(x) = \frac{1}{x - 5} \)
चूँकि \( x = 5 \) पर \( f(x) \) परिभाषित नहीं है।
अतः \( x = 5 \) पर \( f \) संतत नहीं है। लेकिन \( x \in R - \{5\} \) के प्रत्येक बिन्दु पर \( f \) संतत है।
(c) \( f(x) = \frac{x^2 - 25}{x + 5} \)
\( \therefore f(-5) = \frac{25 - 25}{-5 + 5} = \frac{0}{0} \)
(i) \( x = -5 \) पर \( f \) परिभाषित नहीं है।
\( \therefore x = -5 \) पर \( f \) संतत भी नहीं है।
(ii) \( x \neq -5 \),
माना \( x = c \neq -5 \)
\( \lim_{x \to c} f(x) = \lim_{x \to c} \frac{x^2 - 25}{x + 5} \)
\( = \lim_{x \to c} \frac{(x - 5)(x + 5)}{(x + 5)} = \lim_{x \to c} (x - 5) \)
\( = c - 5 \)
अतः \( x \in R - \{-5\} \) के प्रत्येक बिन्दु पर \( f \) संतत है।
(d) \( f(x) = |x - 5| \)
\( x = 5 \) पर, \( \lim_{x \to 5} f(x) = \lim_{x \to 5} |x - 5| = 0 \)
तथा \( f(5) = 0 \)
\( \therefore \lim_{x \to 5} = 0 = f(5) \)
\( \therefore x = 5 \) पर \( f \) संतत है।
पुनः यदि \( x > 5 \),
माना \( x = c > 5 \)
\( \lim_{x \to c} f(x) = \lim_{x \to c} |x - 5| = c - 5 = f(c) \) [\( \because c > 5 \)]
\( \therefore \) जब \( x > 5 \), \( f \) संतत है।
यदि \( x < 5 \),
माना \( x = c < 5 \)
\( \lim_{x \to c} f(x) = \lim_{x \to c} |x - 5| = -(c - 5) = 5 - c = f(c) \)
\( \therefore \) जब \( x < 5 \) तब \( f \) संतत है।
अतः \( x \in R \), प्रत्येक बिन्दु पर \( f \) संतत है।
Answer:
(a) \( f(x) = x-5 \) is continuous everywhere.
(b) \( f(x) = \frac{1}{x-5} \) is continuous for all \( x \in R, x \neq 5 \).
(c) \( f(x) = \frac{x^2-25}{x+5} \) is continuous for all \( x \in R, x \neq -5 \).
(d) \( f(x) = |x-5| \) is continuous everywhere.
In simple words: Polynomials and absolute value functions are continuous everywhere. Rational functions are continuous where they are defined. Discontinuities occur where the denominator is zero, or the function is undefined.

🎯 Exam Tip: For piecewise functions, especially those involving absolute values or rational forms, remember to check for continuity at the boundary points and any points where the function might be undefined (e.g., division by zero).

 

Question 4. सिद्ध कीजिए कि फलन \( f(x) = x^n \), \( x = n \) पर संतत है, जहाँ \( n \) एक धन पूर्णाक है।
हल-
दिया है- \( f(x) = x^n \) एक बहुपदीय फलन है।
संतत है, यदि \( x \in R \) तथा \( x \in N \)
यहाँ \( x = n \) एक पूर्णांक है।
अतः \( f(x) = x^n \) संतत फलन है।
Answer: The function \( f(x) = x^n \) is continuous at \( x = n \), where \( n \) is a positive integer.
In simple words: Since \( f(x) = x^n \) is a polynomial function (as \( n \) is a positive integer), it is continuous for all real numbers. Thus, it is continuous at \( x=n \).

🎯 Exam Tip: Recognizing standard continuous functions like polynomials simplifies continuity proofs significantly. Always mention the domain over which the function is continuous.

 

Question 5. क्या \( f(x) = \begin{cases} x, \text{ यदि } x \leq 1 \\ 5, \text{ यदि } x > 1 \end{cases} \) द्वारा परिभाषित फलन \( f, x = 0, x = 1 \) तथा \( x = 2 \) पर संतत है।
हल-
\( x = 0 \) तथा \( x = 2 \) पर फलन एक बहुपद है।
अतः \( x = 0 \) तथा \( x = 2 \) पर फलन सतत है।
\( x = 1 \) पर, \( f(x) = x, x < 1 \) के लिए, \( f(x) = 5, x > 1 \) के लिए
L.H.L. \( = \lim_{x \to 1^-} f(x) = \lim_{x \to 1^-} (x) = 1 \)
तथा R.H.L. \( = \lim_{x \to 1^+} f(x) = \lim_{x \to 1^+} (5) = 5 \)
\( \therefore \lim_{x \to 1^-} f(x) \neq \lim_{x \to 1^+} f(x) \)
Answer: The function \( f(x) \) is continuous at \( x=0 \) and \( x=2 \), but it is not continuous at \( x=1 \).
In simple words: For \( x=0 \) and \( x=2 \), the function behaves like a polynomial, hence continuous. At \( x=1 \), the left-hand limit is 1, and the right-hand limit is 5. Since these are not equal, the function is discontinuous at \( x=1 \).

🎯 Exam Tip: For piecewise functions, always test the continuity at the boundary points (where the function definition changes) by comparing the left-hand limit, right-hand limit, and the function's value at that point.

 

Question 6. \( f \) के सभी असातत्य बिन्दुओं को ज्ञात कीजिए जबकि \( f \) निम्नलिखित प्रकार से परिभाषित है।
\( f(x) = \begin{cases} 2x + 3, \text{ यदि } x \leq 2 \\ 2x - 3, \text{ यदि } x > 2 \end{cases} \)
हल-
ज्ञात है- \( f(x) = \begin{cases} 2x + 3, \text{ यदि } x \leq 2 \\ 2x - 3, \text{ यदि } x > 2 \end{cases} \)
(i) \( x = 2 \) पर तथा जब \( x < 2 \) हो तब \( f(x) = 2x + 3 \)
L.H.L. \( = \lim_{x \to 2^-} f(x) = \lim_{h \to 0} f(2 - h) \)
\( = \lim_{h \to 0} [2(2 - h) + 3] = \lim_{h \to 0} [-2h + 7] = 7 \)
R.H.L. \( = \lim_{x \to 2^+} f(x) = \lim_{h \to 0} f(2 + h) \)
\( = \lim_{h \to 0} [2(2 + h) - 3] = \lim_{h \to 0} [4 + 2h - 3] = 1 \)
\( \implies \) L.H.L. \( \neq \) R.H.L.
अतः \( x = 2 \) पर फलन \( f \) सतत् नहीं है।
\( \because x \neq 2 \) पर फलन बहुपदी द्वारा परिभाषित है
अतः \( x \neq 2 \) के अतिरिक्त फलन सभी बिन्दुओं पर सतत् है।
Answer: The function \( f \) is discontinuous at \( x = 2 \).
In simple words: The function changes definition at \( x=2 \). We check the left and right limits. Since \( \lim_{x \to 2^-} f(x) = 7 \) and \( \lim_{x \to 2^+} f(x) = 1 \), the limits are not equal, so the function is discontinuous at \( x=2 \). For all other points, it's a polynomial, hence continuous.

🎯 Exam Tip: Discontinuity for piecewise functions often occurs at the points where the definition changes. Always calculate LHL, RHL, and \( f(a) \) at these critical points to determine continuity.

 

Question 7. \( f \) के सभी असातत्य बिन्दुओं को ज्ञात कीजिए जबकि \( f \) निम्नलिखित प्रकार से परिभाषित है।
\( f(x) = \begin{cases} |x| + 3, \text{ यदि } x \leq -3 \\ -2x, \text{ यदि } -3 < x < 3 \\ 6x + 2, \text{ यदि } x \geq 3 \end{cases} \)
हल-
ज्ञात है- \( f(x) = \begin{cases} |x| + 3, \text{ यदि } x \leq -3 \\ -2x, \text{ यदि } -3 < x < 3 \\ 6x + 2, \text{ यदि } x \geq 3 \end{cases} \)
दिया गया फलन वास्तविक संख्या -3 व 3 के अतिरिक्त सभी बिन्दुओं पर सतत् है। माना वास्तविक संख्या रेखा पर कोई बिन्दु नहीं है।
-3 व 3 पर सातत्यता की जाँच करते हैं-
दशा (i) \( x = -3 \) पर,
\( f(x) = |x| + 3, \text{ यदि } x \leq -3 \)
L.H.L. \( = \lim_{x \to -3^-} f(x) = \lim_{h \to 0} f(-3 - h) = \lim_{h \to 0} (|-3 - h| + 3) \)
\( = \lim_{h \to 0} (3 + h) + 3 = 3 + 0 + 3 = 6 \)
जब \( x > -3 \),
\( f(x) = -2x \)
R.H.L. \( = \lim_{x \to -3^+} f(x) = \lim_{h \to 0} f(-3 + h) = \lim_{h \to 0} [-2(-3 + h)] \)
\( = \lim_{h \to 0} (6 - 2h) = 6 - 0 = 6 \)
जब \( x = -3 \),
\( f(x) = |x| + 3 \)
\( f(-3) = |-3| + 3 = 3 + 3 = 6 \)
इस प्रकार, \( \lim_{x \to -3^-} f(x) = \lim_{x \to -3^+} f(x) = f(-3) \)
अतः L.H.L. = R.H.L. = \( f(-3) \)
अतः \( x = -3 \) पर फलन \( f \) संतत है।
दशा (ii) \( x = 3 \) पर, जब \( x < 3 \), \( f(x) = -2x \)
L.H.L. \( = \lim_{x \to 3^-} f(x) = \lim_{h \to 0} f(3 - h) = \lim_{h \to 0} [-2(3 - h)] \)
\( = \lim_{h \to 0} (-6 + 2h) = -6 + 0 = -6 \)
जब \( x > 3 \),
\( f(x) = 6x + 2 \)
R.H.L. \( = \lim_{x \to 3^+} f(x) = \lim_{h \to 0} f(3 + h) = \lim_{h \to 0} [6(3 + h) + 2] \)
\( = 6(3 \times 0) + 2 = 20 \)
अतः \( \lim_{x \to 3^-} f(x) \neq \lim_{x \to 3^+} f(x) \implies \) L.H.L. \( \neq \) R.H.L.
अतः \( x = 3 \) पर फलन \( f \) सतत् नहीं है।
दशा (iii) \( x = c < -3 \), तब \( f(c) = |-c| + 3 = -c + 3 \)
\( \lim_{x \to c} f(x) = \lim_{x \to c} (|x| + 3) = |-c| + 3 = -c + 3 = f(c) \)
\( \therefore x < -3 \) से छोटी सभी बिन्दुओं पर सतत् है।
दशा (iv) यदि \( -3 < c < 3 \), तब \( f(c) = -2c \)
\( \implies \lim_{x \to c} f(x) = \lim_{x \to c} (-2x) = -2c = f(c) \)
अतः \( x = c, -3 < c < 3 \) पर फलन \( f \) सतत् है।
दशा (v) \( x = c > 3 \), तब \( f(c) = 6c + 2 \)
और \( \lim_{x \to c} f(x) = \lim_{x \to c} (6x + 2) = (6c + 2) \)
अतः \( x = c > 3 \) पर फलन \( f \) सतत् है।
अतः फलन केवल \( x = 3 \) पर असतत् है।
Answer: The function \( f \) is discontinuous at \( x=3 \).
In simple words: We check continuity at the boundary points \( x=-3 \) and \( x=3 \). At \( x=-3 \), LHL, RHL, and \( f(-3) \) are all 6, so it's continuous. At \( x=3 \), LHL is -6 and RHL is 20, so it's discontinuous. For other regions, it's a polynomial or absolute value, hence continuous.

🎯 Exam Tip: Pay close attention to the definition of absolute value functions: \( |x| = x \) if \( x \geq 0 \) and \( |x| = -x \) if \( x < 0 \). Apply this carefully when evaluating limits involving absolute values.

 

Question 8. \( f \) के सभी असातत्य बिन्दुओं को ज्ञात कीजिए जबकि \( f \) निम्नलिखित प्रकार से परिभाषित है।
\( f(x) = \begin{cases} \frac{x}{|x|}, \text{ यदि } x < 0 \\ -1, \text{ यदि } x \geq 0 \end{cases} \)
हल-
ज्ञात है, \( f(x) = \begin{cases} \frac{x}{|x|}, \text{ यदि } x < 0 \\ -1, \text{ यदि } x \geq 0 \end{cases} \)
(i) \( x = 0 \) पर, जब \( x < 0 \), \( f(x) = \frac{x}{|x|} \) [\( \because |x| = -x \) जब \( x < 0 \)]
L.H.L. \( = \lim_{x \to 0^-} f(x) = \lim_{x \to 0^-} \frac{x}{|x|} = \lim_{x \to 0^-} \frac{x}{-x} = -1 \)
जब \( x > 0 \),
\( f(x) = -1 \)
R.H.L. \( = \lim_{x \to 0^+} f(x) = \lim_{x \to 0^+} (-1) = -1 \)
तथा \( f(0) = -1 \)
\( \therefore \lim_{x \to 0^-} f(x) = \lim_{x \to 0^+} f(x) = f(0) \)
अतः \( x = 0 \) पर फलन \( f \) संतत है।
(ii) \( x = c < 0 \) पर,
\( f(x) = \frac{x}{|x|} = \frac{x}{-x} = -1 \)
अतः \( \lim_{x \to c} f(x) = \lim_{x \to c} (-1) = -1 = f(c) \)
(iii) \( x = c > 0 \),
\( f(x) = -1 \)
\( \lim_{x \to c} f(x) = \lim_{x \to c} (-1) = -1 = f(c) \)
\( \therefore x = c > 0 \) पर फलन \( f \) संतत है।
अतः दिया हुआ फलन \( x \in R \), सभी बिन्दुओं पर संतत हैं।
Answer: The function \( f(x) \) is continuous at all points \( x \in R \). There are no points of discontinuity.
In simple words: At \( x=0 \), the left-hand limit is -1 (since \( |x| = -x \) for \( x<0 \)), the right-hand limit is -1, and \( f(0) = -1 \). All are equal, so it's continuous at \( x=0 \). For \( x<0 \), \( f(x) = x/(-x) = -1 \), which is continuous. For \( x>0 \), \( f(x)=-1 \), which is continuous. Therefore, the function is continuous everywhere.

🎯 Exam Tip: When simplifying functions involving \( |x| \), remember to define \( |x| \) as \( x \) for \( x \geq 0 \) and \( -x \) for \( x < 0 \). This is crucial for correctly evaluating limits from different sides.

 

Question 9. \( f(x) = \begin{cases} x+1, \text{ यदि } x \geq 1 \\ x^2+1, \text{ यदि } x < 1 \end{cases} \)
हल-
ज्ञात है- \( f(x) = \begin{cases} x+1, \text{ यदि } x \geq 1 \\ x^2+1, \text{ यदि } x < 1 \end{cases} \)
\( \therefore \) फलन \( x > 1 \) तथा \( x < 1 \) के लिए बहुपदीय फलन है
\( \therefore \) यह \( x > 1 \) तथा \( x < 1 \) के \( R \) के प्रत्येक मान के लिए सतत है।
(i) \( x = 1 \) पर, \( f(x) = x^2 + 1 \), जब \( x < 1 \)
L.H.L. \( = \lim_{x \to 1^-} f(x) = \lim_{x \to 1^-} (x^2 + 1) = 1^2 + 1 = 2 \)
जब \( x > 1 \)
R.H.L. \( = \lim_{x \to 1^+} f(x) = \lim_{x \to 1^+} (x + 1) = 1 + 1 = 2 \)
जब \( x = 1 \),
\( f(1) = 1 + 1 = 2 \)
\( \therefore \lim_{x \to 1^-} f(x) = \lim_{x \to 1^+} f(x) = f(1) \)
अतः \( x = 1 \) पर फलन \( f \) संतत है
अतः \( x \in R \), \( f \) के सभी बिन्दुओं पर संतत है।
Answer: The function \( f(x) \) is continuous at all points \( x \in R \).
In simple words: The function changes definition at \( x=1 \). The left-hand limit is \( 1^2+1=2 \), the right-hand limit is \( 1+1=2 \), and \( f(1)=1+1=2 \). Since all three values are equal, the function is continuous at \( x=1 \). For all other points, it is a polynomial, thus continuous.

🎯 Exam Tip: When the function definition changes, explicitly calculate and compare LHL, RHL, and the function value at that point. If they are equal, the function is continuous; otherwise, it's discontinuous at that point.

 

Question 10. \( f(x) = \begin{cases} x^3 - 3, \text{ यदि } x \leq 2 \\ x^2 + 1, \text{ यदि } x > 2 \end{cases} \)
हल-
ज्ञात है- \( f(x) = \begin{cases} x^3 - 3, \text{ यदि } x \leq 2 \\ x^2 + 1, \text{ यदि } x > 2 \end{cases} \)
\( \therefore \) फलन \( x > 2 \) तथा \( x < 2 \) के लिए बहुपदीय फलन है।
अतः \( x < 2 \) तथा \( x > 2 \in R \) के प्रत्येक मान के लिए फलन सतत है।
(i) \( x = 2 \) पर, \( f(x) = x^3 - 3 \), जब \( x < 2 \)
L.H.L. \( = \lim_{x \to 2^-} f(x) = \lim_{x \to 2^-} (x^3 - 3) = 2^3 - 3 = 8 - 3 = 5 \)
जब \( x > 2 \),
R.H.L. \( = \lim_{x \to 2^+} f(x) = \lim_{x \to 2^+} (x^2 + 1) = 2^2 + 1 = 4 + 1 = 5 \)
जब \( x = 2 \),
\( f(2) = (2)^3 - 3 = 8 - 3 = 5 \)
\( \therefore \lim_{x \to 2^-} f(x) = \lim_{x \to 2^+} f(x) = f(2) \)
अतः \( x = 2 \) पर फलन \( f \) संतत है।
अतः फलन \( f, x \in R \) सभी बिन्दुओं पर संतत है।
Answer: The function \( f(x) \) is continuous at all points \( x \in R \).
In simple words: The function changes its definition at \( x=2 \). We calculate the left-hand limit as \( 2^3-3=5 \), the right-hand limit as \( 2^2+1=5 \), and \( f(2)=2^3-3=5 \). Since all three values are equal, the function is continuous at \( x=2 \). In all other regions, it's a polynomial, hence continuous.

🎯 Exam Tip: For piecewise functions, when the LHL, RHL, and function value at the transition point are all equal, the function is continuous at that point. If the function is composed of continuous parts (like polynomials) elsewhere, then it's continuous over its entire domain.

 

Question 11. \( f(x) = \begin{cases} x^{10} - 1, \text{ यदि } x \leq 1 \\ x^2, \text{ यदि } x > 1 \end{cases} \)
हल-
ज्ञात है- \( f(x) = \begin{cases} x^{10} - 1, \text{ यदि } x \leq 1 \\ x^2, \text{ यदि } x > 1 \end{cases} \)
दिया गया फलन वास्तविक संख्या रेखा के सभी बिन्दुओं पर परिभाषित है। माना वास्तविक संख्या रेखा पर कोई बिन्दु \( c \) है।
दशा I \( x = 1 \) पर, \( f(x) = x^{10} - 1 \), जब \( x < 1 \)
L.H.L. \( = \lim_{x \to 1^-} f(x) = \lim_{x \to 1^-} (x^{10} - 1) = 1^{10} - 1 = 0 \)
\( f(x) = x^2 \), जब \( x > 1 \)
R.H.L. \( = \lim_{x \to 1^+} f(x) = \lim_{x \to 1^+} (x^2) = 1^2 = 1 \)
\( x = 1 \) पर,
\( \therefore \) L.H.L. \( \neq \) R.H.L.
अतः \( x = 1 \) पर फलन \( f \) असतत् है।
दशा II \( x = c < 1 \) पर,
\( f(x) = x^{10} - 1 \)
\( \lim_{x \to c} f(x) = \lim_{x \to c} (x^{10} - 1) = c^{10} - 1 = f(c) \)
\( \therefore \) फलन \( x < 1 \) के सभी बिन्दुओं के लिए सतत् है।
दशा III \( x = c > 1 \) पर,
\( f(x) = x^2 \)
\( \lim_{x \to c} f(x) = \lim_{x \to c} (x^2) = c^2 = f(c) \)
\( \therefore x = c > 1 \) पर \( f \) सतत् है।
\( \implies x \in R - \{1\} \) के लिए फलन \( f \) सतत् है।
अतः \( x = 1 \) पर फलन \( f \) सतत् नहीं है।
Answer: The function \( f(x) \) is discontinuous at \( x=1 \).
In simple words: At \( x=1 \), the left-hand limit is \( 1^{10}-1=0 \), but the right-hand limit is \( 1^2=1 \). Since these are not equal, the function is discontinuous at \( x=1 \). For any other point, the function behaves like a polynomial, making it continuous.

🎯 Exam Tip: Remember to calculate the function value at the critical point \( f(a) \) as well. If LHL \( \neq \) RHL, then \( f(x) \) is discontinuous, regardless of \( f(a) \).

 

Question 12. क्या \( f(x) = \begin{cases} x+5, \text{ यदि } x \leq 1 \\ x-5, \text{ यदि } x > 1 \end{cases} \) द्वारा परिभाषित फलन, एक संतत फलन है?
हल-
ज्ञात है- \( f(x) = \begin{cases} x+5, \text{ यदि } x \leq 1 \\ x-5, \text{ यदि } x > 1 \end{cases} \)
\( \therefore \) फलन \( x < 1 \) तथा \( x > 1 \) के लिए बहुपद द्वारा परिभाषित है।
\( \therefore x < 1 \) व \( x > 1 \) के लिए फलन सतत है।
\( x = 1 \) पर,
\( f(x) = x+5 \), जब \( x < 1 \)
L.H.L. \( = \lim_{x \to 1^-} f(x) = \lim_{x \to 1^-} (x+5) = (1+5) = 6 \)
\( f(x) = x-5 \), जब \( x > 1 \)
R.H.L. \( = \lim_{x \to 1^+} f(x) = \lim_{x \to 1^+} (x-5) = (1-5) = -4 \)
\( \implies \) L.H.L. \( \neq \) R.H.L.
अतः \( x = 1 \) पर फलन \( f \) असंतत है।
अतः \( x \in R - \{1\} \) के सभी बिन्दुओं पर फलन \( f \) संतत है।
Answer: No, the function is not continuous. It is discontinuous at \( x=1 \).
In simple words: The function changes its definition at \( x=1 \). The left-hand limit is \( 1+5=6 \), but the right-hand limit is \( 1-5=-4 \). Since the limits are not equal, the function is discontinuous at \( x=1 \).

🎯 Exam Tip: A function is considered discontinuous if its left-hand limit, right-hand limit, or function value at a point are not all equal. Here, just the LHL and RHL not being equal is enough to conclude discontinuity.

 

Question 13. क्या \( f(x) = \begin{cases} 3, \text{ यदि } 0 \leq x \leq 1 \\ 4, \text{ यदि } 1 < x < 3 \\ 5, \text{ यदि } 3 \leq x \leq 10 \end{cases} \) द्वारा परिभाषित फलन, एक संतत फलन है?
हल-
ज्ञात है, \( f(x) = \begin{cases} 3, \text{ यदि } 0 \leq x \leq 1 \\ 4, \text{ यदि } 1 < x < 3 \\ 5, \text{ यदि } 3 \leq x \leq 10 \end{cases} \)
दशा I \( x = 1 \) पर,
\( f(x) = 3 \), जब \( x \leq 1 \)
\( \therefore x = 1 \) पर,
L.H.L. \( = \lim_{x \to 1^-} f(x) = 3 \), \( f(1) = 3 \)
\( f(x) = 4 \), जब \( x > 1 \)
R.H.L. \( = \lim_{x \to 1^+} f(x) = 4 \)
\( \therefore \lim_{x \to 1^-} f(x) \neq \lim_{x \to 1^+} f(x) \)
अतः \( x = 1 \) पर फलन \( f \) सतत् नहीं है।
दशा II \( x = 3 \) पर,
\( f(x) = 4 \), जब \( 1 < x < 3 \)
\( \therefore x = 3 \) पर,
L.H.L. \( = \lim_{x \to 3^-} f(x) = 4 \)
\( f(x) = 5 \), जब \( x \geq 3 \)
\( \therefore x = 3 \) पर,
R.H.L. \( = \lim_{x \to 3^+} f(x) = 5 \)
\( \implies \) L.H.L. \( \neq \) R.H.L.
अतः \( x = 3 \) पर फलन \( f \) सतत् नहीं है।
दशा III \( x = c, 0 \leq c \leq 1 \) पर, \( f(x) = 3 \), जब \( x < 1 \)
\( \lim_{x \to c} f(x) = 3 \), \( f(c) = 3 \)
अतः \( x = c < 1 \) पर फलन \( f \) सतत् है।
दशा IV \( x = c, 1 < c < 3 \),
\( f(x) = 4 \)
\( \lim_{x \to c} f(x) = 4 \), \( f(c) = 4 \)
अतः \( x = c, 1 < c < 3 \) पर फलन \( f \) सतत् है।
दशा V \( x = c, 3 \leq x \leq 10 \) पर,
\( f(x) = 5 \), जब \( x > 3 \)
\( \lim_{x \to c} f(x) = 5 = f(c) \)
अतः \( x = c, 3 < x \leq 10 \) पर फलन \( f \) सतत् है।
इस प्रकार \( x = 1, x = 3 \) पर फलन \( f \) असतत् है तथा \( x \in R - \{1,3\} \), पर फलन \( f \) सतत् है।
Answer: No, the function is not continuous. It is discontinuous at \( x=1 \) and \( x=3 \).
In simple words: The function changes its definition at \( x=1 \) (LHL=3, RHL=4) and at \( x=3 \) (LHL=4, RHL=5). Since the left and right limits are not equal at these points, the function is discontinuous at \( x=1 \) and \( x=3 \). It is continuous in the open intervals \((0,1)\), \((1,3)\), and \((3,10)\).

🎯 Exam Tip: When a piecewise function has multiple transition points, check the continuity at each point separately. A function is only continuous over its entire domain if it is continuous at every single point within that domain.

 

Question 14. \( f(x) = \begin{cases} 2x, \text{ यदि } x < 0 \\ 0, \text{ यदि } 0 \leq x \leq 1 \\ 4x, \text{ यदि } x > 1 \end{cases} \)
हल-
ज्ञात है- \( f(x) = \begin{cases} 2x, \text{ यदि } x < 0 \\ 0, \text{ यदि } 0 \leq x \leq 1 \\ 4x, \text{ यदि } x > 1 \end{cases} \)
दशा I \( x = 0 \) पर,
\( f(x) = 2x \),
L.H.L. \( = \lim_{x \to 0^-} f(x) = \lim_{x \to 0^-} (2x) = 0 \)
\( f(x) = 0 \), जब \( x \geq 0 \)
R.H.L. \( = \lim_{x \to 0^+} f(x) = 0 \)
तथा \( f(0) = 0 \)
\( \therefore \) R.H.L. = L.H.L. = \( f(0) \)
अतः \( x = 0 \) पर फलन \( f \) सतत् है।
दशा II \( x = 1 \) पर,
\( f(x) = 0 \), जब \( x \leq 1 \),
L.H.L. \( = \lim_{x \to 1^-} f(x) = 0 \)
\( f(x) = 4x \), जब \( x > 1 \)
R.H.L. \( = \lim_{x \to 1^+} f(x) = \lim_{x \to 1^+} (4x) = (4 \times 1) = 4 \)
\( \therefore \) L.H.L. \( \neq \) R.H.L.
अतः \( x = 1 \) पर फलन \( f \) असतत् है।
दशा III \( x = c < 0 \) पर,
\( f(x) = 2x \), जब \( x < 0 \)
\( \lim_{x \to c} f(x) = \lim_{x \to c} (2x) = 2c = f(c) \)
अतः \( x = c < 0 \) पर फलन \( f \) सतत् है।
दशा IV \( x = c_1, 0 < c_1 < 1 \), \( f(x) = 0 \), जब \( 0 < x < 1 \)
\( \lim_{x \to c_1} f(x) = 0 \), और \( f(c_1) = 0 \)
अतः \( x = c_1, 0 < c_1 < 1 \) पर फलन \( f \) सतत् है।
दशा V \( x = c_2 > 1 \),
\( f(x) = 4x \), जब \( x > 1 \)
\( \lim_{x \to c_2} f(x) = \lim_{x \to c_2} (4x) = 4c_2 \), \( f(c_2) = 4c_2 \)
\( \therefore x = c_2 > 1 \) पर फलन \( f \) सतत् है।
अतः \( x = 1 \), पर फलन \( f \) असतत् है।
\( \implies x \in R - \{1\} \) पर फलन \( f \) सतत् है।
Answer: The function \( f(x) \) is discontinuous at \( x=1 \).
In simple words: The function is continuous at \( x=0 \) because LHL, RHL, and \( f(0) \) are all 0. However, at \( x=1 \), the LHL is 0 and the RHL is 4, making it discontinuous. For intervals where the function is defined by a single polynomial, it is continuous.

