UP Board Solutions Class 11 Maths Chapter 13 Limits and Derivatives

UP Board Solutions For Class 11 Maths Chapter 13 Limits And Derivatives (सीमा और अवकलज)

Exercise 13.1

Seema Aur Avkalan Class 11 प्रश्न 1 से 22 तक निम्नलिखित सीमाओं के मान प्राप्त कीजिए:

Question 1. lim x + 3.
\( \lim_{x \to 3} x + 3 \)
Answer: \( \lim_{x \to 3} (x + 3) = 3 + 3 = 6. \)
In simple words: To find the limit of \(x+3\) as \(x\) approaches 3, simply substitute \(x=3\) into the expression, which gives \(3+3=6\).

🎯 Exam Tip: Direct substitution is the first approach for limits. If it yields a definite value, that's the limit. If it results in an indeterminate form (like 0/0), further algebraic manipulation is required.

 

Question 2. lim (x-22)
\( \lim_{x \to 7} (x - \frac{22}{7}) \)
Answer: हल : \( \lim_{x \to 7} (x - \frac{22}{7}) = 7 - \frac{22}{7} = \frac{49 - 22}{7} = \frac{27}{7} \)
In simple words: Substitute \(x=7\) into the expression to find the limit, resulting in \(7 - 22/7 = 27/7\).

🎯 Exam Tip: Ensure basic arithmetic and fraction operations are correct during substitution to avoid calculation errors.

 

Question 3. lim \( \pi r^2 \).
\( \lim_{r \to 1} (\pi r^2) \)
Answer: हल. \( \lim_{r \to 1} (\pi r^2) = \pi . (1)^2 = \pi \)
In simple words: For the limit of \( \pi r^2 \) as \(r\) approaches 1, replace \(r\) with 1, which gives \( \pi (1)^2 = \pi \).

🎯 Exam Tip: Variables in the limit operator indicate the term to be substituted. Constants or other variables remain as they are.

 

Question 4. lim 4x +3
\( \lim_{x \to 4} \frac{4x + 3}{x - 2} \)
Answer: हल : \( \lim_{x \to 4} \frac{4x + 3}{x - 2} = \frac{4 \times 4 + 3}{4 - 2} \)
\( = \frac{16 + 3}{2} \)
\( = \frac{19}{2} \)
In simple words: Substitute \(x=4\) into the numerator and denominator. The numerator becomes \(4 \times 4 + 3 = 19\) and the denominator becomes \(4 - 2 = 2\), so the limit is \(19/2\).

🎯 Exam Tip: Always check if direct substitution results in a non-zero denominator. If so, it's a valid limit calculation.

 

Question 5. lim \( \frac{x^{10} + x^5 + 1}{x - 1} \)
\( \lim_{x \to -1} \frac{x^{10} + x^5 + 1}{x - 1} \)
Answer: हल : \( \lim_{x \to -1} \frac{x^{10} + x^5 + 1}{x - 1} = \frac{(-1)^{10} + (-1)^5 + 1}{-1 - 1} \)
\( = \frac{1 - 1 + 1}{-2} \)
\( = \frac{1}{-2} \)
In simple words: Substitute \(x = -1\) into the expression. The numerator becomes \(1 - 1 + 1 = 1\) and the denominator becomes \(-1 - 1 = -2\), giving a limit of \(-1/2\).

🎯 Exam Tip: Be careful with signs when substituting negative values into powers. Even powers result in positive, odd powers retain the negative sign.

 

Question 6. lim \( \frac{(x + 1)^5 - 1}{x} \)
\( \lim_{x \to 0} \frac{(x + 1)^5 - 1}{x} \)
Answer: हल : \( \lim_{x \to 0} \frac{(x + 1)^5 - 1}{x} \)
\( = \lim_{x \to 0} \frac{(1 + 5x + 10x^2 + 10x^3 + 5x^4 + x^5) - 1}{x} \)
\( = \lim_{x \to 0} \frac{5x + 10x^2 + 10x^3 + 5x^4 + x^5}{x} \)
\( = \lim_{x \to 0} (5 + 10x + 10x^2 + 5x^3 + x^4) \)
\( = 5 \)
वैकल्पिक विधि : हम जानते हैं :
\( \lim_{x \to a} \frac{x^n - a^n}{x - a} = n a^{n - 1} \)
\( \lim_{x \to 0} \frac{(x + 1)^5 - 1}{x} = \lim_{x \to 0} \frac{(x + 1)^5 - 1^5}{(x + 1) - 1} \)
\( = 5 (1)^{5 - 1} \)
\( = 5 (1)^4 \)
\( = 5 \)
In simple words: This limit is in the indeterminate form 0/0. Using the binomial expansion of \((x+1)^5\), simplify the expression and then factor out \(x\) from the numerator to cancel with the denominator, allowing for direct substitution to get 5. Alternatively, recognize it as a standard limit form \( \lim_{x \to a} \frac{x^n - a^n}{x - a} = n a^{n - 1} \), with \(y = x+1\) and \(a=1\), which directly yields \(5 \times 1^{4} = 5\).