🎯 Exam Tip: When evaluating continuity at transition points, clearly state the function definition used for LHL, RHL, and the function value itself. This avoids confusion and potential errors.

 

Question 15. \( f(x) = \begin{cases} -2, \text{ यदि } x \leq -1 \\ 2x, \text{ यदि } -1 < x \leq 1 \\ 2, \text{ यदि } x > 1 \end{cases} \)
हल-
ज्ञात है- \( f(x) = \begin{cases} -2, \text{ यदि } x \leq -1 \\ 2x, \text{ यदि } -1 < x \leq 1 \\ 2, \text{ यदि } x > 1 \end{cases} \)
दशा I \( x = -1 \) पर,
\( f(x) = -2 \), जब \( x \leq -1 \)
\( \therefore x = -1 \) पर,
L.H.L. \( = \lim_{x \to -1^-} f(x) = \lim_{x \to -1^-} (-2) = -2 \)
\( \implies f(-1) = -2 \)
\( f(x) = 2x \), जब \( x > -1 \)
R.H.L. \( = \lim_{x \to -1^+} f(x) = \lim_{x \to -1^+} (2x) = -2 \)
\( \therefore \) L.H.L. = R.H.L. = \( f(-1) \)
अतः \( x = -1 \) पर, फलन \( f \) सतत् है।
दशा II \( x = 1 \) पर,
\( f(x) = 2x \), जब \( x \leq 1 \)
\( \therefore x = 1 \) पर
L.H.L. \( = \lim_{x \to 1^-} f(x) = \lim_{x \to 1^-} (2x) = 2 \)
\( f(1) = 2 \)
जब \( x > 1 \)
R.H.L. \( = \lim_{x \to 1^+} f(x) = \lim_{x \to 1^+} (2) = 2 \)
\( \therefore \) L.H.L. = R.H.L.
अतः \( x = 1 \) पर फलन \( f \) सतत् है।
दशा III \( x = c_1 < -1 \),
\( f(x) = -2 \), जब \( x < -1 \)
\( \lim_{x \to c_1} f(x) = -2 = f(c_1) \)
अतः \( x = c_1 < -1 \) पर फलन \( f \) सतत् है।
दशा IV \( x = c_2, -1 < c_2 < 1 \),
\( f(x) = 2x \)
\( \lim_{x \to c_2} f(x) = \lim_{x \to c_2} (2x) = 2c_2 \), \( f(c_2) = 2c_2 \)
\( \therefore x = c_2, -1 < c_2 < 1 \) पर फलन \( f \) सतत् है।
दशा V \( x = c_3 > 1 \),
\( f(x) = 2 \), जब \( x \geq 1 \)
\( \lim_{x \to c_3} f(x) = 2 = f(c_3) \)
\( \therefore x = c_3 > 1 \) पर फलन \( f \) सतत् है।
अतः प्रत्येक बिन्दु \( x \in R \) पर फलन \( f \) सतत् है।
Answer: The function \( f(x) \) is continuous at all points \( x \in R \). There are no points of discontinuity.
In simple words: We check continuity at the boundary points \( x=-1 \) and \( x=1 \). At \( x=-1 \), LHL, RHL, and \( f(-1) \) are all -2, so it's continuous. At \( x=1 \), LHL, RHL, and \( f(1) \) are all 2, so it's continuous. For any other regions, the function is defined by a constant or a polynomial, which are continuous.

🎯 Exam Tip: This problem demonstrates a function that is continuous despite having multiple piecewise definitions, as long as the pieces "meet" smoothly at their transition points.

 

Question 17. a और b के उन मानों को ज्ञात कीजिए जिनके लिए द्वारा परिभाषित फलन x = 3 पर संतत है।
Answer: हल-
ज्ञात है- \( f(x) = \begin{cases} ax + 1, & \text{यदि } x \le 3 \\ bx + 3, & \text{यदि } x > 3 \end{cases} \)
x = 3 पर,
\( f(x) = ax + 1 \), जब \( x \le 3 \)
L.H.L. = \( \lim_{x \to 3^-} f(x) = \lim_{x \to 3^-} (ax + 1) = 3a + 1 \)
और
\( f(3) = 3a + 1 \)
\( f(x) = bx + 3 \), जब \( x > 3 \)
R.H.L. = \( \lim_{x \to 3^+} f(x) = \lim_{x \to 3^+} (b+3)=3b+3 \)
x = 3 पर यदि f संतत है तब L.H.L. = R.H.L.
\( \implies 3a+1=3b+3 \)
\( \implies 3a = 3b+2 \)
\( a = b + \frac{2}{3} \)
के लिए a का मान ज्ञात किया जा सकता हैं।
In simple words: फलन के संतत होने की शर्त का उपयोग करते हुए, बाएँ हाथ की सीमा, दाहिने हाथ की सीमा और फलन के मान को x = 3 पर बराबर किया जाता है, जिससे a और b के बीच संबंध \( a = b + \frac{2}{3} \) प्राप्त होता है।

🎯 Exam Tip: संतत फलन की परिभाषा (LHL = RHL = f(a)) का उपयोग करना और सीमाएं सही ढंग से हल करना महत्वपूर्ण है, खासकर जब फलन को टुकड़ों में परिभाषित किया गया हो।

 

Question 18. λ के किस मान के लिए। यदि x = 0 द्वारा परिभाषित फलन x = 0 पर संतत है। x = 1 पर इसके सातत्य पर विचार कीजिए।
Answer: हल-
ज्ञात है- \( f(x) = \begin{cases} \lambda(x^2 - 2x), & \text{यदि } x \le 0 \\ 4x + 1, & \text{यदि } x > 0 \end{cases} \)
(i) x = 0 रखने पर,
\( f(x) = \lambda(x^2 - 2x) \), जब \( x \le 0 \)
L.H.L. = \( \lim_{x \to 0^-} f(x) = \lim_{x \to 0^-} \lambda(x^2 - 2x) = 0 \)
तथा \( f(0) = 0 \)
R.H.L. = \( \lim_{x \to 0^+} f(x) = \lim_{x \to 0^+} (4x + 1) = 1 \)
\( \implies \lim_{x \to 0^-} f(x) \neq \lim_{x \to 0^+} f(x) \)
L.H.L \( \neq \) R.H.L.
अतः x = 0 पर \( \lambda \) के किसी भी मान के लिए f संतत नहीं है।
(ii) x = 1 पर,
\( f(x) = 4x + 1 \), जब \( x > 0 \)
\( \lim_{x \to 1} f(x) = \lim_{x \to 1} (4x + 1) = 4 \times 1 + 1 = 5 = f(1) \)
अतः x = 1 पर, f संतत है।
अत: x = 0 पर \( \lambda \) के किसी के मान के लिए f संतत नहीं है।
x = 1 पर f संतत है।
In simple words: फलन को x=0 पर संतत होने के लिए बाईं और दाईं सीमाएँ बराबर होनी चाहिए, लेकिन यहाँ वे बराबर नहीं हैं (0 और 1), इसलिए फलन λ के किसी भी मान के लिए संतत नहीं है। x=1 पर फलन \( f(x) = 4x+1 \) है जो एक बहुपद है, इसलिए यह x=1 पर संतत है।

🎯 Exam Tip: टुकड़ों में परिभाषित फलनों की संततता की जाँच करते समय, विभाजन बिंदु पर बाईं और दाहिनी सीमाओं की सावधानीपूर्वक गणना करें। एक बहुपद फलन हमेशा अपने डोमेन में संतत होता है।

 

Question 19. दर्शाइए कि g(x) = x - [x] द्वारा परिभाषित फलन समस्त पूर्णाक बिन्दुओं पर असंतत है। यहाँ [x] उस महत्तम पूर्णांक निरूपित करता है, जो x के बराबर अथवी x से कम है।
Answer: हल- x = c पूर्णांक पर, \( g(x) = x - [x] \)
L.H.L. = \( \lim_{x \to c^-} g(x) = \lim_{x \to c^-} (x-[x]) \)
= \( \lim_{h \to 0} ((c-h) - [c-h]) \) [x = (c - h) रखने पर]
= \( \lim_{h \to 0} (c-h - (c-1)) \) [\( \because [c-h] = c -1 \)]
= \( c-0-c+1=1 \)
R.H.L. = \( \lim_{x \to c^+} g(x) = \lim_{x \to c^+} (x-[x]) \) [x = c + h रखने पर] अतः x = c
= \( \lim_{h \to 0} ((c+h-[c+ h]) \)
= \( \lim_{h \to 0} [c+h-c] = 0 \)
\( f(c) = c - [c] = c - c = 0 \)
\( \therefore \)
L.H.L. \( \neq \) R.H.L. = \( f(c) \)
पूर्णाक पर, f संतत नहीं है।
In simple words: महत्तम पूर्णांक फलन [x] के कारण, किसी भी पूर्णांक c पर, बाईं हाथ की सीमा (1) और दाईं हाथ की सीमा (0) भिन्न होती हैं। चूँकि सीमाएँ समान नहीं हैं, फलन पूर्णांक बिंदुओं पर असंतत है।

🎯 Exam Tip: ग्रेटेस्ट इंटीजर फंक्शन [x] की असंततता पूर्णांक बिंदुओं पर होती है। ऐसे प्रश्नों में, LHL और RHL की गणना करते समय \( [c-h] = c-1 \) और \( [c+h] = c \) नियमों का सही ढंग से उपयोग करें।

 

Question 20. क्या f(x) = x² – sin x + 5 द्वारा परिभाषित फलन x = \( \pi \) पर संतत है?
Answer: हल- माना \( f(x) = x^2 - \sin x + 5 \)
x = \( \pi \) पर,
L.H.L. = \( \lim_{x \to \pi^-} f(x) = \lim_{x \to \pi^-} (x^2 - \sin x + 5) \)
= \( \lim_{h \to 0} [(\pi-h)^2 - \sin (\pi-h) + 5] \) (x = \( \pi \) - h रखने पर)
= \( \lim_{h \to 0} [\pi^2 - 2\pi h + h^2 - \sin h + 5] \) [\( \sin (\pi-h) = \sin h \)]
= \( \pi^2 - 0 + 0 - 0 + 5 = \pi^2 + 5 \)
R.H.L. = \( \lim_{x \to \pi^+} f(x) = \lim_{x \to \pi^+} (x^2 - \sin x + 5) \)
= \( \lim_{h \to 0} [(\pi+h)^2 - \sin (\pi+h) + 5] \) (x = \( \pi \) + h रखने पर)
= \( \lim_{h \to 0} [\pi^2 + 2\pi h + h^2 + \sin h + 5] = \pi^2 + 5 \) [\( \sin (\pi+h) = -\sin h \)]
\( f(\pi) = \pi^2 - \sin \pi + 5 = \pi^2 - 0 + 5 = \pi^2 + 5 \)
\( \lim_{x \to \pi^-} f(x) = \lim_{x \to \pi^+} f(x) = f(\pi) \)
अतः x = \( \pi \) पर f संतत है।
In simple words: x = \( \pi \) पर फलन \( f(x) = x^2 - \sin x + 5 \) एक बहुपद और साइन फलन का संयोजन है। चूँकि ये दोनों फलन अपने डोमेन में संतत होते हैं, और x = \( \pi \) पर बाएँ हाथ की सीमा, दाएँ हाथ की सीमा और फलन का मान सभी बराबर \( \pi^2 + 5 \) हैं, इसलिए फलन x = \( \pi \) पर संतत है।

🎯 Exam Tip: त्रिकोणमितीय फलनों के संततता की जाँच करते समय, सीमाओं का मूल्यांकन करते समय कोणों के रूपांतरण (जैसे \( \sin(\pi-h) = \sin h \)) पर ध्यान दें।

 

Question 21. निम्नलिखित फलनों के सांतत्य पर विचार कीजिए (a) f(x) = sin x + cos x (b) f(x) = sin x – cos x (c) f(x) = sin x.cos x
Answer: हल-
(a) \( f(x) = \sin x + \cos x = \sqrt{2} \left( \frac{1}{\sqrt{2}} \sin x + \frac{1}{\sqrt{2}} \cos x \right) \) [\( \sqrt{2} \) से गुणा तथा भाग करने पर]
= \( \sqrt{2} \left( \sin x \cos \frac{\pi}{4} + \cos x \sin \frac{\pi}{4} \right) \)
= \( \sqrt{2} \sin \left(x + \frac{\pi}{4} \right) \)
x = c \( \in \) R पर,
L.H.L. = \( \lim_{x \to c^-} \sqrt{2} \sin \left(x + \frac{\pi}{4} \right) \)
x = c - h रखने पर,
L.H.L. = \( \lim_{h \to 0} \sqrt{2} \sin \left( (c-h) + \frac{\pi}{4} \right) = \sqrt{2} \sin \left(c + \frac{\pi}{4} \right) \)
इसी प्रकार,
R.H.L. = \( \lim_{x \to c^+} \sqrt{2} \sin \left(x + \frac{\pi}{4} \right) \)
= \( \lim_{h \to 0} \sqrt{2} \sin \left( (c+h) + \frac{\pi}{4} \right) \) [x = (c + h) रखने पर]
= \( \sqrt{2} \sin \left(c + \frac{\pi}{4} \right) \)
\( f(c) = \sqrt{2} \sin \left(c + \frac{\pi}{4} \right) \)
\( \implies \lim_{x \to c^-} f(x) = \lim_{x \to c^+} f(x) = f(c) \)
L.H.L. = R.H.L.
अत: x = c \( \in \) R पर f संतत है।
(b) \( f(x)=\sin x - \cos x \)
= \( \sqrt{2} \left( \frac{1}{\sqrt{2}} \sin x - \frac{1}{\sqrt{2}} \cos x \right) \)
= \( \sqrt{2} \left( \sin x \cos \frac{\pi}{4} - \cos x \sin \frac{\pi}{4} \right) \)
= \( \sqrt{2} \sin \left(x - \frac{\pi}{4} \right) \)
At x=c, L.H.L.
= \( \lim_{x \to c^-} \sqrt{2} \sin \left(x - \frac{\pi}{4} \right) \)
= \( \sqrt{2} \sin \left(c - \frac{\pi}{4} \right) \)
R.H.L. = \( \lim_{x \to c^+} \sqrt{2} \sin \left(x - \frac{\pi}{4} \right) \)
= \( \sqrt{2} \sin \left(c - \frac{\pi}{4} \right) \)
= \( f(c) \)
\( \therefore \) f is continuous for all x \( \in \) R.
(c) माना \( f(x) = \sin x \cos x = \frac{1}{2} (2 \sin x \cos x) \) (2 से गुणा तथा भाग करने पर)
= \( \frac{1}{2} \sin 2x \)
x = c \( \in \) R पर,
L.H.L. = \( \lim_{x \to c^-} f(x) = \lim_{h \to 0} \frac{1}{2} \sin 2 (c - h) = \frac{1}{2} \sin 2c \) [x = (c - h) रखने पर]
R.H.L. = \( \lim_{x \to c^+} f(x) = \lim_{h \to 0} \frac{1}{2} \sin 2 (c + h) = \frac{1}{2} \sin 2c \) [x = (c + h) रखने पर]
\( f(c) = \frac{1}{2} \sin 2c \)
\( \implies \lim_{x \to c^-} f(x) = \lim_{x \to c^+} f(x) = f(c) \)
L.H.L. = R.H.L.
अत: x = c \( \in \) R पर f संतत है।
In simple words:
(a) \( \sin x \) और \( \cos x \) दोनों संतत फलन हैं, इसलिए उनका योग भी संतत होगा। इसे त्रिकोणमितीय पहचान \( \sin(A+B) \) का उपयोग करके \( \sqrt{2} \sin(x+\frac{\pi}{4}) \) के रूप में व्यक्त किया जा सकता है, जो एक संतत फलन है।
(b) इसी तरह, \( \sin x \) और \( \cos x \) का अंतर भी संतत होगा। इसे \( \sqrt{2} \sin(x-\frac{\pi}{4}) \) के रूप में व्यक्त किया जा सकता है, जो एक संतत फलन है।
(c) \( \sin x \cos x \) को \( \frac{1}{2} \sin 2x \) के रूप में लिखा जा सकता है, जो एक संतत फलन है क्योंकि \( \sin 2x \) संतत है और इसे एक स्थिरांक से गुणा किया गया है।

🎯 Exam Tip: दो संतत फलनों का योग, अंतर और गुणनफल भी संतत होता है। त्रिकोणमितीय फलनों के सांतत्य की जाँच के लिए LHL, RHL और f(c) तीनों की गणना करना आवश्यक है।

 

प्रश्नावली 5.3

 

Question 22. cosine, cosecant, secant और cotangent फलनों के सातत्य पर विचार कीजिए।
Answer: हल- (a) माना \( x = c \in R \),
\( \lim_{x \to c} \cos x = \cos c = f(c) \)
अतः \( x = c \in R \) पर f संतत है।
(b) माना \( f(x) = \cosec x \)
\( \sin x = 0 \) पर f परिभाषित नहीं है, अर्थात \( x = n\pi \), पर f परिभाषित नहीं है।
\( x = c \in R - \{n\pi\}, n \in Z \), पर \( f(x) = \cosec x \)
\( \lim_{x \to c} \cosec x = \cosec c = f(c) \)
अतः \( x = c \in R - \{n\pi\}, n \in Z \) पर f संतत है।
(c) माना \( y = \sec x \)
\( \cos x = 0 \) पर f परिभाषित नहीं है। अर्थात् \( x = (2n + 1)\frac{\pi}{2} \), पर f परिभाषित नहीं है। जबकि \( n \in Z \)
\( x = c \in R - \{(2n + 1)\frac{\pi}{2}\}, n \in Z \), पर \( f(x) = \sec x \)
\( \lim_{x \to c} \sec x = \sec c = f(c) \)
अतः \( R - \{(2n + 1)\frac{\pi}{2}\} \) पर संतत है।
(d) माना \( f(x) = \cot x \)
\( \sin x = 0 \) पर f परिभाषित नहीं है, अर्थात \( x = n\pi \), पर f परिभाषित नहीं है। जबकि \( n \in Z \)
\( x = c \in R - \{n\pi\}, n \in Z \), पर \( f(x) = \cot x \)
\( \lim_{x \to c} \cot x = \cot c = f(c) \)
अतः \( x = c \in R - \{n\pi\}, n \in Z \) पर f संतत है।
In simple words:
(a) कोसाइन फलन अपने पूरे डोमेन (सभी वास्तविक संख्याओं) में संतत होता है।
(b) कोसेकेंट फलन \( \sin x = 0 \) वाले बिंदुओं पर परिभाषित नहीं होता है, यानी \( n\pi \) पर, जहाँ n एक पूर्णांक है, इसलिए यह उन बिंदुओं को छोड़कर सभी जगह संतत है।
(c) सेकेंट फलन \( \cos x = 0 \) वाले बिंदुओं पर परिभाषित नहीं होता है, यानी \( (2n+1)\frac{\pi}{2} \) पर, जहाँ n एक पूर्णांक है, इसलिए यह उन बिंदुओं को छोड़कर सभी जगह संतत है।
(d) कोटेन्जेंट फलन \( \sin x = 0 \) वाले बिंदुओं पर परिभाषित नहीं होता है, यानी \( n\pi \) पर, जहाँ n एक पूर्णांक है, इसलिए यह उन बिंदुओं को छोड़कर सभी जगह संतत है।

🎯 Exam Tip: त्रिकोणमितीय फलनों के सांतत्य की जाँच करते समय, उनके डोमेन पर विशेष ध्यान दें, खासकर वे बिंदु जहाँ हर शून्य होता है, क्योंकि फलन उन बिंदुओं पर परिभाषित नहीं होते और इसलिए असंतत होते हैं।

 

Question 23. f के सभी असातत्यता के बिन्दुओं को ज्ञात कीजिए, जहाँ \( f(x) = \begin{cases} \frac{\sin x}{x}, & \text{यदि } x < 0 \\ x+1, & \text{यदि } x \ge 0 \end{cases} \)
Answer: हल-
x = 0 पर,
\( f(x) = \frac{\sin x}{x} \), जब \( x < 0 \)
L.H.L. = \( \lim_{x \to 0^-} f(x) = \lim_{h \to 0} f(0-h) = \lim_{h \to 0} \frac{\sin (-h)}{(-h)} = 1 \)
\( f(x) = x+1 \), जब \( x \ge 0 \)
R.H.L. = \( \lim_{x \to 0^+} f(x) = \lim_{h \to 0} f(h) = \lim_{h \to 0} (h + 1) = 1 \)
तथा \( f(0) = 0 + 1 = 1 \)
\( \therefore \lim_{x \to 0^-} f(x) = \lim_{x \to 0^+} f(x) = f(0) \)
अतः x = 0 पर f संतत है।
x = \( c_1 < 0 \) पर,
\( f(x) = \frac{\sin x}{x} \)
\( \lim_{x \to c_1} f(x) = \lim_{x \to c_1} \left(\frac{\sin x}{x}\right) = \frac{\sin c_1}{c_1} = f(c_1) \)
\( \implies \) x = \( c_1 < 0 \) पर f संतत है।
x = \( c_2 > 0 \) पर,
\( f(x) = x+1 \)
\( \lim_{x \to c_2} f(x) = \lim_{x \to c_2} (x + 1) = c_2 + 1 = f(c_2) \)
\( \therefore \) x = \( c_2 > 0 \) पर f संतत है।
अतः स्पष्ट है कि f किसी भी बिन्दु पर असंतत नहीं है।
In simple words: फलन को विभाजन बिंदु x = 0 पर जाँचने पर, बाईं और दाहिनी सीमाएं और फलन का मान सभी 1 के बराबर आते हैं, इसलिए यह x = 0 पर संतत है। x < 0 के लिए, \( \frac{\sin x}{x} \) परिभाषित और संतत है, और x > 0 के लिए, \( x+1 \) एक बहुपद है जो संतत है। इसलिए, फलन के असातत्यता का कोई बिंदु नहीं है।

🎯 Exam Tip: \( \lim_{x \to 0} \frac{\sin x}{x} = 1 \) एक महत्वपूर्ण सीमा है जिसका उपयोग ऐसे प्रश्नों में किया जाता है। सुनिश्चित करें कि आप सीमाओं, फलन के मान और डोमेन को सही ढंग से जाँचें।

 

Question 24. निम्न प्रकार से परिभाषित फलन की सातत्यता की जाँच x = 0 पर कीजिए \( f(x) = \begin{cases} x^2 \sin \frac{1}{x}, & \text{यदि } x \neq 0 \\ 0, & \text{यदि } x = 0 \end{cases} \)
Answer: हल-
x = 0 पर,
वाम पक्ष सीमा L.H.L. = \( \lim_{x \to 0^-} f(x) \)
= \( \lim_{h \to 0} f(0-h) \)
\( \implies \)
= \( \lim_{h \to 0} (-h)^2 \sin \frac{1}{(-h)} \)
= \( \lim_{h \to 0} h^2 \sin \left(-\frac{1}{h}\right) \)
= \( \lim_{h \to 0} -h^2 \sin \frac{1}{h} \)
= \( 0 \)
[ \( \because h \to 0, -1 < \sin \frac{1}{h} < 1 \) जैसे ही \( h \to 0, h^2 \sin \frac{1}{h} \to 0 \) ]
दक्षिण पक्ष सीमा R.H.L. = \( \lim_{x \to 0^+} f(x) \)
= \( \lim_{h \to 0} f(0+h) \)
= \( \lim_{h \to 0} (0+h)^2 \sin \frac{1}{(0+h)} \)
= \( \lim_{h \to 0} h^2 \sin \frac{1}{h} = 0 \)
[ \( \because h \to 0, -1 < \sin \frac{1}{h} < 1 \)]
\( f(0) = 0 \)
\( \therefore \)
L.H.L. = R.H.L. = \( f(0) \)
\( \therefore \) x = 0 पर f संतत है।
In simple words: फलन के x = 0 पर संततता की जाँच करने के लिए, हमें बाईं और दाहिनी सीमाओं और फलन के मान की तुलना करनी होगी। \( x \sin \frac{1}{x} \) प्रकार के फलनों में, \( \sin \frac{1}{x} \) का मान -1 और 1 के बीच रहता है, और जब \( x \to 0 \), \( x^2 \) शून्य हो जाता है, जिससे पूरी सीमा शून्य हो जाती है। चूँकि LHL = RHL = f(0) = 0, फलन x = 0 पर संतत है।

🎯 Exam Tip: ऐसे प्रश्नों में जहाँ \( \sin \frac{1}{x} \) या \( \cos \frac{1}{x} \) पद शामिल हों, याद रखें कि \( -1 \le \sin(\theta) \le 1 \) होता है। इस तथ्य का उपयोग करके, आप \( \lim_{x \to 0} x^2 \sin \frac{1}{x} = 0 \) सिद्ध करने के लिए सैंडविच प्रमेय का उपयोग कर सकते हैं।

 

Question 25. f के सभी असातत्य बिन्दुओं को ज्ञात कीजिए, जहाँ \( f(x) = \begin{cases} \sin x - \cos x, & \text{यदि } x \neq 0 \\ -1, & \text{यदि } x = 0 \end{cases} \)
Answer: हल-
x = 0 पर,
L.H.L. = \( \lim_{x \to 0^-} f(x) = \lim_{h \to 0} f(0-h) = \lim_{h \to 0} (\sin (-h) - \cos (-h)) \)
= \( \lim_{h \to 0} (-\sin h - \cos h) = (0 - 1) = -1 \)
R.H.L. = \( \lim_{x \to 0^+} f(x) = \lim_{h \to 0} f(0+h) = \lim_{h \to 0} (\sin h - \cos h) \)
= \( (0 - 1) = -1 \)
\( f(0) = -1 \)
\( \therefore \lim_{x \to 0^-} f(x) = \lim_{x \to 0^+} f(x) = f(0) \)
अतः x = 0 पर f संतत है।
x = c \( \neq 0 \) पर,
\( f(x) = \sin x - \cos x \)
\( \lim_{x \to c} (\sin x - \cos x) = \sin c - \cos c = f(c) \)
\( \therefore \) x = c \( \neq 0 \) पर f संतत है।
अतः \( x \in R \), सभी बिन्दुओं पर फलन f संतत है।
In simple words: फलन को x = 0 पर संतत होने के लिए बाईं हाथ की सीमा, दाहिनी हाथ की सीमा और फलन का मान बराबर होना चाहिए। इस मामले में, वे सभी -1 के बराबर हैं, इसलिए फलन x = 0 पर संतत है। x \( \neq \) 0 के लिए, \( \sin x - \cos x \) एक संतत फलन है क्योंकि \( \sin x \) और \( \cos x \) दोनों संतत फलन हैं। इस प्रकार, फलन सभी वास्तविक संख्याओं के लिए संतत है और इसमें कोई असातत्य बिंदु नहीं है।

🎯 Exam Tip: साइन और कोसाइन फलन पूरे डोमेन में संतत होते हैं। उनका योग या अंतर भी संतत होता है। विभाजन बिंदु पर LHL, RHL और f(a) की जाँच करना महत्वपूर्ण है।

 

Question 26. k का मान ज्ञात कीजिए ताकि फलन \( f(x) = \begin{cases} \frac{k \cos x}{\pi - 2x}, & \text{यदि } x \neq \frac{\pi}{2} \\ 3, & \text{यदि } x = \frac{\pi}{2} \end{cases} \) द्वारा परिभाषित फलन x = \( \frac{\pi}{2} \) पर संतत हो।
Answer: हल-
x = \( \frac{\pi}{2} \) पर,
L.H.L. = \( \lim_{x \to \frac{\pi}{2}^-} f(x) = \lim_{h \to 0} f\left(\frac{\pi}{2}-h\right) \)
= \( \lim_{h \to 0} \frac{k \cos \left(\frac{\pi}{2}-h\right)}{\pi - 2\left(\frac{\pi}{2}-h\right)} \)
= \( \lim_{h \to 0} \frac{k \sin h}{\pi - \pi + 2h} = \lim_{h \to 0} \frac{k \sin h}{2h} = \frac{k}{2} \lim_{h \to 0} \frac{\sin h}{h} = \frac{k}{2} \times 1 = \frac{k}{2} \)
R.H.L. = \( \lim_{x \to \frac{\pi}{2}^+} f(x) = \lim_{h \to 0} f\left(\frac{\pi}{2}+h\right) \)
= \( \lim_{h \to 0} \frac{k \cos \left(\frac{\pi}{2}+h\right)}{\pi - 2\left(\frac{\pi}{2}+h\right)} \)
= \( \lim_{h \to 0} \frac{-k \sin h}{\pi - \pi - 2h} = \lim_{h \to 0} \frac{-k \sin h}{-2h} = \frac{k}{2} \lim_{h \to 0} \frac{\sin h}{h} = \frac{k}{2} \times 1 = \frac{k}{2} \)
\( f\left(\frac{\pi}{2}\right) = 3 \)
फलन f संतत \( x = \frac{\pi}{2} \) होगा यदि
L.H.L. = R.H.L. = \( f\left(\frac{\pi}{2}\right) \)
\( \frac{k}{2} = 3 \)
\( \implies k = 6 \)
In simple words: फलन को x = \( \frac{\pi}{2} \) पर संतत होने के लिए, बाईं हाथ की सीमा, दाहिनी हाथ की सीमा और फलन का मान x = \( \frac{\pi}{2} \) पर बराबर होना चाहिए। इन सीमाओं की गणना करते समय, \( \cos(\frac{\pi}{2}-h) = \sin h \) और \( \cos(\frac{\pi}{2}+h) = -\sin h \) जैसे त्रिकोणमितीय रूपांतरणों का उपयोग किया जाता है, साथ ही \( \lim_{h \to 0} \frac{\sin h}{h} = 1 \) का भी उपयोग होता है। इन सभी को बराबर करने पर, k का मान 6 प्राप्त होता है।

🎯 Exam Tip: \( \lim_{\theta \to 0} \frac{\sin \theta}{\theta} = 1 \) और \( \cos(\frac{\pi}{2} - \theta) = \sin \theta \), \( \cos(\frac{\pi}{2} + \theta) = -\sin \theta \) जैसी महत्वपूर्ण त्रिकोणमितीय सीमाएँ और पहचानें याद रखें। संतत फलन की परिभाषा का उपयोग करते हुए अज्ञात स्थिरांक ज्ञात किए जा सकते हैं।

 

Question 27. k का मान ज्ञात कीजिए ताकि फलन \( f(x) = \begin{cases} kx^2, & \text{यदि } x \le 2 \\ 3, & \text{यदि } x > 2 \end{cases} \) द्वारा परिभाषित फलन x = 2 पर संतत हो।
Answer: हल-
x = 2 पर,
L.H.L = \( \lim_{x \to 2^-} f(x) = \lim_{x \to 2^-} (kx^2) = k(2)^2 = 4k \)
R.H.L. = \( \lim_{x \to 2^+} f(x) = \lim_{x \to 2^+} 3 = 3 \)
\( f(2) = k(2)^2 = 4k \)
फलन f संतत होगा यदि x = 2 पर,
L.H.L. = R.H.L. = \( f(2) \)
\( \implies 4k = 3 \)
\( \implies k = \frac{3}{4} \)
In simple words: फलन को x = 2 पर संतत होने के लिए, बाईं हाथ की सीमा, दाहिनी हाथ की सीमा और फलन का मान बराबर होना चाहिए। यहाँ, ये तीनों मान \( 4k \) और \( 3 \) हैं। इन मानों को बराबर करने पर \( 4k = 3 \) प्राप्त होता है, जिससे k का मान \( \frac{3}{4} \) आता है।

🎯 Exam Tip: खंडशः परिभाषित फलनों की संततता की जाँच करते समय, विभाजन बिंदु पर LHL, RHL और f(a) की गणना करना महत्वपूर्ण है। इन तीनों को बराबर करके अज्ञात स्थिरांक का मान ज्ञात किया जा सकता है।

 

Question 28. k का मान ज्ञात कीजिए यदि फलन \( f(x) = \begin{cases} kx+1, & \text{यदि } x \le \pi \\ \cos x, & \text{यदि } x > \pi \end{cases} \) द्वारा परिभाषित फलन x = \( \pi \) पर संतत है।
Answer: हल-
x = \( \pi \) पर,
L.H.L. = \( \lim_{x \to \pi^-} f(x) = \lim_{x \to \pi^-} (kx + 1) = k\pi + 1 \)
R.H.L. = \( \lim_{x \to \pi^+} f(x) = \lim_{x \to \pi^+} \cos x = \cos \pi = -1 \)
\( f(\pi) = k\pi + 1 \)
फलन f x = \( \pi \) पर संतत होगा यदि
L.H.L. = R.H.L. = \( f(\pi) \)
\( \implies k\pi + 1 = -1 \)
\( \implies k\pi = -2 \)
\( k = -\frac{2}{\pi} \)
In simple words: फलन को x = \( \pi \) पर संतत होने के लिए, बाईं हाथ की सीमा, दाहिनी हाथ की सीमा और फलन का मान x = \( \pi \) पर बराबर होना चाहिए। बाईं सीमा \( k\pi+1 \) है और दाहिनी सीमा \( \cos \pi = -1 \) है। इन दोनों को बराबर करने पर \( k\pi+1 = -1 \) प्राप्त होता है, जिससे k का मान \( -\frac{2}{\pi} \) आता है।

🎯 Exam Tip: त्रिकोणमितीय फलनों के मानों को सही ढंग से याद रखना महत्वपूर्ण है, जैसे \( \cos \pi = -1 \)। सीमा की गणना करते समय, \( h \to 0 \) पर \( \cos(\pi-h) = -\cos h \approx -1 \) और \( \cos(\pi+h) = -\cos h \approx -1 \) का उपयोग किया जा सकता है।

 

Question 29. k का मान ज्ञात कीजिए यदि फलन \( f(x) = \begin{cases} kx+1, & \text{यदि } x \le 5 \\ 3x-5, & \text{यदि } x > 5 \end{cases} \) द्वारा परिभाषित फलन x = 5 पर संतत हो।
Answer: हल-
x = 5 पर,
L.H.L. = \( \lim_{x \to 5^-} f(x) = \lim_{x \to 5^-} (kx + 1) = k(5) + 1 = 5k + 1 \)
R.H.L. = \( \lim_{x \to 5^+} f(x) = \lim_{x \to 5^+} (3x - 5) = 3(5) - 5 = 15 - 5 = 10 \)
\( f(5) = k(5) + 1 = 5k + 1 \)
फलन f x = 5 पर संतत होगा यदि
L.H.L. = R.H.L. = \( f(5) \)
\( \implies 5k + 1 = 10 \)
\( \implies 5k = 9 \)
\( k = \frac{9}{5} \)
In simple words: फलन को x = 5 पर संतत होने के लिए, बाएँ हाथ की सीमा, दाएँ हाथ की सीमा और फलन का मान x = 5 पर बराबर होना चाहिए। यहाँ, ये मान \( 5k+1 \) और \( 10 \) हैं। इन दोनों को बराबर करने पर \( 5k+1 = 10 \) प्राप्त होता है, जिससे k का मान \( \frac{9}{5} \) आता है।

🎯 Exam Tip: विभाजन बिंदु पर LHL, RHL और f(a) की गणना में सावधानी बरतें। सीमा का मूल्यांकन करते समय, फलन की सही परिभाषा का उपयोग करें जो x के मान के लिए उपयुक्त हो।

 

Question 30. a तथा b के मानों को ज्ञात कीजिए ताकि फलन \( f(x) = \begin{cases} 5, & \text{यदि } x \le 2 \\ ax+b, & \text{यदि } 2 < x < 10 \\ 21, & \text{यदि } x \ge 10 \end{cases} \) द्वारा परिभाषित फलन एक संतत फलन हो।
Answer: हल-
x = 2 पर,
L.H.L. = \( \lim_{x \to 2^-} f(x) = \lim_{x \to 2^-} 5 = 5 \)
R.H.L. = \( \lim_{x \to 2^+} f(x) = \lim_{x \to 2^+} (ax + b) = 2a + b \)
\( f(2) = 5 \)
x = 2 पर फलन f संतत होगा यदि
L.H.L. = R.H.L. = \( f(2) \)
\( \implies 2a + b = 5 \) ...(1)
x = 10 पर,
L.H.L. = \( \lim_{x \to 10^-} f(x) = \lim_{x \to 10^-} (ax + b) = 10a + b \)
R.H.L. = \( \lim_{x \to 10^+} f(x) = \lim_{x \to 10^+} 21 = 21 \)
\( f(10) = 21 \)
x = 10 पर फलन f संतत होगा यदि
L.H.L. = R.H.L. = \( f(10) \)
\( \implies 10a + b = 21 \) ...(2)
समीकरण (2) में से समीकरण (1) को घटाने पर,
\( (10a + b) - (2a + b) = 21 - 5 \)
\( 8a = 16 \implies a = 2 \)
समीकरण (1) से,
\( 2(2) + b = 5 \implies 4 + b = 5 \implies b = 1 \)
अतः f एक संतत फलन होगा यदि \( a = 2, b = 1 \)
In simple words: फलन के संतत होने के लिए, दोनों विभाजन बिंदुओं (x=2 और x=10) पर LHL, RHL और फलन के मान बराबर होने चाहिए। इससे हमें a और b के लिए दो समीकरण मिलते हैं: \( 2a+b=5 \) और \( 10a+b=21 \)। इन समीकरणों को हल करने पर a=2 और b=1 प्राप्त होता है।

🎯 Exam Tip: टुकड़ों में परिभाषित फलनों के लिए, प्रत्येक विभाजन बिंदु पर संततता की जाँच करें। प्रत्येक बिंदु पर LHL, RHL और फलन के मान को बराबर करके समीकरणों का एक सेट प्राप्त करें, और फिर उन समीकरणों को हल करके अज्ञात स्थिरांक ज्ञात करें।

 

Question 31. दर्शाइए कि f(x) = cos x² द्वारा परिभाषित फलन एक संतत फलन है।
Answer: हल- ज्ञात है- \( f(x) = \cos x^2 \)
माना \( g(x) = x^2 \) और \( h(x) = \cos x \)
तो \( f(x) = h(g(x)) \)
\( g(x) = x^2 \) एक बहुपद फलन है, इसलिए यह संतत है।
\( h(x) = \cos x \) एक कोसाइन फलन है, इसलिए यह भी संतत है।
चूंकि दो संतत फलनों का संयोजन भी संतत होता है, अतः \( f(x) = \cos x^2 \) एक संतत फलन है। इति सिद्धम्
In simple words: फलन \( f(x) = \cos x^2 \) दो संतत फलनों का संयोजन है: \( g(x) = x^2 \) (एक बहुपद) और \( h(x) = \cos x \) (एक त्रिकोणमितीय फलन)। चूँकि दो संतत फलनों का संयोजन हमेशा संतत होता है, इसलिए \( f(x) \) भी संतत है।

🎯 Exam Tip: संयोजन फलनों की संततता सिद्ध करने के लिए, बाहरी और आंतरिक दोनों फलनों की संततता का उल्लेख करें। यदि \( f(x) = h(g(x)) \) और g संतत है, तथा h भी संतत है, तो f भी संतत होगा।

 

Question 32. दर्शाइए कि f(x) = | cos x| द्वारा परिभाषित फलन एक संतत फलन है।
Answer: हल- ज्ञात है- \( f(x) = |\cos x| \)
माना \( g(x) = \cos x \) और \( h(x) = |x| \)
तो \( f(x) = h(g(x)) \)
\( g(x) = \cos x \) एक कोसाइन फलन है, इसलिए यह संतत है।
\( h(x) = |x| \) एक मापांक फलन है, जो अपने पूरे डोमेन में संतत होता है।
चूंकि दो संतत फलनों का संयोजन भी संतत होता है, अतः \( f(x) = |\cos x| \) एक संतत फलन है। इति सिद्धम्
In simple words: फलन \( f(x) = |\cos x| \) को \( h(g(x)) \) के रूप में देखा जा सकता है, जहाँ \( g(x) = \cos x \) और \( h(x) = |x| \)। चूंकि कोसाइन फलन और निरपेक्ष मान फलन दोनों संतत होते हैं, इसलिए उनका संयोजन भी संतत होता है।

🎯 Exam Tip: मापांक फलन \( |x| \) सभी वास्तविक संख्याओं के लिए संतत होता है। संयोजन फलन के संततता को सिद्ध करने के लिए घटक फलनों की संततता को पहचानना महत्वपूर्ण है।

 

Question 33. जाँचिए कि क्या sin|x| एक संतत फलन है।
Answer: हल- माना \( f(x) = \sin |x| \)
माना \( g(x) = |x| \) और \( h(x) = \sin x \)
तो \( f(x) = h(g(x)) \)
\( g(x) = |x| \) एक मापांक फलन है, जो अपने पूरे डोमेन में संतत होता है।
\( h(x) = \sin x \) एक साइन फलन है, इसलिए यह भी संतत है।
चूंकि दो संतत फलनों का संयोजन भी संतत होता है, अतः \( f(x) = \sin |x| \) एक संतत फलन है।
In simple words: फलन \( \sin|x| \) दो संतत फलनों का संयोजन है: निरपेक्ष मान फलन \( |x| \) और साइन फलन \( \sin x \)। चूंकि दोनों घटक फलन संतत हैं, इसलिए उनका संयोजन भी एक संतत फलन होगा।

🎯 Exam Tip: संयोजन फलनों की संततता सिद्ध करने के लिए, बाहरी और आंतरिक दोनों फलनों की संततता का उल्लेख करें। \( |x| \) और \( \sin x \) दोनों संतत फलन हैं, इसलिए \( \sin|x| \) भी संतत है।

 

Question 34. f(x) = |x|-|x + 1| द्वारा परिभाषित फलन f के सभी असातत्यता के बिन्दुओं को ज्ञात कीजिए।
Answer: हल- ज्ञात है- \( f(x) = |x|-|x + 1| \)
हम फलन को टुकड़ों में परिभाषित करते हैं:
जब \( x < -1 \):
\( |x| = -x \)
\( |x+1| = -(x+1) \)
तो \( f(x) = -x - (-(x+1)) = -x + x + 1 = 1 \)
जब \( -1 \le x < 0 \):
\( |x| = -x \)
\( |x+1| = x+1 \)
तो \( f(x) = -x - (x+1) = -2x - 1 \)
जब \( x \ge 0 \):
\( |x| = x \)
\( |x+1| = x+1 \)
तो \( f(x) = x - (x+1) = -1 \)
अतः फलन को इस प्रकार लिखा जा सकता है:
\( f(x) = \begin{cases} 1, & \text{यदि } x < -1 \\ -2x-1, & \text{यदि } -1 \le x < 0 \\ -1, & \text{यदि } x \ge 0 \end{cases} \)
हम विभाजन बिंदुओं x = -1 और x = 0 पर संततता की जाँच करेंगे।
**x = -1 पर:**
L.H.L. = \( \lim_{x \to -1^-} f(x) = \lim_{x \to -1^-} 1 = 1 \)
R.H.L. = \( \lim_{x \to -1^+} f(x) = \lim_{x \to -1^+} (-2x-1) = -2(-1)-1 = 2-1 = 1 \)
\( f(-1) = -2(-1)-1 = 2-1 = 1 \)
चूंकि L.H.L. = R.H.L. = \( f(-1) \), फलन x = -1 पर संतत है।
**x = 0 पर:**
L.H.L. = \( \lim_{x \to 0^-} f(x) = \lim_{x \to 0^-} (-2x-1) = -2(0)-1 = -1 \)
R.H.L. = \( \lim_{x \to 0^+} f(x) = \lim_{x \to 0^+} -1 = -1 \)
\( f(0) = -1 \)
चूंकि L.H.L. = R.H.L. = \( f(0) \), फलन x = 0 पर संतत है।
प्रत्येक अंतराल में, फलन या तो एक स्थिरांक फलन है या एक बहुपद फलन है, जो अपने-अपने अंतरालों में संतत होते हैं।
चूंकि फलन विभाजन बिंदुओं पर भी संतत है, इसलिए फलन के असातत्यता का कोई बिंदु नहीं है।
In simple words: फलन \( f(x) = |x| - |x+1| \) को निरपेक्ष मान की परिभाषा के आधार पर तीन टुकड़ों में परिभाषित किया जाता है: \( x < -1 \), \( -1 \le x < 0 \), और \( x \ge 0 \)। प्रत्येक अंतराल में, फलन या तो एक स्थिरांक या एक रैखिक बहुपद है, जो संतत होते हैं। विभाजन बिंदुओं x = -1 और x = 0 पर भी LHL, RHL और फलन का मान बराबर होता है, इसलिए फलन सभी वास्तविक संख्याओं के लिए संतत है और इसका कोई असातत्य बिंदु नहीं है।

🎯 Exam Tip: मापांक फलनों के साथ व्यवहार करते समय, हमेशा उन बिंदुओं को पहचानकर फलन को टुकड़ों में परिभाषित करें जहाँ मापांक के अंदर के व्यंजक शून्य होते हैं। फिर प्रत्येक विभाजन बिंदु पर संततता की जाँच करें।

 

प्रश्नावली 5.2

 

Question 1. sin(x²+5)
Answer: हल- माना \( y = \sin(x^2+5) \)
x के सापेक्ष अवकलन करने पर,
\( \frac{dy}{dx} = \cos(x^2+5) \frac{d}{dx}(x^2+5) \)
= \( \cos(x^2+5)(2x+0) \)
= \( 2x \cos(x^2+5) \)
In simple words: श्रृंखला नियम का उपयोग करके, पहले \( \sin \) फलन का अवकलन करें (जो \( \cos \) बन जाता है), फिर इसके अंदर के \( x^2+5 \) का अवकलन करें (जो \( 2x \) बन जाता है)। इन दोनों को गुणा करने पर अंतिम उत्तर मिलता है।

🎯 Exam Tip: श्रृंखला नियम \( \frac{d}{dx}f(g(x)) = f'(g(x))g'(x) \) का सही अनुप्रयोग महत्वपूर्ण है। \( \sin \) का अवकलन \( \cos \) है और \( x^2+5 \) का अवकलन \( 2x \) है।

 

Question 2. cos (sin x)
Answer: हल- माना \( y = \cos(\sin x) \)
माना \( t = \sin x \)
तब \( y = \cos t \)
t के सापेक्ष अवकलन करने पर,
\( \frac{dy}{dt} = -\sin t \)
x के सापेक्ष t का अवकलन करने पर,
\( \frac{dt}{dx} = \cos x \)
श्रृंखला नियम से,
\( \frac{dy}{dx} = \frac{dy}{dt} \cdot \frac{dt}{dx} \)
= \( (-\sin t) \cdot (\cos x) \)
t का मान वापस रखने पर,
= \( -\sin(\sin x) \cos x \)
In simple words: श्रृंखला नियम का उपयोग करें। पहले \( \cos \) का अवकलन \( -\sin \) में करें, उसके अंदर का \( \sin x \) वैसे ही रखें। फिर अंदर के \( \sin x \) का अवकलन \( \cos x \) में करें। इन दोनों परिणामों को गुणा करें।

🎯 Exam Tip: श्रृंखला नियम को हमेशा चरण-दर-चरण लागू करें: बाहरी फलन का अवकलन करें, फिर आंतरिक फलन का अवकलन करें, और परिणामों को गुणा करें। \( \cos x \) का अवकलन \( -\sin x \) है और \( \sin x \) का अवकलन \( \cos x \) है।

प्रश्नावली 5.2

Question 3. sin (ax+b) हल- माना \( y = \sin(ax + b) \) \( ax + b = t \) रखने पर, \( y = \sin t \) दोनों पक्षों का \( x \) के सापेक्ष अवकलन करने पर,
\( \frac{dy}{dt} = \frac{d}{dt} (\sin t) \)
\( \implies \frac{dy}{dt} = \cos t \)
तथा \( t = ax + b \)
दोनों पक्षों का \( x \) के सापेक्ष अवकलन करने पर,
\( \frac{dt}{dx} = \frac{d}{dx} (ax + b) = a \)
\( \implies \frac{dy}{dx} = \frac{dy}{dt} . \frac{dt}{dx} = \cos t . a = a \cos t = a \cos (ax + b) \)
In simple words: To differentiate \( \sin(ax+b) \), we use the chain rule. First differentiate \( \sin(\text{something}) \) to get \( \cos(\text{something}) \), then multiply by the derivative of the inside function \( (ax+b) \), which is \( a \).

🎯 Exam Tip: Mastering the chain rule is crucial for these types of differentiation problems, especially when functions are nested. Pay attention to the derivative of the inner function.

 

Question 4. sec (tan√x) हल- माना \( y = \sec (\tan (\sqrt{x})) \) \( x \) के सापेक्ष अवकलन करने पर,
\( \frac{dy}{dx} = \frac{d}{dx} [\sec (\tan \sqrt{x})] \)
\( \implies = \sec (\tan \sqrt{x}) \tan (\tan \sqrt{x}) \frac{d}{dx} (\tan \sqrt{x}) \)
\( \implies = \sec (\tan \sqrt{x}) \tan (\tan \sqrt{x}) \sec^2 \sqrt{x} \frac{d}{dx} (\sqrt{x}) \)
\( \implies = \sec (\tan \sqrt{x}) \tan (\tan \sqrt{x}) (\sec^2 \sqrt{x}) (\frac{1}{2\sqrt{x}}) \)
\( \implies = \frac{1}{2\sqrt{x}} \sec (\tan \sqrt{x}) \tan (\tan \sqrt{x}) (\sec^2 \sqrt{x}) \)
In simple words: This problem involves multiple applications of the chain rule. Differentiate from the outermost function (\( \sec \)) inwards (\( \tan \), then \( \sqrt{x} \)), multiplying the derivatives at each step.

🎯 Exam Tip: For complex nested functions, break down the differentiation into layers. Identify the outermost function first, then its argument, and continue until the innermost function is differentiated.

 

Question 5. sin(ax+b) / cos(cx+d) हल- माना \( y = \frac{\sin(ax+b)}{\cos(cx+d)} \) ...(1)
समी० (1) का \( x \) के सापेक्ष अवकलन करने पर,
\( \frac{dy}{dx} = \frac{d}{dx} \left[ \frac{\sin(ax+b)}{\cos(cx+d)} \right] \)
\( \implies \frac{dy}{dx} = \frac{\cos(cx+d) \frac{d}{dx} \sin(ax+b) - \sin(ax+b) \frac{d}{dx} \cos(cx+d)}{\cos^2(cx+d)} \)
\( \implies = \frac{\cos(cx+d) \cos(ax+b) (a) - \sin(ax+b) [-\sin(cx+d) (c)]}{\cos^2(cx+d)} \)
\( \implies = \frac{a \cos(cx+d) \cos(ax+b) + c \sin(ax+b) \sin(cx+d)}{\cos^2(cx+d)} \)
\( \implies = a \cos(ax+b) \sec(cx+d) + c \sin(ax+b) \tan(cx+d) \sec(cx+d) \)
In simple words: This is a quotient rule problem. Apply the formula \(\frac{d}{dx}(\frac{u}{v}) = \frac{v \frac{du}{dx} - u \frac{dv}{dx}}{v^2}\), and remember to use the chain rule for differentiating \(\sin(ax+b)\) and \(\cos(cx+d)\).

🎯 Exam Tip: The quotient rule is a fundamental differentiation technique. Practice with inverse trigonometric functions and chain rule applications within the quotient rule for higher scores.

 

Question 6. cos x³.sin² x⁵ हल- माना \( y = \cos x^3 . \sin^2 x^5 \)
समी० (1) का \( x \) के सापेक्ष अवकलन करने पर,
\( \frac{dy}{dx} = \frac{d}{dx} [\cos x^3 . \sin^2 x^5] \)
\( \implies = \cos x^3 \frac{d}{dx} \sin^2 x^5 + \sin^2 x^5 \frac{d}{dx} \cos x^3 \)
\( \implies = \cos x^3 . 2 \sin x^5 \frac{d}{dx} \sin x^5 + \sin^2 x^5 (-\sin x^3) \frac{d}{dx} x^3 \)
\( \implies = 2 \cos x^3 \sin x^5 \cos x^5 \frac{d}{dx} x^5 + \sin^2 x^5 (-\sin x^3) 3x^2 \)
\( \implies = 2 \cos x^3 \sin x^5 \cos x^5 (5x^4) - 3 \sin^2 x^5 \sin x^3 . x^2 \)
\( \implies = 10 x^4 \cos x^3 \sin x^5 \cos x^5 - 3 x^2 \sin^2 x^5 \sin x^3 \)
In simple words: This function is a product of two terms, each requiring the chain rule. Use the product rule \((uv)' = u'v + uv'\), and for each term, apply the chain rule carefully.

🎯 Exam Tip: When using the product rule, ensure you differentiate each term completely using the chain rule before combining them. Keep track of the signs and powers, especially with trigonometric functions.

 

Question 7. \( 2\sqrt { \cot\left( { x }^{ 2 } \right) } \) हल- माना \( y = 2 (\cot x^2)^{1/2} \) ...(1)
समी० (1) का \( x \) के सापेक्ष अवकलन करने पर,
\( \frac{dy}{dx} = 2 \cdot \frac{1}{2} (\cot x^2)^{1/2 - 1} \frac{d}{dx} (\cot x^2) \)
\( \implies = (\cot x^2)^{-1/2} (-\operatorname{cosec}^2 x^2) \frac{d}{dx} x^2 \)
\( \implies = (\cot x^2)^{-1/2} (-\operatorname{cosec}^2 x^2) \cdot 2x \)
\( \implies = \frac{-2x \operatorname{cosec}^2 x^2}{\sqrt{\cot x^2}} \)
In simple words: This problem uses the chain rule. First differentiate the outer square root function, then the cotangent function, and finally the inner \( x^2 \) term.

🎯 Exam Tip: Remember that \(\frac{d}{dx}(\sqrt{u}) = \frac{1}{2\sqrt{u}} \frac{du}{dx}\). Also, the derivative of \(\cot u\) is \(-\operatorname{cosec}^2 u \frac{du}{dx}\). Be meticulous with negative signs.

 

Question 8. cos(√x) हल- माना \( y = \cos(\sqrt{x}) \) ...(1)
समी० (1) का \( x \) के सापेक्ष अवकलन करने पर,
\( \frac{dy}{dx} = \frac{d}{dx} (\cos \sqrt{x}) \)
\( \implies = -\sin \sqrt{x} \frac{d}{dx} \sqrt{x} \)
\( \implies = -\sin \sqrt{x} \frac{d}{dx} x^{1/2} \)
\( \implies = -\sin \sqrt{x} \cdot \frac{1}{2} x^{1/2 - 1} \)
\( \implies = -\sin \sqrt{x} \cdot \frac{1}{2} x^{-1/2} \)
\( \implies = -\frac{\sin \sqrt{x}}{2\sqrt{x}} \)
In simple words: Apply the chain rule. First, differentiate the cosine function, then differentiate its argument, which is \(\sqrt{x}\).

🎯 Exam Tip: The derivative of \(\sqrt{x}\) is a common one. Knowing it as \(\frac{1}{2\sqrt{x}}\) can save time. Pay attention to the negative sign when differentiating cosine.

 

Question 9. सिद्ध कीजिए कि फलन \( f(x) = |x - 1|, x \in R, x = 1 \) पर अवकलित नहीं है। हल- दिया है- \( f(x) = |x - 1| \)
या \( f(x) = \begin{cases} x-1, & \text{यदि } x \ge 1 \\ 1-x, & \text{यदि } x < 1 \end{cases} \)
\( x = 1 \) पर,
R.H.D. \( = \lim_{h \to 0} \frac{f(1+h)-f(1)}{h} = \lim_{h \to 0} \frac{[(1+h)-1]-(1-1)}{h} \)
\( \implies = \lim_{h \to 0} \frac{h}{h} = 1 \)
L.H.D. \( = \lim_{h \to 0} \frac{f(1-h)-f(1)}{-h} = \lim_{h \to 0} \frac{[1-(1-h)]-(1-1)}{-h} \)
\( \implies = \lim_{h \to 0} \frac{h}{-h} = -1 \)
R.H.D. \( \ne \) L.H.D. अतः \( x = 1 \) पर \( f \) अवकलनीय नहीं है। इति सिद्धम्
In simple words: A function is differentiable at a point if its left-hand derivative (LHD) equals its right-hand derivative (RHD) at that point. For \( |x-1| \) at \( x=1 \), the LHD is -1 and the RHD is 1, so they are not equal, meaning it's not differentiable.

🎯 Exam Tip: To prove non-differentiability at a point, calculate both the LHD and RHD using the limit definition. If they are unequal, the function is not differentiable at that point.

 

Question 10. सिद्ध कीजिए कि महत्तम पूर्णाक फलन \( f(x) = [x], 0 < x < 3, x = 1 \) तथा \( x = 2 \) पर अवकलित नहीं है। हल- ज्ञात है, \( f(x) = [x] \)
(i) \( x = 1 \) पर,
R\(f' (1) = \lim_{h \to 0} \frac{f(1+h)-f(1)}{h} = \lim_{h \to 0} \frac{[1+h]-[1]}{h} = \lim_{h \to 0} \frac{1-1}{h} = 0 \)
L\(f' (1) = \lim_{h \to 0} \frac{f(1-h)-f(1)}{-h} = \lim_{h \to 0} \frac{[1-h]-[1]}{-h} = \lim_{h \to 0} \frac{0-1}{-h} = \text{परिभाषित नहीं है।} \)
\( \therefore \) R\(f' (1) \ne \) L\(f' (1) \)
\( \implies x = 1 \) पर \( f \) अवकलनीय नहीं है। इति सिद्धम्
(ii) \( x = 2 \) पर,
R\(f' (2) = \lim_{h \to 0} \frac{f(2+h)-f(2)}{h} = \lim_{h \to 0} \frac{[2+h]-[2]}{h} = \lim_{h \to 0} \frac{2-2}{h} = 0 \)
L\(f' (2) = \lim_{h \to 0} \frac{f(2-h)-f(2)}{-h} = \lim_{h \to 0} \frac{[2-h]-[2]}{-h} = \lim_{h \to 0} \frac{1-2}{-h} = \text{परिभाषित नहीं है।} \)
\( \therefore \) R.H.D. \( \ne \) L.H.D.
अतः \( x = 2 \) पर \( f \) अवकलनीय नहीं है। इति सिद्धम्
In simple words: The greatest integer function \( [x] \) has jumps at integer points. This means its left-hand and right-hand derivatives will not be equal at integer points, making it non-differentiable there.

🎯 Exam Tip: The greatest integer function is a classic example for non-differentiability at integer points. Remember that \( [c+h] = c \) for small \( h>0 \) and \( [c-h] = c-1 \) for small \( h>0 \) when \( c \) is an integer.

प्रश्नावली 5.3

Question 1. 2x + 3y = sinx हल- ज्ञात है, \( 2x + 3y = \sin x \) दोनों पक्षों का \( x \) के सापेक्ष अवकलन करने पर,
\( \frac{d}{dx} (2x) + \frac{d}{dx} (3y) = \frac{d}{dx} (\sin x) \)
\( \implies 2 \cdot 1 + 3 \frac{dy}{dx} = \cos x \)
\( \implies 3 \frac{dy}{dx} = \cos x - 2 \)
\( \implies \frac{dy}{dx} = \frac{\cos x - 2}{3} \)
In simple words: To find \( \frac{dy}{dx} \) for an implicit function, differentiate both sides of the equation with respect to \( x \), treating \( y \) as a function of \( x \) and using the chain rule for terms involving \( y \). Then, solve for \( \frac{dy}{dx} \).

🎯 Exam Tip: Implicit differentiation requires careful application of the chain rule. Every term involving \( y \) (e.g., \( y^n \), \( \sin y \)) must be multiplied by \( \frac{dy}{dx} \) after differentiating with respect to \( y \).

 

Question 2. 2x + 3y = siny हल- दिया है- \( 2x + 3y = \sin y \) दोनों पक्षों का \( x \) के सापेक्ष अवकलन करने पर,
\( \frac{d}{dx} (2x) + \frac{d}{dx} (3y) = \frac{d}{dx} (\sin y) \)
\( \implies 2 \cdot 1 + 3 \frac{dy}{dx} = \cos y \frac{dy}{dx} \)
\( \implies 2 = \cos y \frac{dy}{dx} - 3 \frac{dy}{dx} \)
\( \implies 2 = \frac{dy}{dx} (\cos y - 3) \)
\( \implies \frac{dy}{dx} = \frac{2}{\cos y - 3} \)
In simple words: Differentiate both sides of the equation with respect to \( x \). When differentiating \( y \) or a function of \( y \), remember to multiply by \( \frac{dy}{dx} \). Then, rearrange the equation to isolate \( \frac{dy}{dx} \).

🎯 Exam Tip: Pay close attention to the argument of trigonometric functions. When differentiating \( \sin y \) with respect to \( x \), it becomes \( \cos y \cdot \frac{dy}{dx} \), not just \( \cos y \).

 

Question 3. ax + by² = cosy हल- ज्ञात है- \( ax + by^2 = \cos y \) दोनों पक्षों का \( x \) के सापेक्ष अवकलन करने पर,
\( a \frac{d}{dx} (x) + b \frac{d}{dx} (y^2) = \frac{d}{dx} (\cos y) \)
\( \implies a \cdot 1 + b \cdot 2y \frac{dy}{dx} = -\sin y \frac{dy}{dx} \)
\( \implies a + 2by \frac{dy}{dx} = -\sin y \frac{dy}{dx} \)
\( \implies a = -\sin y \frac{dy}{dx} - 2by \frac{dy}{dx} \)
\( \implies a = \frac{dy}{dx} (-2by - \sin y) \)
\( \implies \frac{dy}{dx} = \frac{a}{-(2by + \sin y)} \)
\( \implies \frac{dy}{dx} = -\frac{a}{2by + \sin y} \)
In simple words: Differentiate each term of the equation with respect to \( x \). For terms involving \( y \), remember to use the chain rule and include \( \frac{dy}{dx} \). Then, collect all \( \frac{dy}{dx} \) terms and solve for it.

🎯 Exam Tip: Be careful with signs, especially when differentiating cosine. When isolating \( \frac{dy}{dx} \), make sure to factor it out correctly before dividing.

 

Question 4. xy + y² = tan x + y हल- दिया है \( xy + y^2 = \tan x + y \) दोनों पक्षों का \( x \) के सापेक्ष अवकलन करने पर,
\( \frac{d}{dx} (xy) + \frac{d}{dx} (y^2) = \frac{d}{dx} (\tan x) + \frac{d}{dx} (y) \)
\( \implies (x \frac{dy}{dx} + y \cdot 1) + 2y \frac{dy}{dx} = \sec^2 x + \frac{dy}{dx} \)
\( \implies x \frac{dy}{dx} + 2y \frac{dy}{dx} - \frac{dy}{dx} = \sec^2 x - y \)
\( \implies \frac{dy}{dx} (x + 2y - 1) = \sec^2 x - y \)
\( \implies \frac{dy}{dx} = \frac{\sec^2 x - y}{x + 2y - 1} \)
In simple words: Differentiate each term using appropriate rules: product rule for \( xy \), chain rule for \( y^2 \), and standard derivatives for \( \tan x \). Remember to attach \( \frac{dy}{dx} \) for all derivatives involving \( y \). Finally, group \( \frac{dy}{dx} \) terms and solve.

🎯 Exam Tip: The product rule \((uv)' = u'v + uv'\) is essential for terms like \( xy \). When solving for \( \frac{dy}{dx} \), make sure all terms without \( \frac{dy}{dx} \) are on one side, and all terms with it are on the other.

 

Question 5. x² + xy + y² = 100 हल- दिया है \( x^2 + xy + y^2 = 100 \) दोनों पक्षों का \( x \) के सापेक्ष अवकलन करने पर,
\( \frac{d}{dx} (x^2) + \frac{d}{dx} (xy) + \frac{d}{dx} (y^2) = \frac{d}{dx} (100) \)
\( \implies 2x + (x \frac{dy}{dx} + y \cdot 1) + 2y \frac{dy}{dx} = 0 \)
\( \implies 2x + x \frac{dy}{dx} + y + 2y \frac{dy}{dx} = 0 \)
\( \implies x \frac{dy}{dx} + 2y \frac{dy}{dx} = -(2x + y) \)
\( \implies \frac{dy}{dx} (x + 2y) = -(2x + y) \)
\( \implies \frac{dy}{dx} = -\frac{2x + y}{x + 2y} \)
In simple words: Differentiate term by term. For \( x^2 \), it's \( 2x \). For \( xy \), use the product rule. For \( y^2 \), use the chain rule, resulting in \( 2y \frac{dy}{dx} \). The derivative of a constant (100) is 0. Then, algebraically solve for \( \frac{dy}{dx} \).

🎯 Exam Tip: The key here is correctly applying the product rule for \( xy \), which yields \( x \frac{dy}{dx} + y \), and the chain rule for \( y^2 \), which gives \( 2y \frac{dy}{dx} \). Common errors include forgetting \( \frac{dy}{dx} \) for \( y \) terms.

 

Question 6. x³ + x²y + xy² + y³ = 81 हल- दिया है \( x^3 + x^2y + xy^2 + y^3 = 81 \)
दोनों पक्षों का \( x \) के सापेक्ष अवकलन करने पर,
\( \frac{d}{dx} (x^3) + \frac{d}{dx} (x^2y) + \frac{d}{dx} (xy^2) + \frac{d}{dx} (y^3) = \frac{d}{dx} (81) \)
\( \implies 3x^2 + (x^2 \frac{dy}{dx} + y \cdot 2x) + (x \cdot 2y \frac{dy}{dx} + y^2 \cdot 1) + 3y^2 \frac{dy}{dx} = 0 \)
\( \implies 3x^2 + x^2 \frac{dy}{dx} + 2xy + 2xy \frac{dy}{dx} + y^2 + 3y^2 \frac{dy}{dx} = 0 \)
\( \implies x^2 \frac{dy}{dx} + 2xy \frac{dy}{dx} + 3y^2 \frac{dy}{dx} = -(3x^2 + 2xy + y^2) \)
\( \implies \frac{dy}{dx} (x^2 + 2xy + 3y^2) = -(3x^2 + 2xy + y^2) \)
\( \implies \frac{dy}{dx} = -\frac{3x^2 + 2xy + y^2}{x^2 + 2xy + 3y^2} \)
In simple words: Differentiate each term. For \( x^3 \), it's \( 3x^2 \). For \( x^2y \) and \( xy^2 \), use the product rule and chain rule carefully. For \( y^3 \), use the chain rule. Then, gather all \( \frac{dy}{dx} \) terms and solve.

🎯 Exam Tip: This problem tests multiple applications of the product and chain rules in implicit differentiation. Keep an organized approach to avoid errors in distributing derivatives and collecting terms.

 

Question 7. sin² y + cos xy = k हल- दिया है \( \sin^2 y + \cos xy = k \) दोनों पक्षों का \( x \) के सापेक्ष अवकलन करने पर,
\( \frac{d}{dx} (\sin^2 y) + \frac{d}{dx} (\cos xy) = \frac{d}{dx} (k) \)
\( \implies 2 \sin y \cos y \frac{dy}{dx} + (-\sin xy) \frac{d}{dx} (xy) = 0 \)
\( \implies \sin 2y \frac{dy}{dx} - \sin xy (x \frac{dy}{dx} + y \cdot 1) = 0 \)
\( \implies \sin 2y \frac{dy}{dx} - x \sin xy \frac{dy}{dx} - y \sin xy = 0 \)
\( \implies \sin 2y \frac{dy}{dx} - x \sin xy \frac{dy}{dx} = y \sin xy \)
\( \implies \frac{dy}{dx} (\sin 2y - x \sin xy) = y \sin xy \)
\( \implies \frac{dy}{dx} = \frac{y \sin xy}{\sin 2y - x \sin xy} \)
In simple words: Differentiate \( \sin^2 y \) using the chain rule (outer square, then \( \sin y \)). For \( \cos xy \), use the chain rule, and within that, the product rule for \( xy \). The derivative of the constant \( k \) is 0. Isolate \( \frac{dy}{dx} \) by collecting terms.

🎯 Exam Tip: Recognize \( 2 \sin y \cos y = \sin 2y \) for simplification, though it's not strictly necessary for the derivative. Ensure the product rule is correctly applied within the chain rule for the \( \cos xy \) term.

 

Question 8. sin² x + cos² y = 1 हल- दिया है \( \sin^2 x + \cos^2 y = 1 \) दोनों पक्षों का \( x \) के सापेक्ष अवकलन करने पर,
\( \frac{d}{dx} (\sin^2 x) + \frac{d}{dx} (\cos^2 y) = \frac{d}{dx} (1) \)
\( \implies 2 \sin x \cos x \frac{d}{dx} (x) + 2 \cos y (-\sin y) \frac{d}{dx} (y) = 0 \)
\( \implies 2 \sin x \cos x \cdot 1 - 2 \sin y \cos y \frac{dy}{dx} = 0 \)
\( \implies \sin 2x - \sin 2y \frac{dy}{dx} = 0 \)
\( \implies \sin 2y \frac{dy}{dx} = \sin 2x \)
\( \implies \frac{dy}{dx} = \frac{\sin 2x}{\sin 2y} \)
In simple words: Differentiate each square term using the chain rule. For \( \sin^2 x \), differentiate \( (\sin x)^2 \) and then \( \sin x \). For \( \cos^2 y \), differentiate \( (\cos y)^2 \), then \( \cos y \), and finally \( y \) (which adds \( \frac{dy}{dx} \)). Simplify using trigonometric identities and solve for \( \frac{dy}{dx} \).

🎯 Exam Tip: The identities \( 2 \sin x \cos x = \sin 2x \) and \( 2 \sin y \cos y = \sin 2y \) are useful here. Correct application of the chain rule is paramount, especially for the \( y \) terms.

 

Question 9. \( y = \sin^{-1} \left(\frac{2x}{1+x^2}\right) \) हल- दिया है-
\( y = \sin^{-1} \left(\frac{2x}{1+x^2}\right) \)
माना \( x = \tan \theta \) पर \( \theta = \tan^{-1} x \)
\( \implies y = \sin^{-1} \left(\frac{2 \tan \theta}{1+\tan^2 \theta}\right) \)
\( \implies y = \sin^{-1} (\sin 2\theta) \)
\( \implies y = 2\theta \)
\( \implies y = 2 \tan^{-1} x \)
दोनों पक्षों का \( x \) के सापेक्ष अवकलन करने पर,
\( \frac{dy}{dx} = 2 \frac{d}{dx} (\tan^{-1} x) \)
\( \implies \frac{dy}{dx} = 2 \cdot \frac{1}{1+x^2} \)
\( \implies \frac{dy}{dx} = \frac{2}{1+x^2} \)
In simple words: This problem is simplified by a substitution. Let \( x = \tan \theta \) to transform the expression inside \( \sin^{-1} \) into \( \sin 2\theta \). Then \( y \) simplifies to \( 2\theta \), which is \( 2 \tan^{-1} x \). Differentiating this simple form gives the answer.

🎯 Exam Tip: Recognizing trigonometric substitutions like \( x = \sin \theta \), \( x = \tan \theta \), or \( x = \cos \theta \) can significantly simplify inverse trigonometric differentiation problems. This saves complex chain rule calculations.

 

Question 10. \( y = \tan^{-1} \left(\frac{3x-x^3}{1-3x^2}\right) \) हल- माना \( y = \tan^{-1} \left(\frac{3x-x^3}{1-3x^2}\right) \)
\( x = \tan \theta \) प्रतिस्थापित करने पर, \( \theta = \tan^{-1} x \)
\( \implies y = \tan^{-1} \left(\frac{3 \tan \theta - \tan^3 \theta}{1-3 \tan^2 \theta}\right) \)
\( \implies y = \tan^{-1} (\tan 3\theta) \)
\( \implies y = 3\theta \)
\( \implies y = 3 \tan^{-1} x \)
दोनों पक्षों का \( x \) के सापेक्ष अवकलन करने पर,
\( \frac{dy}{dx} = \frac{d}{dx} (3 \tan^{-1} x) \)
\( \implies \frac{dy}{dx} = 3 \cdot \frac{1}{1+x^2} \)
\( \implies \frac{dy}{dx} = \frac{3}{1+x^2} \)
In simple words: Use the substitution \( x = \tan \theta \). The expression inside the inverse tangent simplifies to \( \tan 3\theta \), a standard trigonometric identity. This makes \( y = 3\theta \), or \( y = 3 \tan^{-1} x \). Differentiating this simplified form is straightforward.

🎯 Exam Tip: Memorize inverse trigonometric identities, especially for \( \tan^{-1} \), \( \sin^{-1} \), and \( \cos^{-1} \), as they are frequently used in simplification before differentiation.

 

Question 11. \( y = \tan^{-1} \left(\frac{2x}{1-x^2}\right) \) हल- माना \( y = \tan^{-1} \left(\frac{2x}{1-x^2}\right) \)
\( x = \tan \theta \) प्रतिस्थापित करने पर, \( \theta = \tan^{-1} x \)
\( \implies y = \tan^{-1} \left(\frac{2 \tan \theta}{1-\tan^2 \theta}\right) \)
\( \implies y = \tan^{-1} (\tan 2\theta) \)
\( \implies y = 2\theta \)
\( \implies y = 2 \tan^{-1} x \)
दोनों पक्षों का \( x \) के सापेक्ष अवकलन करने पर,
\( \frac{dy}{dx} = 2 \frac{d}{dx} (\tan^{-1} x) \)
\( \implies \frac{dy}{dx} = 2 \cdot \frac{1}{1+x^2} \)
\( \implies \frac{dy}{dx} = \frac{2}{1+x^2} \)
In simple words: Substitute \( x = \tan \theta \). The expression \( \frac{2x}{1-x^2} \) becomes \( \tan 2\theta \). So, \( y \) simplifies to \( 2\theta \) or \( 2 \tan^{-1} x \). Differentiate this simple form to get the final answer.

🎯 Exam Tip: The identity \( \tan 2\theta = \frac{2 \tan \theta}{1 - \tan^2 \theta} \) is a critical recognition for solving this problem efficiently.

 

Question 12. \( y = \sin^{-1} \left(\frac{1-x^2}{1+x^2}\right) \) हल- ज्ञात है- \( y = \sin^{-1} \left(\frac{1-x^2}{1+x^2}\right) \)
माना \( x = \tan \theta \) रखने पर, तब \( \theta = \tan^{-1} x \)
\( \implies y = \sin^{-1} \left(\frac{1-\tan^2 \theta}{1+\tan^2 \theta}\right) \)
\( \implies y = \sin^{-1} (\cos 2\theta) \)
\( \implies y = \sin^{-1} \left(\sin \left(\frac{\pi}{2} - 2\theta\right)\right) \)
\( \implies y = \frac{\pi}{2} - 2\theta \)
\( \implies y = \frac{\pi}{2} - 2 \tan^{-1} x \)
दोनों पक्षों \( x \) के सापेक्ष अवकलन करने पर,
\( \frac{dy}{dx} = \frac{d}{dx} \left(\frac{\pi}{2}\right) - 2 \frac{d}{dx} (\tan^{-1} x) \)
\( \implies = 0 - 2 \cdot \frac{1}{1+x^2} \)
\( \implies = -\frac{2}{1+x^2} \)
In simple words: Substitute \( x = \tan \theta \). The expression simplifies to \( \cos 2\theta \). Then, convert \( \cos 2\theta \) to \( \sin(\frac{\pi}{2} - 2\theta) \) so it cancels with \( \sin^{-1} \). This simplifies \( y \) to \( \frac{\pi}{2} - 2 \tan^{-1} x \), which is easy to differentiate.

🎯 Exam Tip: When you have \(\sin^{-1}(\cos \text{A})\), use the identity \(\cos \text{A} = \sin(\frac{\pi}{2} - \text{A})\) to simplify. Similarly, for \(\cos^{-1}(\sin \text{A})\), use \(\sin \text{A} = \cos(\frac{\pi}{2} - \text{A})\).

 

Question 13. \( y = \cos^{-1} \left(\frac{2x}{1+x^2}\right), -1 हल- दिया है- \( y = \cos^{-1} \left(\frac{2x}{1+x^2}\right) \)
माना \( x = \tan \theta \) रखने पर, \( \theta = \tan^{-1} x \)
\( \implies y = \cos^{-1} \left(\frac{2 \tan \theta}{1+\tan^2 \theta}\right) \)
\( \implies y = \cos^{-1} (\sin 2\theta) \)
\( \implies y = \cos^{-1} \left(\cos \left(\frac{\pi}{2} - 2\theta\right)\right) \)
\( \implies y = \frac{\pi}{2} - 2\theta \)
\( \implies y = \frac{\pi}{2} - 2 \tan^{-1} x \)
दोनों पक्षों का \( x \) के सापेक्ष अवकलन करने पर,
\( \frac{dy}{dx} = \frac{d}{dx} \left(\frac{\pi}{2}\right) - 2 \frac{d}{dx} (\tan^{-1} x) \)
\( \implies = 0 - 2 \cdot \frac{1}{1+x^2} \)
\( \implies = -\frac{2}{1+x^2} \)
In simple words: Use the substitution \( x = \tan \theta \), which transforms the argument into \( \sin 2\theta \). Convert \( \sin 2\theta \) to \( \cos(\frac{\pi}{2} - 2\theta) \) to simplify with \( \cos^{-1} \). This leads to \( y = \frac{\pi}{2} - 2 \tan^{-1} x \), which is then differentiated.

🎯 Exam Tip: The condition \( -1

 

Question 14. \( y = \sin^{-1} (2x\sqrt{1-x^2}) \) हल- माना \( y = \sin^{-1} (2x\sqrt{1-x^2}) \)
\( x = \sin \theta \) प्रतिस्थापित करने पर, \( \theta = \sin^{-1} x \)
\( \implies y = \sin^{-1} (2 \sin \theta \sqrt{1-\sin^2 \theta}) \)
\( \implies y = \sin^{-1} (2 \sin \theta \cos \theta) \)
\( \implies y = \sin^{-1} (\sin 2\theta) \)
\( \implies y = 2\theta \)
\( \implies y = 2 \sin^{-1} x \)
दोनों पक्षों का \( x \) के सापेक्ष अवकलन करने पर,
\( \frac{dy}{dx} = 2 \frac{d}{dx} (\sin^{-1} x) \)
\( \implies \frac{dy}{dx} = 2 \cdot \frac{1}{\sqrt{1-x^2}} \)
\( \implies \frac{dy}{dx} = \frac{2}{\sqrt{1-x^2}} \)
In simple words: Substitute \( x = \sin \theta \). The expression inside \( \sin^{-1} \) simplifies to \( 2 \sin \theta \cos \theta \), which is \( \sin 2\theta \). This gives \( y = 2\theta \), or \( y = 2 \sin^{-1} x \). Differentiate this simple form.

🎯 Exam Tip: For expressions involving \( \sqrt{1-x^2} \), a substitution of \( x = \sin \theta \) or \( x = \cos \theta \) is often effective. This leads to common trigonometric identities like \( \sin 2\theta \).

 

Question 15. \( y = \sec^{-1} \left(\frac{1}{2x^2-1}\right) \) हल- माना \( y = \sec^{-1} \left(\frac{1}{2x^2-1}\right) \)
\( x = \cos \theta \) प्रतिस्थापित करने पर, \( \theta = \cos^{-1} x \)
\( \implies y = \sec^{-1} \left(\frac{1}{2 \cos^2 \theta - 1}\right) \)
\( \implies y = \sec^{-1} \left(\frac{1}{\cos 2\theta}\right) \)
\( \implies y = \sec^{-1} (\sec 2\theta) \)
\( \implies y = 2\theta \)
\( \implies y = 2 \cos^{-1} x \)
दोनों पक्षों का \( x \) के सापेक्ष अवकलन करने पर,
\( \frac{dy}{dx} = 2 \frac{d}{dx} (\cos^{-1} x) \)
\( \implies \frac{dy}{dx} = 2 \cdot \left(-\frac{1}{\sqrt{1-x^2}}\right) \)
\( \implies \frac{dy}{dx} = -\frac{2}{\sqrt{1-x^2}} \)
In simple words: Use the substitution \( x = \cos \theta \). The argument of \( \sec^{-1} \) becomes \( \frac{1}{2 \cos^2 \theta - 1} \), which simplifies to \( \frac{1}{\cos 2\theta} \) or \( \sec 2\theta \). This makes \( y = 2\theta \), or \( y = 2 \cos^{-1} x \). Differentiate this simplified form directly.

🎯 Exam Tip: The identity \( \cos 2\theta = 2 \cos^2 \theta - 1 \) is key here. Also, remember that \( \sec^{-1}(\sec A) = A \) in the appropriate domain, and the derivative of \( \cos^{-1} x \) is \( -\frac{1}{\sqrt{1-x^2}} \).

प्रश्नावली 5.4

Question 1. sin x / e x हल- माना \( y = \frac{\sin x}{e^x} \)
दोनों पक्षों का \( x \) के सापेक्ष अवकलन करने पर,
\( \frac{dy}{dx} = \frac{e^x \frac{d}{dx} (\sin x) - \sin x \frac{d}{dx} (e^x)}{(e^x)^2} \)
\( \implies = \frac{e^x \cos x - \sin x e^x}{e^{2x}} \)
\( \implies = \frac{e^x (\cos x - \sin x)}{e^{2x}} \)
\( \implies = \frac{\cos x - \sin x}{e^x} \)
In simple words: This is a quotient rule problem. Apply the formula \(\frac{d}{dx}(\frac{u}{v}) = \frac{v \frac{du}{dx} - u \frac{dv}{dx}}{v^2}\), with \( u = \sin x \) and \( v = e^x \).

🎯 Exam Tip: Remember the basic derivatives: \( \frac{d}{dx}(\sin x) = \cos x \) and \( \frac{d}{dx}(e^x) = e^x \). Simplification of the final expression by factoring out \( e^x \) is good practice.

 

Question 2. \( e^{\sin^{-1} x} \) हल- माना \( y = e^{\sin^{-1} x} \)
दोनों पक्षों का \( x \) के सापेक्ष अवकलन करने पर,
\( \frac{dy}{dx} = \frac{d}{dx} (e^{\sin^{-1} x}) \)
\( \implies = e^{\sin^{-1} x} \frac{d}{dx} (\sin^{-1} x) \)
\( \implies = e^{\sin^{-1} x} \cdot \frac{1}{\sqrt{1-x^2}} \)
\( \implies = \frac{e^{\sin^{-1} x}}{\sqrt{1-x^2}} \)
In simple words: This requires the chain rule. Differentiate the outer exponential function first, which gives \( e^{\text{something}} \). Then, multiply by the derivative of the exponent, which is \( \sin^{-1} x \).

🎯 Exam Tip: The derivative of \( e^u \) is \( e^u \frac{du}{dx} \). Also, remember the derivative of \( \sin^{-1} x \) is \( \frac{1}{\sqrt{1-x^2}} \).

 

Question 3. \( e^{x^3} \) हल- माना \( y = e^{x^3} \)
दोनों पक्षों का \( x \) के सापेक्ष अवकलन करने पर,
\( \frac{dy}{dx} = \frac{d}{dx} (e^{x^3}) \)
\( \implies = e^{x^3} \frac{d}{dx} (x^3) \)
\( \implies = e^{x^3} \cdot 3x^2 \)
\( \implies = 3x^2 e^{x^3} \)
In simple words: Apply the chain rule. Differentiate the exponential function, which is \( e^{\text{something}} \), and then multiply by the derivative of the exponent, which is \( x^3 \).

🎯 Exam Tip: Chain rule is fundamental for derivatives like \( e^{f(x)} \). The derivative is \( e^{f(x)} \cdot f'(x) \). Here, \( f(x) = x^3 \), so \( f'(x) = 3x^2 \).

 

Question 4. sin(tan⁻¹ eˣ) हल- माना \( y = \sin(\tan^{-1} e^x) \)
दोनों पक्षों का \( x \) के सापेक्ष अवकलन करने पर,
\( \frac{dy}{dx} = \frac{d}{dx} (\sin(\tan^{-1} e^x)) \)
\( \implies = \cos(\tan^{-1} e^x) \frac{d}{dx} (\tan^{-1} e^x) \)
\( \implies = \cos(\tan^{-1} e^x) \cdot \frac{1}{1+(e^x)^2} \frac{d}{dx} (e^x) \)
\( \implies = \cos(\tan^{-1} e^x) \cdot \frac{1}{1+e^{2x}} \cdot e^x \)
\( \implies = \frac{e^x \cos(\tan^{-1} e^x)}{1+e^{2x}} \)
In simple words: This requires a three-layer chain rule. Differentiate sine, then inverse tangent, then the exponential function, multiplying the results at each step.

🎯 Exam Tip: For nested functions like this, visualize the layers of the chain rule. \(\frac{d}{dx}[\sin(f(g(x)))] = \cos(f(g(x))) \cdot f'(g(x)) \cdot g'(x)\). Here \(f(u) = \tan^{-1} u\) and \(g(x) = e^x\).

 

Question 5. log(cos eˣ) हल- माना \( y = \log(\cos e^x) \)
दोनों पक्षों का \( x \) के सापेक्ष अवकलन करने पर,
\( \frac{dy}{dx} = \frac{d}{dx} (\log(\cos e^x)) \)
\( \implies = \frac{1}{\cos e^x} \frac{d}{dx} (\cos e^x) \)
\( \implies = \frac{1}{\cos e^x} (-\sin e^x) \frac{d}{dx} (e^x) \)
\( \implies = \frac{-\sin e^x}{\cos e^x} \cdot e^x \)
\( \implies = -e^x \tan e^x \)
In simple words: Use the chain rule three times. Differentiate log, then cosine, then exponential, multiplying the derivatives at each stage.

🎯 Exam Tip: Recall that \( \frac{d}{dx}(\log u) = \frac{1}{u} \frac{du}{dx} \) and \( \frac{d}{dx}(\cos u) = -\sin u \frac{du}{dx} \). Be careful with the negative sign from the cosine derivative.

 

Question 6. \( e^x + e^{x^2} + e^{x^3} + e^{x^4} + e^{x^5} \) हल- माना \( y = e^x + e^{x^2} + e^{x^3} + e^{x^4} + e^{x^5} \)
दोनों पक्षों का \( x \) के सापेक्ष अवकलन करने पर,
\( \frac{dy}{dx} = \frac{d}{dx} (e^x) + \frac{d}{dx} (e^{x^2}) + \frac{d}{dx} (e^{x^3}) + \frac{d}{dx} (e^{x^4}) + \frac{d}{dx} (e^{x^5}) \)
\( \implies = e^x + e^{x^2} \frac{d}{dx} (x^2) + e^{x^3} \frac{d}{dx} (x^3) + e^{x^4} \frac{d}{dx} (x^4) + e^{x^5} \frac{d}{dx} (x^5) \)
\( \implies = e^x + e^{x^2} (2x) + e^{x^3} (3x^2) + e^{x^4} (4x^3) + e^{x^5} (5x^4) \)
\( \implies = e^x + 2x e^{x^2} + 3x^2 e^{x^3} + 4x^3 e^{x^4} + 5x^4 e^{x^5} \)
In simple words: Differentiate each term separately. For each \( e^{x^n} \) term, apply the chain rule: differentiate \( e^{\text{power}} \) to get \( e^{\text{power}} \), then multiply by the derivative of the power \( x^n \).

🎯 Exam Tip: The derivative of a sum is the sum of the derivatives. For each term \( e^{x^n} \), remember \( \frac{d}{dx}(e^{x^n}) = e^{x^n} \cdot nx^{n-1} \).

 

Question 7. \( \sqrt{e^{\sqrt{x}}} \) हल- माना \( y = (e^{\sqrt{x}})^{1/2} \)
दोनों पक्षों का \( x \) के सापेक्ष अवकलन करने पर,
\( \frac{dy}{dx} = \frac{d}{dx} ((e^{\sqrt{x}})^{1/2}) \)
\( \implies = \frac{1}{2} (e^{\sqrt{x}})^{1/2 - 1} \frac{d}{dx} (e^{\sqrt{x}}) \)
\( \implies = \frac{1}{2} (e^{\sqrt{x}})^{-1/2} e^{\sqrt{x}} \frac{d}{dx} (\sqrt{x}) \)
\( \implies = \frac{1}{2} (e^{\sqrt{x}})^{-1/2} e^{\sqrt{x}} \frac{1}{2\sqrt{x}} \)
\( \implies = \frac{1}{4\sqrt{x}} e^{\sqrt{x} - \sqrt{x}/2} \)
\( \implies = \frac{1}{4\sqrt{x}} e^{\sqrt{x}/2} \)
\( \implies = \frac{e^{\sqrt{x}}}{4\sqrt{x}e^{\sqrt{x}/2}} \)
\( \implies = \frac{e^{\sqrt{x}}}{4\sqrt{x} \sqrt{e^{\sqrt{x}}}} \)
\( \implies = \frac{\sqrt{e^{\sqrt{x}}}}{4\sqrt{x}} \)
In simple words: Apply the chain rule in layers: first for the outermost square root, then for the exponential function, and finally for the innermost square root of \( x \).

🎯 Exam Tip: This problem involves nested functions and requires careful application of the chain rule. Remember \(\frac{d}{dx}(\sqrt{u}) = \frac{1}{2\sqrt{u}} \frac{du}{dx}\) and \(\frac{d}{dx}(e^u) = e^u \frac{du}{dx}\). Simplifying the exponents is crucial for the final form.

 

Question 8. log(log x), x>1 हल- माना \( y = \log(\log x) \), \( x>1 \)
दोनों पक्षों का \( x \) के सापेक्ष अवकलन करने पर,
\( \frac{dy}{dx} = \frac{d}{dx} (\log(\log x)) \)
\( \implies = \frac{1}{\log x} \frac{d}{dx} (\log x) \)
\( \implies = \frac{1}{\log x} \cdot \frac{1}{x} \)
\( \implies = \frac{1}{x \log x} \)
In simple words: This is a chain rule problem with nested logarithms. Differentiate the outer log function first, then multiply by the derivative of its argument (the inner log function).

🎯 Exam Tip: Remember that \( \frac{d}{dx}(\log u) = \frac{1}{u} \frac{du}{dx} \). Here, \( u = \log x \), so \( \frac{du}{dx} = \frac{1}{x} \).

 

Question 9. (log x - cos x) / (log x)² हल- माना \( y = \frac{\log x - \cos x}{\log x} \)
\( \implies y = 1 - \frac{\cos x}{\log x} \)
दोनों पक्षों का \( x \) के सापेक्ष अवकलन करने पर,
\( \frac{dy}{dx} = \frac{d}{dx} (1) - \frac{d}{dx} \left(\frac{\cos x}{\log x}\right) \)
\( \implies = 0 - \frac{\log x \frac{d}{dx}(\cos x) - \cos x \frac{d}{dx}(\log x)}{(\log x)^2} \)
\( \implies = -\frac{\log x (-\sin x) - \cos x \cdot \frac{1}{x}}{(\log x)^2} \)
\( \implies = -\frac{-x \sin x \log x - \cos x}{x (\log x)^2} \)
\( \implies = \frac{x \sin x \log x + \cos x}{x (\log x)^2} \)
In simple words: Simplify the function first by dividing \( \log x \) by itself, then apply the quotient rule to the remaining fraction \(\frac{\cos x}{\log x}\). Remember that the derivative of a constant is zero.

🎯 Exam Tip: Simplifying the function as \( 1 - \frac{\cos x}{\log x} \) makes differentiation easier. The constant 1 differentiates to 0, and then only the quotient rule is applied to the second term.

 

Question 10. cos(log x + eˣ) हल- माना \( y = \cos(\log x + e^x) \)
दोनों पक्षों का \( x \) के सापेक्ष अवकलन करने पर,
\( \frac{dy}{dx} = \frac{d}{dx} (\cos(\log x + e^x)) \)
\( \implies = -\sin(\log x + e^x) \frac{d}{dx} (\log x + e^x) \)
\( \implies = -\sin(\log x + e^x) \left(\frac{d}{dx}(\log x) + \frac{d}{dx}(e^x)\right) \)
\( \implies = -\sin(\log x + e^x) \left(\frac{1}{x} + e^x\right) \)
\( \implies = -\left(\frac{1}{x} + e^x\right) \sin(\log x + e^x) \)
In simple words: Use the chain rule. Differentiate the outer cosine function, then multiply by the derivative of its argument (\( \log x + e^x \)). Remember to differentiate each term within the argument.

🎯 Exam Tip: The derivative of \( \cos u \) is \( -\sin u \frac{du}{dx} \). Here, \( u = \log x + e^x \), so \( \frac{du}{dx} = \frac{1}{x} + e^x \). Ensure correct application of the sum rule for the inner derivative.

प्रश्नावली 5.5

Question 1. cos x.cos 2x.cos 3x हल- माना \( y = \cos x \cdot \cos 2x \cdot \cos 3x \) ...(1)
दोनों पक्षों को लघुगणक लेने पर,
\( \log y = \log (\cos x \cdot \cos 2x \cdot \cos 3x) \)
\( \implies \log y = \log \cos x + \log \cos 2x + \log \cos 3x \)
दोनों पक्षों में \( x \) के सापेक्ष अवकलन करने पर,
\( \frac{1}{y} \frac{dy}{dx} = \frac{d}{dx} (\log \cos x) + \frac{d}{dx} (\log \cos 2x) + \frac{d}{dx} (\log \cos 3x) \)
\( \implies = \frac{1}{\cos x} (-\sin x) + \frac{1}{\cos 2x} (-\sin 2x) (2) + \frac{1}{\cos 3x} (-\sin 3x) (3) \)
\( \implies = -\tan x - 2 \tan 2x - 3 \tan 3x \)
\( \implies \frac{dy}{dx} = -y (\tan x + 2 \tan 2x + 3 \tan 3x) \)
समीकरण (1) से \( y \) का मान रखने पर
\( \implies \frac{dy}{dx} = -\cos x \cdot \cos 2x \cdot \cos 3x (\tan x + 2 \tan 2x + 3 \tan 3x) \)
In simple words: When dealing with products of multiple functions, taking the logarithm of both sides simplifies the product into a sum of logarithms. Then, differentiate implicitly with respect to \( x \).

🎯 Exam Tip: Logarithmic differentiation is highly effective for products/quotients or functions with variables in exponents. Remember \( \frac{d}{dx}(\log y) = \frac{1}{y} \frac{dy}{dx} \). Also, \( \frac{d}{dx}(\log(\cos u)) = \frac{1}{\cos u} (-\sin u) \frac{du}{dx} = -\tan u \frac{du}{dx} \).

 

Question 2. \( \sqrt{\frac{(x-1)(x-2)}{(x-3)(x-4)(x-5)}} \) हल- माना \( y = \left( \frac{(x-1)(x-2)}{(x-3)(x-4)(x-5)} \right)^{1/2} \) ...(1)
दोनों पक्षों का लघुगणक लेने पर,
\( \log y = \log \left( \frac{(x-1)(x-2)}{(x-3)(x-4)(x-5)} \right)^{1/2} \)
\( \implies \log y = \frac{1}{2} \left[ \log((x-1)(x-2)) - \log((x-3)(x-4)(x-5)) \right] \)
\( \implies \log y = \frac{1}{2} [\log(x-1) + \log(x-2) - (\log(x-3) + \log(x-4) + \log(x-5))] \)
दोनों पक्षों का \( x \) के सापेक्ष अवकलन करने पर,
\( \frac{1}{y} \frac{dy}{dx} = \frac{1}{2} \left[ \frac{d}{dx}(\log(x-1)) + \frac{d}{dx}(\log(x-2)) - \frac{d}{dx}(\log(x-3)) - \frac{d}{dx}(\log(x-4)) - \frac{d}{dx}(\log(x-5)) \right] \)
\( \implies = \frac{1}{2} \left[ \frac{1}{x-1} + \frac{1}{x-2} - \frac{1}{x-3} - \frac{1}{x-4} - \frac{1}{x-5} \right] \)
समीकरण (1) से \( y \) का मान रखने पर,
\( \implies \frac{dy}{dx} = \frac{1}{2} \sqrt{\frac{(x-1)(x-2)}{(x-3)(x-4)(x-5)}} \left[ \frac{1}{x-1} + \frac{1}{x-2} - \frac{1}{x-3} - \frac{1}{x-4} - \frac{1}{x-5} \right] \)
In simple words: Use logarithmic differentiation. Take the natural logarithm of both sides. This transforms the complex fraction with a square root into a sum and difference of simpler logarithms. Then, differentiate implicitly and solve for \( \frac{dy}{dx} \).

🎯 Exam Tip: Remember the log properties: \( \log(a^b) = b \log a \), \( \log(ab) = \log a + \log b \), and \( \log(\frac{a}{b}) = \log a - \log b \). These are crucial for simplifying the expression before differentiation.

 

Question 3. (log x)cos x हल- माना \( y = (\log x)^{\cos x} \)
दोनों पक्षों का लघुगणक लेने पर,
\( \log y = \log ((\log x)^{\cos x}) \)
\( \implies \log y = \cos x \log (\log x) \)
दोनों पक्षों का \( x \) के सापेक्ष अवकलन करने पर,
\( \frac{1}{y} \frac{dy}{dx} = \frac{d}{dx} (\cos x \log (\log x)) \)
\( \implies = \cos x \frac{d}{dx} (\log (\log x)) + \log (\log x) \frac{d}{dx} (\cos x) \)
\( \implies = \cos x \cdot \frac{1}{\log x} \frac{d}{dx} (\log x) + \log (\log x) (-\sin x) \)
\( \implies = \cos x \cdot \frac{1}{\log x} \cdot \frac{1}{x} - \sin x \log (\log x) \)
\( \implies = \frac{\cos x}{x \log x} - \sin x \log (\log x) \)
\( \implies \frac{dy}{dx} = y \left[ \frac{\cos x}{x \log x} - \sin x \log (\log x) \right] \)
\( \implies \frac{dy}{dx} = (\log x)^{\cos x} \left[ \frac{\cos x}{x \log x} - \sin x \log (\log x) \right] \)
In simple words: This is a function with a variable in the exponent. Use logarithmic differentiation. Take the natural logarithm of both sides, then apply the product rule for \( \cos x \log (\log x) \), and finally differentiate implicitly.

🎯 Exam Tip: For functions of the form \( [f(x)]^{g(x)} \), logarithmic differentiation is the standard method. Remember to apply the chain rule carefully when differentiating \( \log(\log x) \).

 

Question 4. xˣ - 2sin x हल- माना \( y = x^x - 2^{\sin x} \)
पुनः माना \( u = x^x, v = 2^{\sin x} \)
\( \implies y = u - v \)
दोनों पक्षों का \( x \) के सापेक्ष अवकलन करने पर,
\( \frac{dy}{dx} = \frac{du}{dx} - \frac{dv}{dx} \) ...(1)
पहले \( u = x^x \) पर विचार करें। दोनों पक्षों का लघुगणक लेने पर, \( \log u = \log x^x = x \log x \)
दोनों पक्षों का \( x \) के सापेक्ष अवकलन करने पर,
\( \frac{1}{u} \frac{du}{dx} = \frac{d}{dx} (x \log x) \)
\( \implies = x \frac{d}{dx} (\log x) + \log x \frac{d}{dx} (x) \)
\( \implies = x \cdot \frac{1}{x} + \log x \cdot 1 \)
\( \implies = 1 + \log x \)
\( \implies \frac{du}{dx} = u(1 + \log x) = x^x(1 + \log x) \) ...(2)
अब, \( v = 2^{\sin x} \) पर विचार करें। दोनों पक्षों का लघुगणक लेने पर, \( \log v = \log (2^{\sin x}) = \sin x \log 2 \)
दोनों पक्षों का \( x \) के सापेक्ष अवकलन करने पर,
\( \frac{1}{v} \frac{dv}{dx} = \frac{d}{dx} (\sin x \log 2) \)
\( \implies = \log 2 \frac{d}{dx} (\sin x) \)
\( \implies = \log 2 \cos x \)
\( \implies \frac{dv}{dx} = v (\cos x \log 2) = 2^{\sin x} (\cos x \log 2) \) ...(3)
समी० (2) तथा (3) से \( \frac{du}{dx} \) तथा \( \frac{dv}{dx} \) के मान समीकरण (1) में रखने पर,
\( \frac{dy}{dx} = x^x (1 + \log x) - 2^{\sin x} (\cos x \log 2) \)
In simple words: Break the function into two parts, \( u = x^x \) and \( v = 2^{\sin x} \). Use logarithmic differentiation for both \( u \) and \( v \) separately to find \( \frac{du}{dx} \) and \( \frac{dv}{dx} \). Finally, combine them as \( \frac{dy}{dx} = \frac{du}{dx} - \frac{dv}{dx} \).

🎯 Exam Tip: When a function is a sum or difference of terms, and each term requires logarithmic differentiation, differentiate each term separately and then combine. Remember \( \frac{d}{dx}(a^u) = a^u \log a \frac{du}{dx} \).

 

Question 5. (x+3)² .(x+4)³ .(x+5)⁴ हल- माना \( y = (x+3)^2 \cdot (x+4)^3 \cdot (x+5)^4 \)
दोनों पक्षों का लघुगणक लेने पर,
\( \log y = \log [(x+3)^2 \cdot (x+4)^3 \cdot (x+5)^4] \)
\( \implies \log y = \log (x+3)^2 + \log (x+4)^3 + \log (x+5)^4 \)
\( \implies \log y = 2 \log (x+3) + 3 \log (x+4) + 4 \log (x+5) \)
दोनों पक्षों का \( x \) के सापेक्ष अवकलन करने पर,
\( \frac{1}{y} \frac{dy}{dx} = 2 \frac{d}{dx} (\log (x+3)) + 3 \frac{d}{dx} (\log (x+4)) + 4 \frac{d}{dx} (\log (x+5)) \)
\( \implies = 2 \cdot \frac{1}{x+3} \cdot 1 + 3 \cdot \frac{1}{x+4} \cdot 1 + 4 \cdot \frac{1}{x+5} \cdot 1 \)
\( \implies = \frac{2}{x+3} + \frac{3}{x+4} + \frac{4}{x+5} \)
\( \implies \frac{dy}{dx} = y \left[ \frac{2(x+4)(x+5) + 3(x+3)(x+5) + 4(x+3)(x+4)}{(x+3)(x+4)(x+5)} \right] \)
\( \implies = (x+3)^2 (x+4)^3 (x+5)^4 \left[ \frac{2(x^2+9x+20) + 3(x^2+8x+15) + 4(x^2+7x+12)}{(x+3)(x+4)(x+5)} \right] \)
\( \implies = (x+3)(x+4)^2 (x+5)^3 [2x^2+18x+40 + 3x^2+24x+45 + 4x^2+28x+48] \)
\( \implies = (x+3)(x+4)^2 (x+5)^3 [9x^2+70x+133] \)
In simple words: Use logarithmic differentiation. Take the natural logarithm of both sides to convert the product into a sum of logarithms, making it easier to differentiate. Then, differentiate implicitly with respect to \( x \).

🎯 Exam Tip: Logarithmic differentiation simplifies complex products. Remember the properties of logarithms \( \log(ab) = \log a + \log b \) and \( \log(a^n) = n \log a \). The final step involves substituting back the original expression for \( y \).

 

Question 5. (x + 3)2 .(x + 4)3 .(x + 5)4

Answer:

हल-

माना \( y = (x + 3)^2 .(x + 4)^3 .(x + 5)^4 \)

दोनों पक्षों का लघुगणक लेने पर,

\( \log y = \log [(x + 3)^2 .(x + 4)^3 (x + 5)^4] \)

\( = \log (x + 3)^2 + \log (x + 4)^3 + \log (x + 5)^4 \)

[::: log mn = log m + log n]

\( = 2 \log (x + 3) + 3 \log (x + 4) + 4 \log (x + 5) \)

[: log \( m^n \) = n log m]

दोनों पक्षों का x के सापेक्ष अवकलन करने पर,

\( \frac{1}{y} \frac{dy}{dx} = 2 \frac{d}{dx} \log (x + 3) + 3 \frac{d}{dx} \log (x + 4) + 4 \frac{d}{dx} \log (x + 5) \)

\( \frac{1}{y} \frac{dy}{dx} = 2 \frac{1}{(x + 3)} \frac{d}{dx} (x + 3) + 3 \frac{1}{(x + 4)} \frac{d}{dx} (x + 4) + 4 \frac{1}{(x + 5)} \frac{d}{dx} (x + 5) \)

\( \frac{1}{y} \frac{dy}{dx} = \frac{2(1+0)}{(x+3)} + \frac{3(1+0)}{(x+4)} + \frac{4(1+0)}{(x+5)} \)

\( \frac{dy}{dx} = y \left[ \frac{2}{x+3} + \frac{3}{x+4} + \frac{4}{x+5} \right] \)

\( = (x + 3)^2 (x + 4)^3 (x + 5)^4 \left[ \frac{2(x + 4)(x+5) + 3(x+3)(x+5) + 4(x+3)(x+4)}{(x + 3) (x + 4) (x + 5)} \right] \)

\( = (x + 3)^2 (x + 4)^3 (x + 5)^4 \left[ \frac{2(x^2 + 9x + 20) + 3 (x^2 + 8x + 15) + 4 (x^2 + 7x + 12)}{(x + 3) (x + 4) (x + 5)} \right] \)

\( = (x + 3) (x + 4)^2 (x + 5)^3 [9x^2 + 70x + 133] \)

In simple words: This question asks for the derivative of a product of functions raised to powers. The solution uses logarithmic differentiation to simplify the product into a sum of logarithms, making it easier to differentiate.

🎯 Exam Tip: Logarithmic differentiation is crucial for products/quotients of multiple functions or functions with variables in both base and exponent. Remember to apply the chain rule correctly when differentiating the logarithm of each factor.

 

Question 6. y=(1+1/x)x+x(1/x)

Answer:

हल-

माना \( y = \left(1+\frac{1}{x}\right)^x + x^{\frac{1}{x}} \)

पुनः माना \( y_1 = \left(1+\frac{1}{x}\right)^x \) तथा \( y_2 = x^{\frac{1}{x}} \)

तब \( y = y_1 + y_2 \)

\( \frac{dy}{dx} = \frac{dy_1}{dx} + \frac{dy_2}{dx} \) ...(1)

अब \( y_1 = \left(1+\frac{1}{x}\right)^x \)

दोनों ओर का log लेने पर, \( \log y_1 = x \log \left(1+\frac{1}{x}\right) \)

दोनों ओर का x के सापेक्ष अवकलन करने पर,

\( \frac{1}{y_1} \frac{dy_1}{dx} = x \frac{d}{dx} \left[ \log \left(1+\frac{1}{x}\right) \right] + \log \left(1+\frac{1}{x}\right) \frac{d}{dx} (x) \)

\( \frac{dy_1}{dx} = y_1 \left[ x \cdot \frac{1}{\left(1+\frac{1}{x}\right)} \cdot \frac{d}{dx} \left(1+\frac{1}{x}\right) + \log \left(1+\frac{1}{x}\right) \cdot 1 \right] \)

\( = y_1 \left[ x \cdot \frac{x}{(x+1)} \cdot \left(-\frac{1}{x^2}\right) + \log \left(1+\frac{1}{x}\right) \right] \)

\( = y_1 \left[ -\frac{1}{(x+1)} + \log \left(1+\frac{1}{x}\right) \right] \)

पुनः \( y_1 \) का मान रखने पर,

\( \frac{dy_1}{dx} = \left(1+\frac{1}{x}\right)^x \left[ -\frac{1}{(x+1)} + \log \left(1+\frac{1}{x}\right) \right] \)

पुनः \( y_2 = x^{\frac{1}{x}} \)

दोनों ओर का log लेने पर, \( \log y_2 = \frac{1}{x} \log x \)

दोनों ओर का x के सापेक्ष अवकलन करने पर,

\( \frac{1}{y_2} \frac{dy_2}{dx} = \frac{1}{x} \frac{d}{dx} (\log x) + \log x \frac{d}{dx} \left(\frac{1}{x}\right) \)

\( = \frac{1}{x} \cdot \frac{1}{x} + \log x \cdot \left(-\frac{1}{x^2}\right) \)

\( = \frac{1}{x^2} - \frac{\log x}{x^2} \)

\( \frac{dy_2}{dx} = y_2 \left[ \frac{1-\log x}{x^2} \right] \)

पुनः \( y_2 \) का मान रखने पर,

\( \frac{dy_2}{dx} = x^{\frac{1}{x}} \left[ \frac{1-\log x}{x^2} \right] \)

\( \frac{dy_1}{dx} \) तथा \( \frac{dy_2}{dx} \) के मान समीकरण (1) में रखने पर,

\( \frac{dy}{dx} = \left(1+\frac{1}{x}\right)^x \left[ -\frac{1}{(x+1)} + \log \left(1+\frac{1}{x}\right) \right] + x^{\frac{1}{x}} \left[ \frac{1-\log x}{x^2} \right] \)

In simple words: This problem involves differentiating a sum of two complex functions, each with a variable in both the base and exponent. The strategy is to break the problem into two parts, use logarithmic differentiation for each part, and then add the results to find the total derivative.

🎯 Exam Tip: When dealing with functions of the form \( f(x)^{g(x)} \), always use logarithmic differentiation. Remember to split the function into simpler terms if it's a sum or difference before applying logarithms.

 

Question 7. (log x)x + xlog x

Answer:

हल-

माना \( y = (\log x)^x + x^{\log x} \)

पुनः माना \( y = u + v \)

दोनों पक्षों का x के सापेक्ष अवकलन करने पर,

\( \frac{dy}{dx} = \frac{du}{dx} + \frac{dv}{dx} \) ...(1)

अब, \( u = (\log x)^x \)

दोनों पक्षों का लघुगणक लेने पर,

\( \log u = \log (\log x)^x = x \log (\log x) \)

[log \( m^n \) = n log m]

दोनों पक्षों का x के सापेक्ष अवकलन करने पर,

\( \frac{1}{u} \frac{du}{dx} = x \frac{d}{dx} [\log (\log x)] + \log (\log x) \frac{d}{dx} (x) \)

\( = x \cdot \frac{1}{\log x} \cdot \frac{d}{dx} (\log x) + \log (\log x) \cdot 1 \)

\( = x \cdot \frac{1}{\log x} \cdot \frac{1}{x} + \log (\log x) \)

\( = \frac{1}{\log x} + \log (\log x) \)

\( \frac{du}{dx} = u \left[ \frac{1}{\log x} + \log (\log x) \right] \)

\( = (\log x)^x \left[ \frac{1}{\log x} + \log (\log x) \right] \) ...(2)

तथा \( v = x^{\log x} \)

दोनों पक्षों का लघुगणक लेने पर,

\( \log v = \log x^{\log x} = (\log x) \cdot (\log x) = (\log x)^2 \)

दोनों पक्षों का x के सापेक्ष अवकलन करने पर,

\( \frac{1}{v} \frac{dv}{dx} = \frac{d}{dx} [(\log x)^2] \)

\( = 2 (\log x) \frac{d}{dx} (\log x) \)

\( = 2 (\log x) \cdot \frac{1}{x} \)

\( \frac{dv}{dx} = v \left[ \frac{2 \log x}{x} \right] \)

\( = x^{\log x} \left[ \frac{2 \log x}{x} \right] \) ...(3)

समी० (1) से,

\( \frac{dy}{dx} = \frac{du}{dx} + \frac{dv}{dx} \)

\( = (\log x)^x \left[ \frac{1}{\log x} + \log (\log x) \right] + x^{\log x} \left[ \frac{2 \log x}{x} \right] \)

In simple words: This question requires finding the derivative of a sum of two functions. Each function has a variable base and a variable exponent, necessitating the use of logarithmic differentiation for each term separately. The final derivative is the sum of the derivatives of the individual terms.

🎯 Exam Tip: Always remember to apply the product rule and chain rule correctly during logarithmic differentiation. Pay close attention to nested logarithmic functions, as their derivatives involve multiple steps.

 

Question 8. (sin x)x + sin-1√x

Answer:

हल-

माना \( y = (\sin x)^x + \sin^{-1}\sqrt{x} \)

पुनः माना \( y = u + v \)

दोनों पक्षों का x के सापेक्ष अवकलन करने पर,

\( \frac{dy}{dx} = \frac{du}{dx} + \frac{dv}{dx} \) ...(1)

अब, \( u = (\sin x)^x \)

दोनों पक्षों का लघुगणक लेने पर,

\( \log u = \log (\sin x)^x = x \log \sin x \)

दोनों पक्षों का x के सापेक्ष अवकलन करने पर,

\( \frac{1}{u} \frac{du}{dx} = x \frac{d}{dx} (\log \sin x) + \log \sin x \frac{d}{dx} (x) \)

\( = x \cdot \frac{1}{\sin x} \cdot \frac{d}{dx} (\sin x) + \log \sin x \cdot 1 \)

\( = x \cdot \frac{1}{\sin x} \cdot \cos x + \log \sin x \)

\( = x \cot x + \log \sin x \)

\( \frac{du}{dx} = u (x \cot x + \log \sin x) \)

\( = (\sin x)^x [x \cot x + \log \sin x] \) ...(2)

पुनः \( v = \sin^{-1}\sqrt{x} \)

दोनों पक्षों का x के सापेक्ष अवकलन करने पर,

\( \frac{dv}{dx} = \frac{d}{dx} (\sin^{-1}\sqrt{x}) \)

\( = \frac{1}{\sqrt{1-(\sqrt{x})^2}} \cdot \frac{d}{dx} (\sqrt{x}) \)

\( = \frac{1}{\sqrt{1-x}} \cdot \frac{1}{2\sqrt{x}} \)

\( = \frac{1}{2\sqrt{x}\sqrt{1-x}} \) ...(3)

समी० (2) तथा (3) से \( \frac{du}{dx} \) तथा \( \frac{dv}{dx} \) का मान समी० (1) में रखने पर,

\( \frac{dy}{dx} = (\sin x)^x [x \cot x + \log \sin x] + \frac{1}{2\sqrt{x}\sqrt{1-x}} \)

In simple words: This problem involves differentiating a sum of two functions: one with a variable base and exponent, and another an inverse trigonometric function. Logarithmic differentiation is applied to the first term, and the chain rule is used for the inverse function. The final derivative is the sum of these two results.

🎯 Exam Tip: For inverse trigonometric functions like \( \sin^{-1}f(x) \), remember the derivative formula \( \frac{d}{dx}(\sin^{-1}u) = \frac{1}{\sqrt{1-u^2}}\frac{du}{dx} \). Be careful with the chain rule when u is a function of x, like \( \sqrt{x} \).

 

Question 9. xsin x + (sin x)cos x

Answer:

हल-

माना \( y = x^{\sin x} + (\sin x)^{\cos x} \)

पुनः माना \( y = u + v \)

दोनों पक्षों का x के सापेक्ष अवकलन करने पर,

\( \frac{dy}{dx} = \frac{du}{dx} + \frac{dv}{dx} \) ...(1)

अब, \( u = x^{\sin x} \)

दोनों पक्षों का लघुगणक लेने पर,

\( \log u = \log x^{\sin x} = \sin x \log x \)

दोनों पक्षों का x के सापेक्ष अवकलन करने पर,

\( \frac{1}{u} \frac{du}{dx} = \sin x \frac{d}{dx} (\log x) + \log x \frac{d}{dx} (\sin x) \)

\( = \sin x \cdot \frac{1}{x} + \log x \cdot \cos x \)

\( = \frac{\sin x}{x} + \cos x \log x \)

\( \frac{du}{dx} = u \left[ \frac{\sin x}{x} + \cos x \log x \right] \)

\( = x^{\sin x} \left[ \frac{\sin x}{x} + \cos x \log x \right] \) ...(2)

तथा \( v = (\sin x)^{\cos x} \)

दोनों पक्षों का लघुगणक लेने पर,

\( \log v = \log (\sin x)^{\cos x} = \cos x \log \sin x \)

दोनों पक्षों का x के सापेक्ष अवकलन करने पर,

\( \frac{1}{v} \frac{dv}{dx} = \cos x \frac{d}{dx} (\log \sin x) + \log \sin x \frac{d}{dx} (\cos x) \)

\( = \cos x \cdot \frac{1}{\sin x} \cdot \frac{d}{dx} (\sin x) + \log \sin x \cdot (-\sin x) \)

\( = \cos x \cdot \frac{1}{\sin x} \cdot \cos x - \sin x \log \sin x \)

\( = \cot x \cos x - \sin x \log \sin x \)

\( \frac{dv}{dx} = v [\cot x \cos x - \sin x \log \sin x] \)

\( = (\sin x)^{\cos x} [\cot x \cos x - \sin x \log \sin x] \) ...(3)

समी० (2) तथा (3) से \( \frac{du}{dx} \) तथा \( \frac{dv}{dx} \) के मान समी० (1) में रखने पर,

\( \frac{dy}{dx} = x^{\sin x} \left[ \frac{\sin x}{x} + \cos x \log x \right] + (\sin x)^{\cos x} [\cot x \cos x - \sin x \log \sin x] \)

In simple words: This problem asks for the derivative of a sum of two functions, each having both base and exponent as variables. Logarithmic differentiation is applied to each term separately, and their results are combined to get the final derivative.

🎯 Exam Tip: When differentiating \( \log(\sin x) \), remember the chain rule: \( \frac{d}{dx}(\log u) = \frac{1}{u}\frac{du}{dx} \). So, \( \frac{d}{dx}(\log \sin x) = \frac{1}{\sin x} \cos x = \cot x \).

 

Question 10. x(x cos x) + (x2+1)/(x2-1)

Answer:

हल-

माना \( y = x^{(x \cos x)} + \frac{(x^2 + 1)}{(x^2 - 1)} \)

पुनः माना \( y = u + v \)

दोनों पक्षों का x के सापेक्ष अवकलन करने पर,

\( \frac{dy}{dx} = \frac{du}{dx} + \frac{dv}{dx} \) ...(1)

अब, \( u = x^{(x \cos x)} \)

दोनों पक्षों में लघुगणक लेने पर,

\( \log u = \log x^{(x \cos x)} = x \cos x \log x \)

दोनों पक्षों का x के सापेक्ष अवकलन करने पर,

\( \frac{1}{u} \frac{du}{dx} = \frac{d}{dx} (x \cos x \log x) \)

\( = (x \cos x) \frac{d}{dx} (\log x) + \log x \frac{d}{dx} (x \cos x) \)

\( = (x \cos x) \cdot \frac{1}{x} + \log x [x \frac{d}{dx} (\cos x) + \cos x \frac{d}{dx} (x)] \)

\( = \cos x + \log x [x (-\sin x) + \cos x \cdot 1] \)

\( = \cos x + \log x \cos x - x \sin x \log x \)

\( \frac{du}{dx} = u [\cos x + \cos x \log x - x \sin x \log x] \)

\( = x^{(x \cos x)} [\cos x + \cos x \log x - x \sin x \log x] \) ...(2)

पुनः \( v = \frac{x^2 + 1}{x^2 - 1} \)

दोनों पक्षों का x के सापेक्ष अवकलन करने पर,

\( \frac{dv}{dx} = \frac{(x^2 - 1) \frac{d}{dx} (x^2 + 1) - (x^2 + 1) \frac{d}{dx} (x^2 - 1)}{(x^2 - 1)^2} \)

\( = \frac{(x^2 - 1) (2x) - (x^2 + 1) (2x)}{(x^2 - 1)^2} \)

\( = \frac{2x^3 - 2x - 2x^3 - 2x}{(x^2 - 1)^2} \)

\( = \frac{-4x}{(x^2 - 1)^2} \) ...(3)

समीकरण (2) तथा (3) से \( \frac{du}{dx} \) तथा \( \frac{dv}{dx} \) के मान समीकरण (1) में रखने पर,

\( \frac{dy}{dx} = x^{(x \cos x)} [\cos x + \cos x \log x - x \sin x \log x] - \frac{4x}{(x^2 - 1)^2} \)

In simple words: This problem involves differentiating a sum of two distinct types of functions: one with a variable in both the base and exponent, requiring logarithmic differentiation, and another a rational function, which is differentiated using the quotient rule. The derivatives of both parts are then added together.

🎯 Exam Tip: When a function is a sum or difference of terms, differentiate each term separately. For rational functions, the quotient rule \( \frac{d}{dx}\left(\frac{u}{v}\right) = \frac{v\frac{du}{dx} - u\frac{dv}{dx}}{v^2} \) is essential.

 

Question 11. (x cos x)x + (xsin x)1/x

Answer:

हल-

माना \( y = (x \cos x)^x + (x \sin x)^{1/x} \)

पुनः माना \( y = u + v \)

दोनों पक्षों का x के सापेक्ष अवकलन करने पर,

\( \frac{dy}{dx} = \frac{du}{dx} + \frac{dv}{dx} \) ...(1)

अब, \( u = (x \cos x)^x \)

दोनों पक्षों का लघुगणक लेने पर,

\( \log u = \log (x \cos x)^x = x \log (x \cos x) \)

दोनों पक्षों का x के सापेक्ष अवकलन करने पर,

\( \frac{1}{u} \frac{du}{dx} = x \frac{d}{dx} [\log (x \cos x)] + \log (x \cos x) \frac{d}{dx} (x) \)

\( = x \cdot \frac{1}{x \cos x} \cdot \frac{d}{dx} (x \cos x) + \log (x \cos x) \cdot 1 \)

\( = \frac{1}{\cos x} [x \frac{d}{dx} (\cos x) + \cos x \frac{d}{dx} (x)] + \log (x \cos x) \)

\( = \frac{1}{\cos x} [x (-\sin x) + \cos x \cdot 1] + \log (x \cos x) \)

\( = -x \tan x + 1 + \log (x \cos x) \)

\( \frac{du}{dx} = u [1 - x \tan x + \log (x \cos x)] \)

\( = (x \cos x)^x [1 - x \tan x + \log (x \cos x)] \) ...(2)

तथा \( v = (x \sin x)^{1/x} \)

दोनों पक्षों का लघुगणक लेने पर,

\( \log v = \log (x \sin x)^{1/x} = \frac{1}{x} \log (x \sin x) \)

दोनों पक्षों का x के सापेक्ष अवकलन करने पर,

\( \frac{1}{v} \frac{dv}{dx} = \frac{1}{x} \frac{d}{dx} [\log (x \sin x)] + \log (x \sin x) \frac{d}{dx} \left(\frac{1}{x}\right) \)

\( = \frac{1}{x} \cdot \frac{1}{x \sin x} \cdot \frac{d}{dx} (x \sin x) + \log (x \sin x) \cdot \left(-\frac{1}{x^2}\right) \)

\( = \frac{1}{x^2 \sin x} [x \frac{d}{dx} (\sin x) + \sin x \frac{d}{dx} (x)] - \frac{\log (x \sin x)}{x^2} \)

\( = \frac{1}{x^2 \sin x} [x \cos x + \sin x \cdot 1] - \frac{\log (x \sin x)}{x^2} \)

\( = \frac{x \cos x + \sin x}{x^2 \sin x} - \frac{\log (x \sin x)}{x^2} \)

\( = \frac{x \cos x}{x^2 \sin x} + \frac{\sin x}{x^2 \sin x} - \frac{\log (x \sin x)}{x^2} \)

\( = \frac{\cot x}{x} + \frac{1}{x^2} - \frac{\log (x \sin x)}{x^2} \)

\( = \frac{x \cot x + 1 - \log (x \sin x)}{x^2} \)

\( \frac{dv}{dx} = v \left[ \frac{x \cot x + 1 - \log (x \sin x)}{x^2} \right] \)

\( = (x \sin x)^{1/x} \left[ \frac{x \cot x + 1 - \log (x \sin x)}{x^2} \right] \) ...(3)

समी० (2) से \( \frac{du}{dx} \) तथा (3) से \( \frac{dv}{dx} \) का मान समी० (1) में रखने पर,

\( \frac{dy}{dx} = (x \cos x)^x [1 - x \tan x + \log (x \cos x)] + (x \sin x)^{1/x} \left[ \frac{x \cot x + 1 - \log (x \sin x)}{x^2} \right] \)

In simple words: This problem asks for the derivative of a sum of two complex functions, each requiring logarithmic differentiation due to variables in both the base and exponent. The problem is solved by differentiating each term separately and then adding their derivatives.

🎯 Exam Tip: Remember to apply the product rule when differentiating expressions like \( x \log(x \cos x) \). Also, be cautious with the chain rule for terms like \( \log(x \cos x) \) where the argument is a product of functions.

 

Question 12. xy + yx = 1

Answer:

हल-

दिया है, \( x^y + y^x = 1 \)

माना \( u = x^y, v = y^x \).:. \( u + v = 1 \)

दोनों पक्षों का x के सापेक्ष अवकलन करने पर,

\( \frac{du}{dx} + \frac{dv}{dx} = 0 \) ...(1)

अब, \( u = x^y \)

दोनों पक्षों का लघुगणक लेने पर, \( \log u = \log x^y = y \log x \)

दोनों पक्षों का x के सापेक्ष अवकलन करने पर,

\( \frac{1}{u} \frac{du}{dx} = y \frac{d}{dx} (\log x) + \log x \frac{d}{dx} (y) \)

\( = y \cdot \frac{1}{x} + \log x \frac{dy}{dx} \)

\( = \frac{y}{x} + (\log x) \frac{dy}{dx} \)

\( \frac{du}{dx} = u \left[ \frac{y}{x} + (\log x) \frac{dy}{dx} \right] \)

\( = x^y \left[ \frac{y}{x} + (\log x) \frac{dy}{dx} \right] \) ...(2)

तथा \( v = y^x \)

दोनों पक्षों का लघुगणक लेने पर, \( \log v = \log y^x = x \log y \)

दोनों पक्षों का x के सापेक्ष अवकलन करने पर,

\( \frac{1}{v} \frac{dv}{dx} = x \frac{d}{dx} (\log y) + \log y \frac{d}{dx} (x) \)

\( = x \cdot \frac{1}{y} \cdot \frac{dy}{dx} + \log y \cdot 1 \)

\( = \frac{x}{y} \frac{dy}{dx} + \log y \)

\( \frac{dv}{dx} = v \left[ \frac{x}{y} \frac{dy}{dx} + \log y \right] \)

\( = y^x \left[ \frac{x}{y} \frac{dy}{dx} + \log y \right] \) ...(3)

समीकरण (2) तथा (3) से, \( \frac{du}{dx} \) तथा \( \frac{dv}{dx} \) का मान समीकरण (1) में रखने पर,

\( x^y \left[ \frac{y}{x} + (\log x) \frac{dy}{dx} \right] + y^x \left[ \frac{x}{y} \frac{dy}{dx} + \log y \right] = 0 \)

\( x^y \frac{y}{x} + x^y (\log x) \frac{dy}{dx} + y^x \frac{x}{y} \frac{dy}{dx} + y^x (\log y) = 0 \)

\( x^y \frac{y}{x} + y^x \log y + \left[ x^y \log x + y^x \frac{x}{y} \right] \frac{dy}{dx} = 0 \)

\( \left[ x^y \log x + x y^{x-1} \right] \frac{dy}{dx} = - \left[ y x^{y-1} + y^x \log y \right] \)

\( \frac{dy}{dx} = - \frac{y x^{y-1} + y^x \log y}{x^y \log x + x y^{x-1}} \)

In simple words: This problem requires implicit differentiation for an equation involving two terms, each with a variable base and exponent. Both terms are differentiated using logarithmic differentiation, and then the equation is rearranged to solve for \( \frac{dy}{dx} \).

🎯 Exam Tip: When implicitly differentiating, remember that \( \frac{dy}{dx} \) will appear on both sides of the equation. Collect all terms with \( \frac{dy}{dx} \) on one side and the remaining terms on the other to solve for it.

 

Question 13. yx = xy

Answer:

हल-

दिया है, \( y^x = x^y \)

दोनों पक्षों को लघुगणक लेने पर, \( \log y^x = \log x^y \)

\( x \log y = y \log x \)

दोनों पक्षों में x के सापेक्ष अवकलन करने पर,

\( x \frac{d}{dx} (\log y) + \log y \frac{d}{dx} (x) = y \frac{d}{dx} (\log x) + \log x \frac{d}{dx} (y) \)

\( x \cdot \frac{1}{y} \cdot \frac{dy}{dx} + \log y \cdot 1 = y \cdot \frac{1}{x} + \log x \cdot \frac{dy}{dx} \)

\( \frac{x}{y} \frac{dy}{dx} + \log y = \frac{y}{x} + (\log x) \frac{dy}{dx} \)

\( \frac{x}{y} \frac{dy}{dx} - (\log x) \frac{dy}{dx} = \frac{y}{x} - \log y \)

\( \frac{dy}{dx} \left( \frac{x}{y} - \log x \right) = \frac{y}{x} - \log y \)

\( \frac{dy}{dx} \left( \frac{x - y \log x}{y} \right) = \frac{y - x \log y}{x} \)

\( \frac{dy}{dx} = \frac{y}{x} \frac{(y - x \log y)}{(x - y \log x)} \)

\( \frac{dy}{dx} = \frac{y^2 - xy \log y}{x^2 - xy \log x} \)

In simple words: This problem involves implicit differentiation for an equation where variables appear in both the base and exponent. Taking the logarithm of both sides simplifies the exponents, allowing the use of product rule and chain rule to find the derivative.

🎯 Exam Tip: When solving \( f(y)^x = g(x)^y \) type problems, always apply `log` on both sides first. Remember to use the product rule for terms like \( x \log y \) and \( y \log x \), and the chain rule for \( \frac{d}{dx}(\log y) = \frac{1}{y}\frac{dy}{dx} \).

 

Question 14. (cos x)y = (cos y)x

Answer:

हल-

दिया है, \( (\cos x)^y = (\cos y)^x \)

दोनों पक्षों का लघुगणक लेने पर,

\( \log (\cos x)^y = \log (\cos y)^x \)

\( y \log \cos x = x \log \cos y \)

दोनों पक्षों का x के सापेक्ष अवकलन करने पर,

\( y \frac{d}{dx} (\log \cos x) + \log \cos x \frac{d}{dx} (y) = x \frac{d}{dx} (\log \cos y) + \log \cos y \frac{d}{dx} (x) \)

\( y \cdot \frac{1}{\cos x} \cdot \frac{d}{dx} (\cos x) + \log \cos x \cdot \frac{dy}{dx} = x \cdot \frac{1}{\cos y} \cdot \frac{d}{dx} (\cos y) + \log \cos y \cdot 1 \)

\( y \cdot \frac{1}{\cos x} \cdot (-\sin x) + \log \cos x \cdot \frac{dy}{dx} = x \cdot \frac{1}{\cos y} \cdot (-\sin y) \cdot \frac{dy}{dx} + \log \cos y \)

\( -y \tan x + (\log \cos x) \frac{dy}{dx} = -x \tan y \frac{dy}{dx} + \log \cos y \)

\( (\log \cos x + x \tan y) \frac{dy}{dx} = \log \cos y + y \tan x \)

\( \frac{dy}{dx} = \frac{\log \cos y + y \tan x}{\log \cos x + x \tan y} \)

In simple words: This problem involves implicit differentiation for an equation where both sides have variable bases and exponents. Logarithmic differentiation is applied to transform the exponential forms into products, after which the equation is differentiated with respect to \( x \) and rearranged to find \( \frac{dy}{dx} \).

🎯 Exam Tip: For expressions like \( \frac{d}{dx}(\log \cos y) \), use the chain rule: \( \frac{1}{\cos y}(-\sin y)\frac{dy}{dx} = -\tan y \frac{dy}{dx} \). Keep track of \( \frac{dy}{dx} \) terms to isolate them correctly.

 

Question 15. xy = e(x-y)

Answer:

हल-

दिया है, \( xy = e^{(x-y)} \)

दोनों पक्षों को लघुगणक लेने पर,

\( \log (xy) = \log e^{(x-y)} \)

\( \log x + \log y = (x - y) \log_e e \)

[::: log xy = log x + log y]

\( \log x + \log y = x - y \)

[: \( \log_e e = 1 \)]

दोनों पक्षों का x के सापेक्ष अवकलन करने पर,

\( \frac{d}{dx} (\log x) + \frac{d}{dx} (\log y) = \frac{d}{dx} (x) - \frac{d}{dx} (y) \)

\( \frac{1}{x} + \frac{1}{y} \frac{dy}{dx} = 1 - \frac{dy}{dx} \)

\( \frac{1}{y} \frac{dy}{dx} + \frac{dy}{dx} = 1 - \frac{1}{x} \)

\( \frac{dy}{dx} \left( \frac{1}{y} + 1 \right) = \frac{x - 1}{x} \)

\( \frac{dy}{dx} \left( \frac{1 + y}{y} \right) = \frac{x - 1}{x} \)

\( \frac{dy}{dx} = \frac{y}{x} \left( \frac{x - 1}{1 + y} \right) \)

In simple words: This problem asks for implicit differentiation of an exponential equation. Taking the natural logarithm of both sides simplifies the equation into a more manageable algebraic form, which is then differentiated term by term to solve for \( \frac{dy}{dx} \).

🎯 Exam Tip: When you see an equation with 'e' as a base, taking the natural logarithm is often the easiest first step. Remember that \( \log e^A = A \log e = A \). Be careful with signs when rearranging terms to isolate \( \frac{dy}{dx} \).

 

Question 16. f(x) = (1 + x) (1 + x2) (1 + x4) (1 + x8)

Answer:

हल-

दिया है, \( f(x) = (1 + x) (1 + x^2) (1 + x^4) (1 + x^8) \)

दोनों पक्षों का लघुगणक लेने पर,

\( \log f(x) = \log [(1 + x) (1 + x^2) (1 + x^4) (1 + x^8)] \)

\( \log f(x) = \log (1 + x) + \log (1 + x^2) + \log (1 + x^4) + \log (1 + x^8) \)

[: log mn = log m + log n]

दोनों पक्षों का x के सापेक्ष अवकलन करने पर,

\( \frac{1}{f(x)} f'(x) = \frac{d}{dx} \log (1 + x) + \frac{d}{dx} \log (1 + x^2) + \frac{d}{dx} \log (1 + x^4) + \frac{d}{dx} \log (1 + x^8) \)

\( \frac{f'(x)}{f(x)} = \frac{1}{(1 + x)} \frac{d}{dx} (1 + x) + \frac{1}{(1 + x^2)} \frac{d}{dx} (1 + x^2) + \frac{1}{(1 + x^4)} \frac{d}{dx} (1 + x^4) + \frac{1}{(1 + x^8)} \frac{d}{dx} (1 + x^8) \)

\( \frac{f'(x)}{f(x)} = \frac{1}{(1 + x)} \cdot 1 + \frac{1}{(1 + x^2)} \cdot 2x + \frac{1}{(1 + x^4)} \cdot 4x^3 + \frac{1}{(1 + x^8)} \cdot 8x^7 \)

\( f'(x) = f(x) \left[ \frac{1}{1 + x} + \frac{2x}{1 + x^2} + \frac{4x^3}{1 + x^4} + \frac{8x^7}{1 + x^8} \right] \)

\( f'(x) = (1 + x) (1 + x^2) (1 + x^4) (1 + x^8) \left[ \frac{1}{1 + x} + \frac{2x}{1 + x^2} + \frac{4x^3}{1 + x^4} + \frac{8x^7}{1 + x^8} \right] \)

x = 1 रखने पर,

\( f'(1) = (1 + 1) (1 + 1^2) (1 + 1^4) (1 + 1^8) \left[ \frac{1}{1 + 1} + \frac{2 \cdot 1}{1 + 1^2} + \frac{4 \cdot 1^3}{1 + 1^4} + \frac{8 \cdot 1^7}{1 + 1^8} \right] \)

\( = (2) (2) (2) (2) \left[ \frac{1}{2} + \frac{2}{2} + \frac{4}{2} + \frac{8}{2} \right] \)

\( = 16 \left[ \frac{1}{2} + 1 + 2 + 4 \right] \)

\( = 16 \left[ \frac{1}{2} + 7 \right] \)

\( = 16 \left[ \frac{1 + 14}{2} \right] \)

\( = 16 \cdot \frac{15}{2} \)

\( = 8 \cdot 15 = 120 \)In simple words: This problem requires finding the derivative of a product of four functions. Logarithmic differentiation is the most efficient method, converting the product into a sum of logarithms before differentiating each term and then multiplying back by the original function. Finally, the derivative is evaluated at \( x=1 \).

🎯 Exam Tip: Logarithmic differentiation is highly recommended for complex products. Remember the property \( \log(abc) = \log a + \log b + \log c \), which simplifies the derivative significantly. Don't forget to substitute the original function \( f(x) \) back at the end and simplify the numerical calculation.

 

Question 17. (x2 – 5x + 8) (x3 + 7x + 9) का अवकलन निम्नलिखित तीन प्रकार से कीजिए
(i) गुणनफल नियम का प्रयोग करके
(ii) गुणनफल के विस्तारण द्वारा एक एकल बहुपद प्राप्त करके
(iii) लघुगणकीय अवकलन द्वारा
यह भी सत्यापित कीजिए कि इस प्रकार प्राप्त तीनों उत्तर समान हैं।

Answer:

हल-

(i) गुणनफल नियम के प्रयोग द्वारा ।

माना \( y = (x^2 – 5x + 8) \cdot (x^3 + 7x + 9) \)

दोनों पक्षों का x के सापेक्ष अवकलन करने पर,

\( \frac{dy}{dx} = (x^2 – 5x + 8) \frac{d}{dx} (x^3 + 7x + 9) + (x^3 + 7x + 9) \frac{d}{dx} (x^2 – 5x + 8) \)

\( \frac{dy}{dx} = (x^2 – 5x + 8) (3x^2 + 7) + (x^3 + 7x + 9) (2x – 5) \)

\( = (3x^2 (x^2 – 5x + 8) + 7 (x^2 – 5x + 8)) + (2x (x^3 + 7x + 9) – 5(x^3 + 7x + 9)) \)

\( = (3x^4 – 15x^3 + 24x^2 + 7x^2 – 35x + 56) + (2x^4 + 14x^2 – 5x^3 – 35x – 45) \)

\( = 3x^4 – 15x^3 + 31x^2 – 35x + 56 + 2x^4 – 5x^3 + 14x^2 – 35x – 45 \)

\( = 5x^4 – 20x^3 + 45x^2 – 70x + 11 \) ...(1)

(ii) गुणनफल के विस्तारण द्वारा हल

\( y = (x^2 – 5x + 8) (x^3 + 7x + 9) \)

\( = x^2 (x^3 + 7x + 9) - 5x (x^3 + 7x + 9) + 8(x^3 + 7x + 9) \)

\( = x^5 + 7x^3 + 9x^2 - 5x^4 - 35x^2 - 45x + 8x^3 + 56x + 72 \)

\( = x^5 - 5x^4 + 15x^3 - 26x^2 + 11x + 72 \)

दोनों पक्षों का x के सापेक्ष अवकलन करने पर,

\( \frac{dy}{dx} = 5x^4 - 20x^3 + 45x^2 - 52x + 11 \) ...(2)

(iii) लघुगणकीय अवकलन द्वारा हल

माना \( y = (x^2 – 5x + 8) (x^3 + 7x + 9) \)

दोनों पक्षों का लघुगणक लेने पर,

\( \log y = \log (x^2 – 5x + 8) + \log (x^3 + 7x + 9) \)

[:: log (mn) = log m + log n]

दोनों पक्षों का x के सापेक्ष अवकलन करने पर,

\( \frac{1}{y} \frac{dy}{dx} = \frac{1}{(x^2 – 5x + 8)} \frac{d}{dx} (x^2 – 5x + 8) + \frac{1}{(x^3 + 7x + 9)} \frac{d}{dx} (x^3 + 7x + 9) \)

\( \frac{dy}{dx} = y \left[ \frac{(2x – 5)}{(x^2 – 5x + 8)} + \frac{(3x^2 + 7)}{(x^3 + 7x + 9)} \right] \)

\( = (x^2 – 5x + 8) (x^3 + 7x + 9) \left[ \frac{(2x – 5)(x^3 + 7x + 9) + (3x^2 + 7)(x^2 – 5x + 8)}{(x^2 – 5x + 8) (x^3 + 7x + 9)} \right] \)

\( = (2x – 5)(x^3 + 7x + 9) + (3x^2 + 7)(x^2 – 5x + 8) \)

\( = (2x^4 + 14x^2 + 18x – 5x^3 – 35x – 45) + (3x^4 – 15x^3 + 24x^2 + 7x^2 – 35x + 56) \)

\( = 5x^4 – 20x^3 + 45x^2 – 70x + 11 \) ...(3)

समीकरण (1), (2) और (3) से स्पष्ट है कि \( \frac{dy}{dx} \) के मान समान हैं।

In simple words: This question demonstrates three different methods to find the derivative of a product of two polynomial functions: the product rule, expansion into a single polynomial, and logarithmic differentiation. All three methods yield the same result, verifying their equivalence.

🎯 Exam Tip: This question is excellent for practicing various differentiation techniques. For polynomial products, direct expansion is often quicker if the polynomials are small. However, for more complex products, the product rule or logarithmic differentiation becomes essential. Always simplify your answers to compare them easily.

 

Question 18. यदि u, v और w, x के फलन हैं तो दो विधियों अर्थात् प्रथम गुणनफल नियम की पुनरावृत्ति द्वारा, द्वितीय- लघुगणकीय अवकलन द्वारा दर्शाइए कि ।

Answer:

हल-

(i) माना \( y = u \cdot v \cdot w = u \cdot (v \cdot w) \)

दोनों पक्षों का x के सापेक्ष अवकलन करने पर,

\( \frac{dy}{dx} = \frac{d}{dx} (u (v \cdot w)) \)

\( = u \frac{d}{dx} (v \cdot w) + (v \cdot w) \frac{d}{dx} (u) \)

\( = u \left[ v \frac{d}{dx} (w) + w \frac{d}{dx} (v) \right] + v w \frac{d}{dx} (u) \)

\( = u v \frac{dw}{dx} + u w \frac{dv}{dx} + v w \frac{du}{dx} \)

(ii) माना \( y = u v w \)

दोनों पक्षों का लघुगणक लेने पर,

\( \log y = \log (u v w) = \log u + \log v + \log w \)

दोनों पक्षों का x के सापेक्ष अवकलन करने पर,

\( \frac{1}{y} \frac{dy}{dx} = \frac{1}{u} \frac{du}{dx} + \frac{1}{v} \frac{dv}{dx} + \frac{1}{w} \frac{dw}{dx} \)

\( \frac{dy}{dx} = y \left[ \frac{1}{u} \frac{du}{dx} + \frac{1}{v} \frac{dv}{dx} + \frac{1}{w} \frac{dw}{dx} \right] \)

\( = (u v w) \left[ \frac{1}{u} \frac{du}{dx} + \frac{1}{v} \frac{dv}{dx} + \frac{1}{w} \frac{dw}{dx} \right] \)

\( = v w \frac{du}{dx} + u w \frac{dv}{dx} + u v \frac{dw}{dx} \)

In simple words: This problem demonstrates two ways to find the derivative of a product of three functions: first, by repeatedly applying the product rule for two functions, and second, by using logarithmic differentiation. Both methods prove that the derivative involves the sum of three terms, where each term is the derivative of one function multiplied by the other two original functions.

🎯 Exam Tip: The extended product rule for three functions is \( (uvw)' = u'vw + uv'w + uvw' \). Logarithmic differentiation provides an elegant way to derive this by converting products into sums, simplifying the differentiation process significantly.

प्रश्नावली 5.6

 

Question 1. यदि प्रश्न संख्या 1 से 10 तक में तथाy के लिए समीकरणों द्वारा, एक-दूसरे से प्राचलिक रूप में सम्बन्धित हों तो प्राचलों का विलोपन किए बिना \( \frac{dy}{dx} \) ज्ञात कीजिए।
x = 2at², y = at⁴
Answer:
हल-
दिया है, \( x = 2at² \) तथा \( y = at⁴ \)
दोनों पक्षों का t के सापेक्ष अवकलन करने पर,
\( \frac{dx}{dt} = \frac{d}{dt} (2at²) = 2a \frac{d}{dt} (t²) = 2a (2t) = 4at \)
तथा \( \frac{dy}{dt} = \frac{d}{dt} (at⁴) = a \frac{d}{dt} (t⁴) = a (4t³) = 4at³ \)
अतः \( \frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{4at³}{4at} = t² \)
In simple words: To find \( dy/dx \) for parametric equations, we first find \( dx/dt \) and \( dy/dt \) and then divide \( dy/dt \) by \( dx/dt \). In this case, it simplifies to \( t² \).

🎯 Exam Tip: Remember to express \( dy/dx \) in terms of the parameter (t in this case) or any other variables as required. Keep track of differentiation rules for powers.

 

Question 2. x = a cos θ, y = b cos θ
Answer:
हल- दिया है : \( x = a \cos \theta \) तथा \( y = b \cos \theta \)
दोनों पक्षों का \( \theta \) के सापेक्ष अवकलन करने पर,
\( \frac{dx}{d\theta} = \frac{d}{d\theta} (a \cos \theta) = - a \sin \theta \)
तथा \( \frac{dy}{d\theta} = \frac{d}{d\theta} (b \cos \theta) = - b \sin \theta \)
अतः \( \frac{dy}{dx} = \frac{dy/d\theta}{dx/d\theta} = \frac{- b \sin \theta}{- a \sin \theta} = \frac{b}{a} \)
In simple words: For parametric equations, differentiate both x and y with respect to the parameter (here, \( \theta \)) and then divide \( dy/d\theta \) by \( dx/d\theta \) to find \( dy/dx \).

🎯 Exam Tip: Always state the differentiation parameter (e.g., 'with respect to \( \theta \)') clearly. Be careful with signs during differentiation.

 

Question 3. x = sin t, y = cos 2t
Answer:
हल- दिया है, \( x = \sin t \) तथा \( y = \cos 2t \)
दोनों पक्षों का t के सापेक्ष अवकलन करने पर,
\( \frac{dx}{dt} = \frac{d}{dt} (\sin t) = \cos t \)
तथा \( \frac{dy}{dt} = \frac{d}{dt} (\cos 2t) = - \sin 2t \frac{d}{dt} (2t) = - \sin 2t (2) = - 2 \sin 2t \)
\( \implies \)
\( - 2 (2 \sin t \cos t) = - 4 \sin t \cos t \)
अतः \( \frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{- 4 \sin t \cos t}{\cos t} = - 4 \sin t \)
In simple words: First find the derivatives of x and y with respect to t, then divide them. Use the double angle identity for sine to simplify the expression.

🎯 Exam Tip: Remember chain rule for functions like \( \cos 2t \). Trigonometric identities can often simplify the final answer, so look for opportunities to apply them.

 

Question 4. x = 4t, y = 4/t
Answer:
हल- दिया है, \( x = 4t \) तथा \( y = \frac{4}{t} \)
दोनों पक्षों का t के सापेक्ष अवकलन करने पर,
\( \frac{dx}{dt} = \frac{d}{dt} (4t) = 4 \)
तथा \( \frac{dy}{dt} = \frac{d}{dt} (\frac{4}{t}) = 4 \frac{d}{dt} (t^{-1}) = 4 (-1 t^{-2}) = - \frac{4}{t²} \)
अतः \( \frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{- 4/t²}{4} = - \frac{1}{t²} \)
In simple words: Differentiate x and y separately with respect to t. Then, divide the derivative of y by the derivative of x to get the final answer.

🎯 Exam Tip: Be careful with negative exponents when differentiating. Ensure proper simplification of the fraction.

 

Question 5. x = cos θ - cos 2θ, y = sin θ - sin 2θ
Answer:
हल- दिया है, \( x = \cos \theta - \cos 2\theta \) तथा \( y = \sin \theta - \sin 2\theta \)
दोनों पक्षों का \( \theta \) के सापेक्ष अवकलन करने पर,
\( \frac{dx}{d\theta} = \frac{d}{d\theta} (\cos \theta - \cos 2\theta) = - \sin \theta - (-\sin 2\theta) \frac{d}{d\theta} (2\theta) = - \sin \theta + 2 \sin 2\theta \)
तथा \( \frac{dy}{d\theta} = \frac{d}{d\theta} (\sin \theta - \sin 2\theta) = \cos \theta - \cos 2\theta \frac{d}{d\theta} (2\theta) = \cos \theta - 2 \cos 2\theta \)
अतः \( \frac{dy}{dx} = \frac{dy/d\theta}{dx/d\theta} = \frac{\cos \theta - 2 \cos 2\theta}{- \sin \theta + 2 \sin 2\theta} = \frac{\cos \theta - 2 \cos 2\theta}{2 \sin 2\theta - \sin \theta} \)
In simple words: Differentiate x and y with respect to \( \theta \), applying the chain rule where necessary. Then, form the ratio \( dy/d\theta \) divided by \( dx/d\theta \) to find the result.

🎯 Exam Tip: The chain rule is crucial for terms like \( \cos 2\theta \) and \( \sin 2\theta \). Ensure careful algebraic manipulation and sign management.

 

Question 6. x = a(θ – sin θ), y = a(1 + cos θ)
Answer:
हल-
\( \frac{dx}{d\theta} = a \frac{d}{d\theta} (\theta - \sin \theta) = a (1 - \cos \theta) \)
\( \implies \)
\( \frac{dy}{d\theta} = a \frac{d}{d\theta} (1 + \cos \theta) = a (0 - \sin \theta) = - a \sin \theta \)
\( \implies \)
\( \frac{dy}{dx} = \frac{dy/d\theta}{dx/d\theta} = \frac{- a \sin \theta}{a (1 - \cos \theta)} = \frac{- \sin \theta}{1 - \cos \theta} \)
\( \implies \)
\( \frac{- (2 \sin (\theta/2) \cos (\theta/2))}{2 \sin² (\theta/2)} = \frac{- \cos (\theta/2)}{\sin (\theta/2)} = - \cot (\theta/2) \)
In simple words: Differentiate x and y with respect to \( \theta \), then divide the results. Use trigonometric half-angle formulas to simplify the expression to a cotangent function.

🎯 Exam Tip: Double-check differentiation of trigonometric functions. Applying half-angle identities \( \sin \theta = 2 \sin (\theta/2) \cos (\theta/2) \) and \( 1 - \cos \theta = 2 \sin² (\theta/2) \) is key for simplification.

 

Question 7. x = sin³t/√cos 2t, y = cos³t/√cos 2t
Answer:
हल-
दिया है, \( x = \frac{\sin³ t}{\sqrt{\cos 2t}} \) तथा \( y = \frac{\cos³ t}{\sqrt{\cos 2t}} \)
दोनों पक्षों का t के सापेक्ष अवकलन करने पर,
\( \frac{dx}{dt} = \frac{\sqrt{\cos 2t} \frac{d}{dt} (\sin³ t) - \sin³ t \frac{d}{dt} (\sqrt{\cos 2t})}{(\sqrt{\cos 2t})²} \)
\( = \frac{\sqrt{\cos 2t} (3 \sin² t \cos t) - \sin³ t (\frac{1}{2\sqrt{\cos 2t}}) (- \sin 2t) (2)}{\cos 2t} \)
\( = \frac{3 \sin² t \cos t \cos 2t + \sin³ t \sin 2t}{\cos 2t \sqrt{\cos 2t}} = \frac{\sin² t (3 \cos t \cos 2t + \sin t \sin 2t)}{(\cos 2t)^{3/2}} \)
\( \implies \)
\( \frac{dy}{dt} = \frac{\sqrt{\cos 2t} \frac{d}{dt} (\cos³ t) - \cos³ t \frac{d}{dt} (\sqrt{\cos 2t})}{(\sqrt{\cos 2t})²} \)
\( = \frac{\sqrt{\cos 2t} (3 \cos² t (- \sin t)) - \cos³ t (\frac{1}{2\sqrt{\cos 2t}}) (- \sin 2t) (2)}{\cos 2t} \)
\( = \frac{- 3 \cos² t \sin t \cos 2t + \cos³ t \sin 2t}{\cos 2t \sqrt{\cos 2t}} = \frac{\cos² t (- 3 \sin t \cos 2t + \cos t \sin 2t)}{(\cos 2t)^{3/2}} \)
\( \implies \)
\( \frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{\cos² t (- 3 \sin t \cos 2t + \cos t \sin 2t)}{\sin² t (3 \cos t \cos 2t + \sin t \sin 2t)} \)
\( = \frac{\cos² t (\cos t \sin 2t - 3 \sin t \cos 2t)}{\sin² t (3 \cos t \cos 2t + \sin t \sin 2t)} \)
\( = \frac{\cos² t (\cos t (2 \sin t \cos t) - 3 \sin t (2 \cos² t - 1))}{\sin² t (3 \cos t (2 \cos² t - 1) + \sin t (2 \sin t \cos t))} \)
\( = \frac{\cos² t (2 \sin t \cos² t - 6 \sin t \cos² t + 3 \sin t)}{\sin² t (6 \cos³ t - 3 \cos t + 2 \sin² t \cos t)} \)
\( = \frac{\cos² t \sin t (2 \cos² t - 6 \cos² t + 3)}{\sin² t \cos t (6 \cos² t - 3 + 2 \sin² t)} \)
\( = \frac{\cos t (3 - 4 \cos² t)}{\sin t (4 \cos² t - 3)} = \frac{- \cos t (4 \cos² t - 3)}{\sin t (4 \cos² t - 3)} = - \cot t \)
In simple words: This problem involves quotient rule for differentiation. Differentiate both x and y with respect to t. Then divide \( dy/dt \) by \( dx/dt \). Simplify the expression using trigonometric identities to get the final cotangent result.

🎯 Exam Tip: The quotient rule and chain rule are essential here. Simplify trigonometric expressions at each step to manage complexity. Look for identities like \( \sin 2t = 2 \sin t \cos t \) and \( \cos 2t = 2 \cos² t - 1 \).

 

Question 8. हल- \( \frac{dy}{dx} = - \cot 3t \)
Answer: This appears to be a solution snippet without the full question context, likely related to trigonometric parametric differentiation. Assuming the question was to find \( \frac{dy}{dx} \) for a given parametric function involving 3t.
In simple words: The derivative is \( - \cot 3t \), implying a parametric function where the ratio of derivatives simplified to this trigonometric form.

🎯 Exam Tip: When given only a solution, deduce the type of problem it might be. In parametric equations, the derivative often involves trigonometric functions of the parameter.

 

Question 9. x = a(cost + log tan(t/2)), y = a sint
Answer:
हल- दिया है, \( x = a (\cos t + \log \tan \frac{t}{2}) \)
तथा \( y = a \sin t \)
\( \implies \)
\( \frac{dx}{dt} = a \frac{d}{dt} (\cos t + \log \tan \frac{t}{2}) \)
\( = a [- \sin t + \frac{1}{\tan (t/2)} \frac{d}{dt} (\tan \frac{t}{2})] \)
\( = a [- \sin t + \frac{1}{\tan (t/2)} \sec² \frac{t}{2} \frac{d}{dt} (\frac{t}{2})] \)
\( = a [- \sin t + \frac{\cos (t/2)}{\sin (t/2)} \frac{1}{\cos² (t/2)} \frac{1}{2}] \)
\( = a [- \sin t + \frac{1}{2 \sin (t/2) \cos (t/2)}] \)
\( = a [- \sin t + \frac{1}{\sin t}] = a [\frac{1 - \sin² t}{\sin t}] = a \frac{\cos² t}{\sin t} \)
\( \implies \)
\( \frac{dy}{dt} = a \frac{d}{dt} (\sin t) = a \cos t \)
\( \implies \)
\( \frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{a \cos t}{a \frac{\cos² t}{\sin t}} = \frac{\sin t}{\cos t} = \tan t \)
In simple words: Differentiate x and y with respect to t. For x, use the chain rule for log and tan functions. For y, it's a simple sine derivative. Then, divide \( dy/dt \) by \( dx/dt \) and simplify the trigonometric expression to tan t.

🎯 Exam Tip: Differentiating logarithmic and trigonometric compositions requires careful application of the chain rule. Using identities like \( 2 \sin(t/2) \cos(t/2) = \sin t \) is crucial for simplification.

 

Question 10. x = a(cosθ + θ sinθ), y = a(sinθ – θcosθ)
Answer:
हल- दिया है, \( x = a (\cos \theta + \theta \sin \theta) \)
तथा \( y = a (\sin \theta - \theta \cos \theta) \)
\( \implies \)
\( \frac{dx}{d\theta} = a \frac{d}{d\theta} (\cos \theta + \theta \sin \theta) \)
\( = a [- \sin \theta + (\theta \cos \theta + \sin \theta)] \) (using product rule for \( \theta \sin \theta \))
\( = a [- \sin \theta + \theta \cos \theta + \sin \theta] = a \theta \cos \theta \)
\( \implies \)
\( \frac{dy}{d\theta} = a \frac{d}{d\theta} (\sin \theta - \theta \cos \theta) \)
\( = a [\cos \theta - (\theta (- \sin \theta) + \cos \theta)] \) (using product rule for \( \theta \cos \theta \))
\( = a [\cos \theta + \theta \sin \theta - \cos \theta] = a \theta \sin \theta \)
\( \implies \)
\( \frac{dy}{dx} = \frac{dy/d\theta}{dx/d\theta} = \frac{a \theta \sin \theta}{a \theta \cos \theta} = \tan \theta \)
In simple words: Differentiate x and y with respect to \( \theta \). Use the product rule for terms like \( \theta \sin \theta \) and \( \theta \cos \theta \). Simplify the resulting expressions before dividing \( dy/d\theta \) by \( dx/d\theta \) to find the tangent.

🎯 Exam Tip: Product rule is vital when \( \theta \) is multiplied by trigonometric functions. Pay attention to signs and cancellation of terms for accurate simplification.

 

Question 11. यदि \( x = \sqrt{a^{\sin^{-1}t}} \), \( y = \sqrt{a^{\cos^{-1}t}} \) तो दर्शाइए कि \( \frac{dy}{dx} = - \frac{y}{x} \)
Answer:
हल-
दिया है, \( x = \sqrt{a^{\sin^{-1}t}} \) तथा \( y = \sqrt{a^{\cos^{-1}t}} \)
दोनों पक्षों का t के सापेक्ष अवकलन करने पर,
\( x = (a^{\sin^{-1}t})^{1/2} = a^{\frac{1}{2}\sin^{-1}t} \)
\( \implies \)
\( \frac{dx}{dt} = \frac{d}{dt} (a^{\frac{1}{2}\sin^{-1}t}) \)
\( = a^{\frac{1}{2}\sin^{-1}t} \log_e a \cdot \frac{d}{dt} (\frac{1}{2}\sin^{-1}t) \) (using \( \frac{d}{dx} (a^u) = a^u \log_e a \frac{du}{dx} \))
\( = a^{\frac{1}{2}\sin^{-1}t} \log_e a \cdot \frac{1}{2} \cdot \frac{1}{\sqrt{1 - t²}} \)
\( = x \log_e a \cdot \frac{1}{2\sqrt{1 - t²}} \)
इसी प्रकार, \( y = (a^{\cos^{-1}t})^{1/2} = a^{\frac{1}{2}\cos^{-1}t} \)
\( \implies \)
\( \frac{dy}{dt} = \frac{d}{dt} (a^{\frac{1}{2}\cos^{-1}t}) \)
\( = a^{\frac{1}{2}\cos^{-1}t} \log_e a \cdot \frac{d}{dt} (\frac{1}{2}\cos^{-1}t) \)
\( = a^{\frac{1}{2}\cos^{-1}t} \log_e a \cdot \frac{1}{2} \cdot \frac{-1}{\sqrt{1 - t²}} \)
\( = y \log_e a \cdot \frac{-1}{2\sqrt{1 - t²}} \)
अतः \( \frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{y \log_e a \cdot \frac{-1}{2\sqrt{1 - t²}}}{x \log_e a \cdot \frac{1}{2\sqrt{1 - t²}}} = - \frac{y}{x} \)
In simple words: First, express x and y in exponential form. Differentiate both x and y with respect to t using the derivative rule for \( a^u \). Substitute back x and y into their respective derivatives. Finally, divide \( dy/dt \) by \( dx/dt \) and observe the cancellation to prove the given relationship.

🎯 Exam Tip: Remember the differentiation rule for \( a^u \), which is \( a^u \log_e a \frac{du}{dx} \). The derivatives of inverse trigonometric functions \( \sin^{-1}t \) and \( \cos^{-1}t \) are key here. Look for opportunities to substitute back x and y to simplify the final expression.

 

प्रश्नावली 5.7

 

Question 1. x² + 3x + 2
Answer:
हल- माना \( y = x² + 3x + 2 \)
दोनों पक्षों का x के सापेक्ष अवकलन करने पर,
\( \frac{dy}{dx} = \frac{d}{dx} (x² + 3x + 2) = 2x + 3 \)
दोनों पक्षों का पुनः x के सापेक्ष अवकलन करने पर,
\( \frac{d²y}{dx²} = \frac{d}{dx} (2x + 3) = 2 \)
In simple words: To find the second derivative, differentiate the function once to get the first derivative, then differentiate the first derivative again.

🎯 Exam Tip: Second derivatives involve differentiating the first derivative. Be systematic and apply differentiation rules correctly in each step.

 

Question 2. x²⁰
Answer:
हल-
माना \( y = x^{20} \)
दोनों पक्षों का x के सापेक्ष अवकलन करने पर,
\( \frac{dy}{dx} = \frac{d}{dx} (x^{20}) = 20x^{20-1} = 20x^{19} \)
दोनों पक्षों का पुनः x के सापेक्ष अवकलन करने पर,
\( \frac{d²y}{dx²} = \frac{d}{dx} (20x^{19}) = 20 \frac{d}{dx} (x^{19}) = 20 \times 19x^{19-1} = 380x^{18} \)
In simple words: Differentiate the power function once, then differentiate the result again to find the second derivative.

🎯 Exam Tip: The power rule \( \frac{d}{dx} (x^n) = nx^{n-1} \) is repeatedly applied. Ensure correct exponent calculations in both differentiation steps.

 

Question 3. x cos x
Answer:
हल- माना \( y = x \cos x \)
दोनों पक्षों का x के सापेक्ष अवकलन करने पर,
\( \frac{dy}{dx} = \frac{d}{dx} (x \cos x) \)
\( = x \frac{d}{dx} (\cos x) + \cos x \frac{d}{dx} (x) \) (using product rule)
\( = x (- \sin x) + \cos x (1) = - x \sin x + \cos x \)
दोनों पक्षों का पुनः x के सापेक्ष अवकलन करने पर,
\( \frac{d²y}{dx²} = \frac{d}{dx} (- x \sin x + \cos x) \)
\( = \frac{d}{dx} (- x \sin x) + \frac{d}{dx} (\cos x) \)
\( = - [x \frac{d}{dx} (\sin x) + \sin x \frac{d}{dx} (x)] - \sin x \) (using product rule for \( -x \sin x \))
\( = - [x \cos x + \sin x (1)] - \sin x \)
\( = - x \cos x - \sin x - \sin x = - x \cos x - 2 \sin x \)
In simple words: Use the product rule to find the first derivative of x cos x. Then, differentiate the resulting expression again, applying the product rule for the x sin x term, to get the second derivative.

🎯 Exam Tip: The product rule is applied twice. Pay attention to negative signs from differentiating \( \cos x \) and \( \sin x \). Group terms for clear simplification.

 

Question 4. log x
Answer:
हल- माना \( y = \log x \)
दोनों पक्षों का x के सापेक्ष अवकलन करने पर,
\( \frac{dy}{dx} = \frac{d}{dx} (\log x) = \frac{1}{x} \)
दोनों पक्षों का पुनः x के सापेक्ष अवकलन करने पर,
\( \frac{d²y}{dx²} = \frac{d}{dx} (\frac{1}{x}) = \frac{d}{dx} (x^{-1}) = -1x^{-2} = - \frac{1}{x²} \)
In simple words: Differentiate log x to get 1/x. Then, differentiate 1/x to find the second derivative, which is -1/x².

🎯 Exam Tip: Remember \( \frac{d}{dx} (\log x) = \frac{1}{x} \) and express \( \frac{1}{x} \) as \( x^{-1} \) for easy differentiation using the power rule.

 

Question 5. x³ log x
Answer:
हल-
माना \( y = x³ \log x \)
दोनों पक्षों का x के सापेक्ष अवकलन करने पर,
\( \frac{dy}{dx} = \frac{d}{dx} (x³ \log x) \)
\( = x³ \frac{d}{dx} (\log x) + \log x \frac{d}{dx} (x³) \) (using product rule)
\( = x³ (\frac{1}{x}) + \log x (3x²) = x² + 3x² \log x \)
दोनों पक्षों का पुनः x के सापेक्ष अवकलन करने पर,
\( \frac{d²y}{dx²} = \frac{d}{dx} (x² + 3x² \log x) \)
\( = \frac{d}{dx} (x²) + \frac{d}{dx} (3x² \log x) \)
\( = 2x + [3x² \frac{d}{dx} (\log x) + \log x \frac{d}{dx} (3x²)] \) (using product rule for \( 3x² \log x \))
\( = 2x + [3x² (\frac{1}{x}) + \log x (6x)] \)
\( = 2x + 3x + 6x \log x = 5x + 6x \log x = x (5 + 6 \log x) \)
In simple words: Apply the product rule to find the first derivative of x³ log x. Then, differentiate the resulting expression again, applying the product rule for the 3x² log x term, to find the second derivative.

🎯 Exam Tip: The product rule is applied multiple times. Keep calculations organized to avoid errors. Factor out common terms for a simplified final answer.

 

Question 6. eˣ sin 5x
Answer:
हल-
माना \( y = e^x \sin 5x \)
दोनों पक्षों का x के सापेक्ष अवकलन करने पर,
\( \frac{dy}{dx} = \frac{d}{dx} (e^x \sin 5x) \)
\( = e^x \frac{d}{dx} (\sin 5x) + \sin 5x \frac{d}{dx} (e^x) \) (using product rule)
\( = e^x (\cos 5x \cdot 5) + \sin 5x \cdot e^x = 5e^x \cos 5x + e^x \sin 5x \)
\( = e^x (5 \cos 5x + \sin 5x) \)
दोनों पक्षों का पुनः x के सापेक्ष अवकलन करने पर,
\( \frac{d²y}{dx²} = \frac{d}{dx} (e^x (5 \cos 5x + \sin 5x)) \)
\( = e^x \frac{d}{dx} (5 \cos 5x + \sin 5x) + (5 \cos 5x + \sin 5x) \frac{d}{dx} (e^x) \) (using product rule)
\( = e^x [5 (- \sin 5x \cdot 5) + (\cos 5x \cdot 5)] + (5 \cos 5x + \sin 5x) e^x \)
\( = e^x [- 25 \sin 5x + 5 \cos 5x] + e^x [5 \cos 5x + \sin 5x] \)
\( = e^x [- 25 \sin 5x + 5 \cos 5x + 5 \cos 5x + \sin 5x] \)
\( = e^x [10 \cos 5x - 24 \sin 5x] \)
In simple words: Apply the product rule twice. For the first derivative, differentiate \( e^x \sin 5x \). For the second derivative, differentiate the resulting expression, again using the product rule, and simplify.

🎯 Exam Tip: The chain rule is needed for \( \sin 5x \) and \( \cos 5x \). Be careful with signs and combine like terms accurately. Factor out \( e^x \) for a cleaner final expression.

 

Question 7. e⁶ˣ cos 3x
Answer:
हल-
माना \( y = e^{6x} \cos 3x \)
दोनों पक्षों का x के सापेक्ष अवकलन करने पर,
\( \frac{dy}{dx} = \frac{d}{dx} (e^{6x} \cos 3x) \)
\( = e^{6x} \frac{d}{dx} (\cos 3x) + \cos 3x \frac{d}{dx} (e^{6x}) \) (using product rule)
\( = e^{6x} (- \sin 3x \cdot 3) + \cos 3x (e^{6x} \cdot 6) \)
\( = - 3e^{6x} \sin 3x + 6e^{6x} \cos 3x = e^{6x} (6 \cos 3x - 3 \sin 3x) \)
दोनों पक्षों का पुनः x के सापेक्ष अवकलन करने पर,
\( \frac{d²y}{dx²} = \frac{d}{dx} (e^{6x} (6 \cos 3x - 3 \sin 3x)) \)
\( = e^{6x} \frac{d}{dx} (6 \cos 3x - 3 \sin 3x) + (6 \cos 3x - 3 \sin 3x) \frac{d}{dx} (e^{6x}) \) (using product rule)
\( = e^{6x} [6 (- \sin 3x \cdot 3) - 3 (\cos 3x \cdot 3)] + (6 \cos 3x - 3 \sin 3x) (e^{6x} \cdot 6) \)
\( = e^{6x} [- 18 \sin 3x - 9 \cos 3x] + e^{6x} [36 \cos 3x - 18 \sin 3x] \)
\( = e^{6x} [- 18 \sin 3x - 9 \cos 3x + 36 \cos 3x - 18 \sin 3x] \)
\( = e^{6x} [27 \cos 3x - 36 \sin 3x] = 9e^{6x} (3 \cos 3x - 4 \sin 3x) \)
In simple words: Apply the product rule for both the first and second derivatives. Remember the chain rule for terms like \( e^{6x} \) and \( \cos 3x \), \( \sin 3x \). Combine like terms and factor out \( e^{6x} \) for the final simplified expression.

🎯 Exam Tip: Pay close attention to the chain rule for exponential and trigonometric functions. Organize calculations to prevent errors in combining terms and ensure correct factoring.

 

Question 8. tan⁻¹x
Answer:
हल-
माना \( y = \tan^{-1}x \)
दोनों पक्षों का x के सापेक्ष अवकलन करने पर,
\( \frac{dy}{dx} = \frac{d}{dx} (\tan^{-1}x) = \frac{1}{1 + x²} \)
दोनों पक्षों का पुनः x के सापेक्ष अवकलन करने पर,
\( \frac{d²y}{dx²} = \frac{d}{dx} (\frac{1}{1 + x²}) = \frac{d}{dx} ((1 + x²)^{-1}) \)
\( = -1 (1 + x²)^{-1-1} \frac{d}{dx} (1 + x²) \) (using chain rule)
\( = - (1 + x²)^{-2} (2x) = - \frac{2x}{(1 + x²)²} \)
In simple words: Differentiate tan⁻¹x to get 1/(1+x²). Then, differentiate 1/(1+x²) using the chain rule (by writing it as (1+x²)^(-1)) to find the second derivative.

🎯 Exam Tip: Remember \( \frac{d}{dx} (\tan^{-1}x) = \frac{1}{1 + x²} \). For the second derivative, treat \( \frac{1}{1 + x²} \) as \( (1 + x²)^{-1} \) and apply the chain rule correctly.

 

Question 9. log (log x)
Answer:
हल- माना \( y = \log (\log x) \)
दोनों पक्षों का x के सापेक्ष अवकलन करने पर,
\( \frac{dy}{dx} = \frac{d}{dx} (\log (\log x)) \)
\( = \frac{1}{\log x} \frac{d}{dx} (\log x) \) (using chain rule)
\( = \frac{1}{\log x} \cdot \frac{1}{x} = \frac{1}{x \log x} \)
दोनों पक्षों का पुनः x के सापेक्ष अवकलन करने पर,
\( \frac{d²y}{dx²} = \frac{d}{dx} (\frac{1}{x \log x}) = \frac{d}{dx} ((x \log x)^{-1}) \)
\( = -1 (x \log x)^{-1-1} \frac{d}{dx} (x \log x) \) (using chain rule)
\( = - (x \log x)^{-2} [x \frac{d}{dx} (\log x) + \log x \frac{d}{dx} (x)] \) (using product rule for \( x \log x \))
\( = - \frac{1}{(x \log x)²} [x \cdot \frac{1}{x} + \log x \cdot 1] \)
\( = - \frac{1 + \log x}{(x \log x)²} \)
In simple words: First, apply the chain rule to differentiate log(log x). Then, to find the second derivative, differentiate the result (1/(x log x)) using the chain rule and product rule.

🎯 Exam Tip: Both chain rule and product rule are essential. Write \( \frac{1}{x \log x} \) as \( (x \log x)^{-1} \) for easier differentiation in the second step. Be careful with exponents and algebraic signs.

 

Question 10. sin (log x)
Answer:
हल- माना \( y = \sin (\log x) \)
दोनों पक्षों का x के सापेक्ष अवकलन करने पर,
\( \frac{dy}{dx} = \frac{d}{dx} (\sin (\log x)) \)
\( = \cos (\log x) \frac{d}{dx} (\log x) \) (using chain rule)
\( = \cos (\log x) \cdot \frac{1}{x} = \frac{\cos (\log x)}{x} \)
दोनों पक्षों का पुनः x के सापेक्ष अवकलन करने पर,
\( \frac{d²y}{dx²} = \frac{d}{dx} (\frac{\cos (\log x)}{x}) \)
\( = \frac{x \frac{d}{dx} (\cos (\log x)) - \cos (\log x) \frac{d}{dx} (x)}{x²} \) (using quotient rule)
\( = \frac{x (- \sin (\log x) \cdot \frac{1}{x}) - \cos (\log x) (1)}{x²} \)
\( = \frac{- \sin (\log x) - \cos (\log x)}{x²} \)
\( = - \frac{\sin (\log x) + \cos (\log x)}{x²} \)
In simple words: Differentiate sin(log x) using the chain rule to get the first derivative. Then, apply the quotient rule to differentiate the first derivative, which involves another chain rule for sin(log x) and cos(log x), to get the second derivative.

🎯 Exam Tip: Both chain rule and quotient rule are critical. Be mindful of negative signs from differentiating \( \cos (\log x) \). Ensure proper handling of the derivative of \( \log x \) as \( 1/x \) within the chain rule.

 

Question 11. यदि \( y = 5 \cos x - 3 \sin x \) है तो सिद्ध कीजिए कि \( \frac{d²y}{dx²} + y = 0 \)
Answer:
हल-
दिया है, \( y = 5 \cos x - 3 \sin x \)
दोनों पक्षों का x के सापेक्ष अवकलन करने पर,
\( \frac{dy}{dx} = \frac{d}{dx} (5 \cos x - 3 \sin x) \)
\( = 5 (- \sin x) - 3 (\cos x) = - 5 \sin x - 3 \cos x \)
दोनों पक्षों का पुनः x के सापेक्ष अवकलन करने पर,
\( \frac{d²y}{dx²} = \frac{d}{dx} (- 5 \sin x - 3 \cos x) \)
\( = - 5 (\cos x) - 3 (- \sin x) = - 5 \cos x + 3 \sin x \)
अतः \( \frac{d²y}{dx²} + y = (- 5 \cos x + 3 \sin x) + (5 \cos x - 3 \sin x) = 0 \)
इति सिद्धम्
In simple words: Find the first and second derivatives of y. Then substitute these derivatives and the original function y into the given equation \( \frac{d²y}{dx²} + y = 0 \) to show that it holds true.

🎯 Exam Tip: Differentiate trigonometric functions carefully, noting the sign changes. Algebraic addition and subtraction must be precise to show the desired result.

 

Question 12. यदि \( y = \cos^{-1} x \) है तो \( \frac{d²y}{dx²} \) को केवल y के पदों में ज्ञात कीजिए।
Answer:
हल-
दिया है, \( y = \cos^{-1} x \)
\( \implies x = \cos y \)
दोनों पक्षों का x के सापेक्ष अवकलन करने पर,
\( \frac{d}{dx} (x) = \frac{d}{dx} (\cos y) \)
\( \implies 1 = - \sin y \frac{dy}{dx} \)
\( \implies \frac{dy}{dx} = - \frac{1}{\sin y} = - \csc y \)
दोनों पक्षों का पुनः x के सापेक्ष अवकलन करने पर,
\( \frac{d²y}{dx²} = \frac{d}{dx} (- \csc y) \)
\( = - (- \csc y \cot y) \frac{dy}{dx} \) (using chain rule)
\( = \csc y \cot y \cdot (- \csc y) \) (substituting \( \frac{dy}{dx} = - \csc y \))
\( = - \csc² y \cot y \)
In simple words: First, rewrite y = cos⁻¹x as x = cos y. Differentiate both sides with respect to x to find \( dy/dx \) in terms of y. Then differentiate \( dy/dx \) again with respect to x, using the chain rule, and substitute \( dy/dx \) to express the second derivative solely in terms of y.

🎯 Exam Tip: Implicit differentiation is crucial here. Remember the derivatives of trigonometric functions and their reciprocals, and ensure \( dy/dx \) is substituted correctly in the second differentiation step.

 

Question 13. यदि \( y = 3 \cos (\log x) + 4 \sin (\log x) \) है तो सिद्ध कीजिए कि \( x² \frac{d²y}{dx²} + x \frac{dy}{dx} + y = 0 \)
Answer:
हल-
दिया है, \( y = 3 \cos (\log x) + 4 \sin (\log x) \) ...(1)
दोनों पक्षों का x के सापेक्ष अवकलन करने पर,
\( \frac{dy}{dx} = \frac{d}{dx} (3 \cos (\log x) + 4 \sin (\log x)) \)
\( = 3 (- \sin (\log x)) \frac{d}{dx} (\log x) + 4 (\cos (\log x)) \frac{d}{dx} (\log x) \) (using chain rule)
\( = 3 (- \sin (\log x)) \frac{1}{x} + 4 (\cos (\log x)) \frac{1}{x} \)
\( = \frac{- 3 \sin (\log x) + 4 \cos (\log x)}{x} \)
दोनों पक्षों में x से गुणा करने पर,
\( x \frac{dy}{dx} = - 3 \sin (\log x) + 4 \cos (\log x) \)
दोनों पक्षों का पुनः x के सापेक्ष अवकलन करने पर,
\( \frac{d}{dx} (x \frac{dy}{dx}) = \frac{d}{dx} (- 3 \sin (\log x) + 4 \cos (\log x)) \)
\( x \frac{d²y}{dx²} + \frac{dy}{dx} \cdot 1 = - 3 (\cos (\log x)) \frac{d}{dx} (\log x) + 4 (- \sin (\log x)) \frac{d}{dx} (\log x) \) (using product rule on LHS and chain rule on RHS)
\( x \frac{d²y}{dx²} + \frac{dy}{dx} = - 3 \cos (\log x) \cdot \frac{1}{x} - 4 \sin (\log x) \cdot \frac{1}{x} \)
\( x \frac{d²y}{dx²} + \frac{dy}{dx} = \frac{- 3 \cos (\log x) - 4 \sin (\log x)}{x} \)
दोनों पक्षों में x से गुणा करने पर,
\( x² \frac{d²y}{dx²} + x \frac{dy}{dx} = - [3 \cos (\log x) + 4 \sin (\log x)] \)
From (1), \( y = 3 \cos (\log x) + 4 \sin (\log x) \)
\( x² \frac{d²y}{dx²} + x \frac{dy}{dx} = - y \)
\( \implies x² \frac{d²y}{dx²} + x \frac{dy}{dx} + y = 0 \)
इति सिद्धम्
In simple words: Find the first derivative of y using the chain rule. Multiply by x. Then, find the second derivative of the equation, using the product rule on the left side and the chain rule on the right side. Multiply by x again, and substitute y back to prove the identity.

🎯 Exam Tip: This problem involves careful application of chain rule, product rule, and algebraic manipulation. Keeping terms organized and substituting the original function `y` correctly are crucial for the proof.

 

Question 14. यदि \( y = A e^{mx} + B e^{nx} \) है तो सिद्ध कीजिए कि \( \frac{d²y}{dx²} - (m+n)\frac{dy}{dx} + mny = 0 \)
Answer:
हल-
दिया है, \( y = A e^{mx} + B e^{nx} \) ...(1)
दोनों पक्षों का x के सापेक्ष अवकलन करने पर,
\( \frac{dy}{dx} = \frac{d}{dx} (A e^{mx} + B e^{nx}) \)
\( = A \frac{d}{dx} (e^{mx}) + B \frac{d}{dx} (e^{nx}) \)
\( = A e^{mx} (m) + B e^{nx} (n) = A m e^{mx} + B n e^{nx} \) ...(2)
दोनों पक्षों का पुनः x के सापेक्ष अवकलन करने पर,
\( \frac{d²y}{dx²} = \frac{d}{dx} (A m e^{mx} + B n e^{nx}) \)
\( = A m \frac{d}{dx} (e^{mx}) + B n \frac{d}{dx} (e^{nx}) \)
\( = A m e^{mx} (m) + B n e^{nx} (n) = A m² e^{mx} + B n² e^{nx} \) ...(3)
बायाँ पक्ष = \( \frac{d²y}{dx²} - (m+n)\frac{dy}{dx} + mny \)
\( = (A m² e^{mx} + B n² e^{nx}) - (m+n) (A m e^{mx} + B n e^{nx}) + mn (A e^{mx} + B e^{nx}) \) (from (1), (2), (3))
\( = A m² e^{mx} + B n² e^{nx} - (m A m e^{mx} + m B n e^{nx} + n A m e^{mx} + n B n e^{nx}) + mn A e^{mx} + mn B e^{nx} \)
\( = A m² e^{mx} + B n² e^{nx} - A m² e^{mx} - m n B e^{nx} - m n A e^{mx} - B n² e^{nx} + mn A e^{mx} + mn B e^{nx} \)
\( = (A m² e^{mx} - A m² e^{mx}) + (B n² e^{nx} - B n² e^{nx}) + (- m n B e^{nx} + mn B e^{nx}) + (- m n A e^{mx} + mn A e^{mx}) \)
\( = 0 + 0 + 0 + 0 = 0 \)
दायाँ पक्ष = 0
इति सिद्धम्
In simple words: Find the first and second derivatives of y. Substitute these derivatives and the original function y into the given equation. Expand and combine like terms to show that the entire expression simplifies to zero.

🎯 Exam Tip: Differentiate exponential functions correctly (chain rule is important for \( e^{mx} \) and \( e^{nx} \)). Be meticulous with algebraic expansion and cancellation of terms to prove the identity.

 

Question 15. y = 5e⁷ˣ + 6e⁻⁷ˣ. सिद्ध कीजिए कि \( \frac{d²y}{dx²} = 49y \)
Answer:
हल-
प्रश्नानुसार, \( y = 5e^{7x} + 6e^{-7x} \)
x के सापेक्ष अवकलन करने पर,
\( \frac{dy}{dx} = \frac{d}{dx} (5e^{7x} + 6e^{-7x}) \)
\( = 5 \frac{d}{dx} (e^{7x}) + 6 \frac{d}{dx} (e^{-7x}) \)
\( = 5e^{7x} (7) + 6e^{-7x} (-7) = 35e^{7x} - 42e^{-7x} \)
पुनः x के सापेक्ष अवकलन करने पर,
\( \frac{d²y}{dx²} = \frac{d}{dx} (35e^{7x} - 42e^{-7x}) \)
\( = 35 \frac{d}{dx} (e^{7x}) - 42 \frac{d}{dx} (e^{-7x}) \)
\( = 35e^{7x} (7) - 42e^{-7x} (-7) \)
\( = 245e^{7x} + 294e^{-7x} \)
\( = 49 (5e^{7x} + 6e^{-7x}) \)
\( = 49y \) (using the original function y)
इति सिद्धम्
In simple words: Find the first and second derivatives of y. In the second derivative, factor out 49, and then substitute the original function y back into the expression to prove that \( \frac{d²y}{dx²} = 49y \).

🎯 Exam Tip: The chain rule is essential for differentiating \( e^{7x} \) and \( e^{-7x} \). Look for common factors (like 49) in the second derivative to establish the relationship with the original function.

 

Question 16. यदि \( e^x (x+1) = 1 \) है तो दर्शाइए कि \( \frac{d²y}{dx²} = (\frac{dy}{dx})² \)
Answer:
हल-
दिया है, \( e^x (x + 1) = 1 \) ...(1)
दोनों पक्षों का x के सापेक्ष अवकलन करने पर,
\( \frac{d}{dx} (e^x (x + 1)) = \frac{d}{dx} (1) \)
\( e^x \frac{d}{dx} (x + 1) + (x + 1) \frac{d}{dx} (e^x) = 0 \) (using product rule)
\( e^x (1) + (x + 1) e^x \frac{dy}{dx} = 0 \) (Since \( \frac{d}{dx}(e^x) \) means \( e^x \) not \( e^y \), so the problem is implicitly asking for a relationship between \( \frac{d^2y}{dx^2} \) and \( \frac{dy}{dx} \) where y is an implicit function of x, but the given relation is \( e^x(x+1)=1 \) which is a relation in x only, not y. Let's assume the actual question might be related to a function \( e^y(x+1)=1 \) or the solution is providing a different problem solution)
Let's re-interpret the problem based on the solution:
The solution is for a different problem, which is: If \( e^y(x+1)=1 \), show that \( \frac{d^2y}{dx^2} = (\frac{dy}{dx})^2 \). Let's proceed with this assumed problem.
Given \( e^y(x+1) = 1 \)
\( e^y = \frac{1}{x+1} \)
Differentiating both sides w.r.t x:
\( e^y \frac{dy}{dx} = - \frac{1}{(x+1)²} \)
Substitute \( e^y = \frac{1}{x+1} \):
\( \frac{1}{x+1} \frac{dy}{dx} = - \frac{1}{(x+1)²} \)
\( \frac{dy}{dx} = - \frac{1}{x+1} \)
Now, differentiate \( \frac{dy}{dx} \) again w.r.t x to find \( \frac{d²y}{dx²} \):
\( \frac{d²y}{dx²} = \frac{d}{dx} (- \frac{1}{x+1}) = - (-1)(x+1)^{-2}(1) = \frac{1}{(x+1)²} \)
We need to show \( \frac{d²y}{dx²} = (\frac{dy}{dx})² \)
LHS = \( \frac{1}{(x+1)²} \)
RHS = \( (\frac{dy}{dx})² = (- \frac{1}{x+1})² = \frac{1}{(x+1)²} \)
LHS = RHS.
इति सिद्धम्
In simple words: Assuming the question is \( e^y(x+1)=1 \), first find \( dy/dx \) by implicitly differentiating the equation. Then, find the second derivative \( d²y/dx² \). Compare \( d²y/dx² \) with the square of \( dy/dx \) to show they are equal.

🎯 Exam Tip: Implicit differentiation is key for problems where y is an implicit function of x. Be careful with chain rule when differentiating terms involving y with respect to x. Ensure that the question is correctly interpreted or clarified if ambiguous.

 

Question 17. यदि \( y = (\tan^{-1} x)² \) है तो सिद्ध कीजिए कि \( (x² + 1)² y₂ + 2x (x² + 1) y₁ = 2 \)
Answer:
हल-
दिया है, \( y = (\tan^{-1} x)² \)
दोनों पक्षों का x के सापेक्ष अवकलन करने पर,
\( \frac{dy}{dx} = y₁ = \frac{d}{dx} ((\tan^{-1} x)²) \)
\( = 2 (\tan^{-1} x) \frac{d}{dx} (\tan^{-1} x) \) (using chain rule)
\( = 2 (\tan^{-1} x) \frac{1}{1 + x²} = \frac{2 \tan^{-1} x}{1 + x²} \)
दोनों पक्षों में \( (1 + x²) \) से गुणा करने पर,
\( (1 + x²) y₁ = 2 \tan^{-1} x \)
दोनों पक्षों का पुनः x के सापेक्ष अवकलन करने पर,
\( \frac{d}{dx} ((1 + x²) y₁) = \frac{d}{dx} (2 \tan^{-1} x) \)
\( (1 + x²) \frac{d}{dx} (y₁) + y₁ \frac{d}{dx} (1 + x²) = 2 \frac{d}{dx} (\tan^{-1} x) \) (using product rule on LHS)
\( (1 + x²) y₂ + y₁ (2x) = 2 \frac{1}{1 + x²} \)
\( (1 + x²) y₂ + 2x y₁ = \frac{2}{1 + x²} \)
दोनों पक्षों में \( (1 + x²) \) से गुणा करने पर,
\( (1 + x²)² y₂ + 2x (1 + x²) y₁ = 2 \)
इति सिद्धम्
In simple words: Find the first derivative \( y₁ \) of \( y = (\tan^{-1} x)² \), which involves the chain rule. Multiply by \( (1+x²) \) to simplify. Then, differentiate implicitly again to find \( y₂ \), using the product rule on the left side. Multiply by \( (1+x²) \) one more time to obtain the desired identity.

🎯 Exam Tip: This problem tests multiple differentiation rules, including chain rule and product rule. Careful algebraic manipulation and keeping track of \( y₁ \) and \( y₂ \) are essential for successfully proving the identity.

 

प्रश्नावली 5.8

 

Question 1. फलन f(x) = x² + 2x - 8, x∈[-4,2] के लिए रोले के प्रमेय को सत्यापित कीजिए।
Answer:
हल-
यहाँ \( f(x) = x² + 2x - 8 \) एक बहुपद फलन है।
\( \implies \) यह अपने डोमेन \( x \in R \) में संतत और अवकलनीय है।
इसलिए, यह अन्तराल \( [-4,2] \) में संतत है और अन्तराल \( (-4,2) \) में अवकलनीय है।
\( f(-4) = (-4)² + 2(-4) - 8 = 16 - 8 - 8 = 0 \)
\( f(2) = (2)² + 2(2) - 8 = 4 + 4 - 8 = 0 \)
चूंकि \( f(-4) = f(2) \), रोले के प्रमेय की शर्तें सन्तुष्ट होती हैं।
अब, हम एक \( c \in (-4,2) \) ज्ञात करते हैं जिसके लिए \( f'(c) = 0 \) हो।
\( f'(x) = \frac{d}{dx} (x² + 2x - 8) = 2x + 2 \)
\( f'(c) = 2c + 2 \)
\( f'(c) = 0 \implies 2c + 2 = 0 \implies 2c = -2 \implies c = -1 \)
चूँकि \( c = -1 \in (-4,2) \), रोले का प्रमेय सत्यापित होता है।
In simple words: To verify Rolle's Theorem, check if the function is continuous on the closed interval, differentiable on the open interval, and if \( f(a) = f(b) \). If these conditions hold, find a value 'c' in the open interval where the derivative is zero.

🎯 Exam Tip: Remember the three conditions for Rolle's Theorem: continuity, differentiability, and \( f(a) = f(b) \). Polynomials are always continuous and differentiable. The final step is to find 'c' such that \( f'(c) = 0 \).

 

Question 2. जाँच कीजिए कि रोले का प्रमेय निम्नलिखित फलनों में से किन-किन पर लागू होता है? इन उदाहरणों से क्या आप रोले के प्रमेय के विलोम के बारे में कुछ कह सकते हैं? (i) f(x) = [x] के लिए x∈[5,9] (ii) f(x) = [x] के लिए x∈[-2,2] (iii) f(x) = x² - 1 के लिए x∈[1,2]
Answer:
हल-
(i) \( f(x) = [x] \), \( x \in [5,9] \) के लिए:
\( f(x) = [x] \) पूर्णांक बिन्दुओं पर संतत नहीं है (जैसे \( x = 6, 7, 8 \)) और इसलिए अवकलनीय भी नहीं है।
अतः रोले का प्रमेय लागू नहीं होता है।
(ii) \( f(x) = [x] \), \( x \in [-2,2] \) के लिए:
\( f(x) = [x] \) पूर्णांक बिन्दुओं पर संतत नहीं है (जैसे \( x = -1, 0, 1 \)) और इसलिए अवकलनीय भी नहीं है।
अतः रोले का प्रमेय लागू नहीं होता है।
(iii) \( f(x) = x² - 1 \), \( x \in [1,2] \) के लिए:
यह एक बहुपद फलन है, इसलिए यह अन्तराल \( [1,2] \) में संतत है और \( (1,2) \) में अवकलनीय है।
\( f(1) = (1)² - 1 = 0 \)
\( f(2) = (2)² - 1 = 3 \)
चूँकि \( f(1) \neq f(2) \), रोले का प्रमेय लागू नहीं होता है।
रोले के प्रमेय के विलोम के बारे में: इन उदाहरणों से पता चलता है कि यदि रोले के प्रमेय की शर्तें पूरी नहीं होती हैं, तो भी \( f'(c) = 0 \) संभव हो सकता है, लेकिन यह आवश्यक नहीं है। रोले का प्रमेय एक पर्याप्त शर्त है, आवश्यक नहीं।
In simple words: Rolle's Theorem requires continuity, differentiability, and \( f(a) = f(b) \). For parts (i) and (ii), the greatest integer function is not continuous. For part (iii), while it's continuous and differentiable, \( f(a) \neq f(b) \). Therefore, Rolle's Theorem does not apply to any of these cases.

🎯 Exam Tip: The greatest integer function is a common example for discontinuities. For Rolle's Theorem, all three conditions must be met. If even one condition fails, the theorem cannot be applied.

 

Question 3. यदि f :[-5, 5] → R एक संतत फलन है और यदि f '(x) किसी भी बिन्दु पर शून्य नहीं होता है तो सिद्ध कीजिए कि f(-5) ≠ f(5).
Answer:
हल-
दिया है, फलन f:[-5, 5] → R एक संतत फलन है।
तथा f '(x) किसी भी बिन्दु पर शून्य नहीं होता है, यानी \( f'(x) \neq 0 \) for all \( x \in (-5,5) \).
हमें सिद्ध करना है कि \( f(-5) \neq f(5) \).
माना इसके विपरीत, \( f(-5) = f(5) \).
चूँकि f अन्तराल \( [-5,5] \) में संतत है और अन्तराल \( (-5,5) \) में अवकलनीय है (क्योंकि \( f'(x) \) मौजूद है), और \( f(-5) = f(5) \), तो रोले के प्रमेय के अनुसार, एक ऐसा \( c \in (-5,5) \) मौजूद होना चाहिए जिसके लिए \( f'(c) = 0 \) हो।
लेकिन यह दी गई शर्त के विपरीत है कि \( f'(x) \) किसी भी बिन्दु पर शून्य नहीं होता है।
अतः हमारी मूल धारणा \( f(-5) = f(5) \) गलत है।
इसलिए, \( f(-5) \neq f(5) \).
इति सिद्धम्
In simple words: We prove this by contradiction. Assume \( f(-5) = f(5) \). Since f is continuous and differentiable, by Rolle's Theorem, there must be a point 'c' where \( f'(c) = 0 \). This contradicts the given information that \( f'(x) \) is never zero. Therefore, our assumption is false, meaning \( f(-5) \neq f(5) \).

🎯 Exam Tip: This problem is a direct application of Rolle's Theorem using proof by contradiction. Clearly state the assumptions, apply the theorem, and show the contradiction to conclude the proof.

 

Question 4. माध्यमान प्रमेय सत्यापित कीजिए, यदि अन्तराल [a, b] में f(x) = x² - 4x - 3, जहाँ a = 1 और b = 4 है।
Answer:
हल-
दिया है, \( f(x) = x² - 4x - 3 \), अन्तराल \( [1,4] \) के लिए।
फलन \( f(x) \) एक बहुपद फलन है, इसलिए यह अन्तराल \( [1,4] \) में संतत है और अन्तराल \( (1,4) \) में अवकलनीय है।
माध्यमान प्रमेय की शर्तें सन्तुष्ट होती हैं।
अब, हमें एक \( c \in (1,4) \) ज्ञात करना है जिसके लिए \( f'(c) = \frac{f(b) - f(a)}{b - a} \) हो।
\( f(a) = f(1) = (1)² - 4(1) - 3 = 1 - 4 - 3 = -6 \)
\( f(b) = f(4) = (4)² - 4(4) - 3 = 16 - 16 - 3 = -3 \)
\( f'(x) = \frac{d}{dx} (x² - 4x - 3) = 2x - 4 \)
\( f'(c) = 2c - 4 \)
माध्यमान प्रमेय के अनुसार:
\( f'(c) = \frac{f(4) - f(1)}{4 - 1} \)
\( 2c - 4 = \frac{-3 - (-6)}{3} \)
\( 2c - 4 = \frac{-3 + 6}{3} = \frac{3}{3} = 1 \)
\( 2c = 1 + 4 \)
\( 2c = 5 \)
\( c = \frac{5}{2} \)
चूँकि \( c = \frac{5}{2} = 2.5 \in (1,4) \), माध्यमान प्रमेय सत्यापित होती है।
इति सिद्धम्
In simple words: Verify that the function is continuous and differentiable over the given interval. Then, calculate \( f(a) \) and \( f(b) \), and the derivative \( f'(x) \). Equate \( f'(c) \) to the slope of the secant line \( \frac{f(b) - f(a)}{b - a} \) to find 'c'. If 'c' is within the open interval, the theorem is verified.

🎯 Exam Tip: Mean Value Theorem requires continuity on the closed interval and differentiability on the open interval. The formula \( f'(c) = \frac{f(b) - f(a)}{b - a} \) is central. Ensure algebraic steps are correct when solving for 'c' and check if 'c' lies in the open interval.

 

Question 5. माध्यमान प्रमेय सत्यापित कीजिए, यदि अन्तराल [a, b] में f(x) = x³ - 5x² - 3x, जहाँ a = 1 और b = 3 है। f '(c) = 0 के लिए c∈ (1, 3) को ज्ञात कीजिए।
Answer:
हल-
दिया है, \( f(x) = x³ - 5x² - 3x \), अन्तराल \( [1,3] \) में।
यह एक बहुपद फलन है, इसलिए यह अन्तराल \( [1,3] \) में संतत है और अन्तराल \( (1,3) \) में अवकलनीय है।
माध्यमान प्रमेय की शर्तें सन्तुष्ट होती हैं।
अब, हमें एक \( c \in (1,3) \) ज्ञात करना है जिसके लिए \( f'(c) = \frac{f(b) - f(a)}{b - a} \) हो।
\( f(a) = f(1) = (1)³ - 5(1)² - 3(1) = 1 - 5 - 3 = -7 \)
\( f(b) = f(3) = (3)³ - 5(3)² - 3(3) = 27 - 45 - 9 = -27 \)
\( f'(x) = \frac{d}{dx} (x³ - 5x² - 3x) = 3x² - 10x - 3 \)
\( f'(c) = 3c² - 10c - 3 \)
माध्यमान प्रमेय के अनुसार:
\( f'(c) = \frac{f(3) - f(1)}{3 - 1} \)
\( 3c² - 10c - 3 = \frac{-27 - (-7)}{2} \)
\( 3c² - 10c - 3 = \frac{-27 + 7}{2} = \frac{-20}{2} = -10 \)
\( 3c² - 10c - 3 + 10 = 0 \)
\( 3c² - 10c + 7 = 0 \)
इस द्विघात समीकरण को हल करने पर:
\( c = \frac{- (-10) \pm \sqrt{(-10)² - 4(3)(7)}}{2(3)} \)
\( c = \frac{10 \pm \sqrt{100 - 84}}{6} = \frac{10 \pm \sqrt{16}}{6} = \frac{10 \pm 4}{6} \)
दो मान प्राप्त होते हैं:
\( c₁ = \frac{10 + 4}{6} = \frac{14}{6} = \frac{7}{3} \)
\( c₂ = \frac{10 - 4}{6} = \frac{6}{6} = 1 \)
चूँकि \( c \in (1,3) \), हम \( c = 1 \) को छोड़ देंगे क्योंकि यह अन्तराल में नहीं है।
इसलिए, \( c = \frac{7}{3} \approx 2.33 \) जो अन्तराल \( (1,3) \) में है।
अतः माध्यमान प्रमेय सत्यापित होता है और \( c = \frac{7}{3} \).
इति सिद्धम्
In simple words: Verify continuity and differentiability of the polynomial function. Calculate \( f(a) \), \( f(b) \), and \( f'(x) \). Set \( f'(c) \) equal to the slope of the secant line \( \frac{f(b) - f(a)}{b - a} \) to form a quadratic equation. Solve for 'c' and select the value that lies within the open interval \( (a,b) \).

🎯 Exam Tip: For Mean Value Theorem, ensure the function is continuous and differentiable. Pay attention to calculating \( f(a) \), \( f(b) \), and solving the quadratic equation for 'c'. Only 'c' values strictly within the open interval are valid.

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