🎯 Exam Tip: For indeterminate forms (0/0), use algebraic simplification (factoring, rationalization, binomial expansion) or L'Hôpital's Rule if applicable. Recognizing standard limit formulas can save time.

 

Question 7. lim \( \frac{3x^2 - x - 10}{x^2 - 4} \)
\( \lim_{x \to 2} \frac{3x^2 - x - 10}{x^2 - 4} \)
Answer: हल : \( \lim_{x \to 2} \frac{3x^2 - x - 10}{x^2 - 4} \)
\( = \lim_{x \to 2} \frac{(x - 2)(3x + 5)}{(x - 2)(x + 2)} \)
\( = \lim_{x \to 2} \frac{3x + 5}{x + 2} \)
\( = \frac{3 \times 2 + 5}{2 + 2} \)
\( = \frac{6 + 5}{4} \)
\( = \frac{11}{4} \)
In simple words: This is an indeterminate form (0/0). Factorize both the numerator and denominator, cancel out the common factor \((x-2)\), and then substitute \(x=2\) into the simplified expression to get \(11/4\).

🎯 Exam Tip: When dealing with quadratic or polynomial expressions in limits that yield 0/0, factoring out \((x-a)\) (where \(a\) is the limit point) from both numerator and denominator is a common and effective strategy.

 

Question 8. lim \( \frac{x^4 - 81}{2x^2 - 5x - 3} \)
\( \lim_{x \to 3} \frac{x^4 - 81}{2x^2 - 5x - 3} \)
Answer: हल : \( \lim_{x \to 3} \frac{x^4 - 81}{2x^2 - 5x - 3} \)
\( = \lim_{x \to 3} \frac{(x^2 - 9)(x^2 + 9)}{(x - 3)(2x + 1)} \)
\( = \lim_{x \to 3} \frac{(x - 3)(x + 3)(x^2 + 9)}{(x - 3)(2x + 1)} \)
\( = \lim_{x \to 3} \frac{(x + 3)(x^2 + 9)}{2x + 1} \)
\( = \frac{(3 + 3)(3^2 + 9)}{2 \times 3 + 1} \)
\( = \frac{(6)(9 + 9)}{6 + 1} \)
\( = \frac{6 \times 18}{7} \)
\( = \frac{108}{7} \)
In simple words: This is an indeterminate form (0/0). Factorize the numerator using the difference of squares twice, and factorize the quadratic denominator. Cancel the common factor \((x-3)\) and then substitute \(x=3\) to find the limit, which is \(108/7\).

🎯 Exam Tip: Always look for factoring opportunities, especially difference of squares \((a^2 - b^2) = (a-b)(a+b)\) or factoring quadratics, when dealing with rational functions in limits.

 

Question 9. lim \( \frac{ax + b}{cx + 1} \)
\( \lim_{x \to 0} \frac{ax + b}{cx + 1} \)
Answer: हल : \( \lim_{x \to 0} \frac{ax + b}{cx + 1} = \frac{a \times 0 + b}{c \times 0 + 1} \)
\( = \frac{0 + b}{0 + 1} \)
\( = b. \)
In simple words: Substitute \(x=0\) directly into the expression. The numerator becomes \(a(0)+b = b\) and the denominator becomes \(c(0)+1 = 1\), so the limit is \(b/1 = b\).

🎯 Exam Tip: For rational functions, if direct substitution does not lead to an indeterminate form and the denominator is non-zero, it is the limit.

 

Question 10. lim \( \frac{z^{1/3} - 1}{z^{1/6} - 1} \)
\( \lim_{z \to 1} \frac{z^{1/3} - 1}{z^{1/6} - 1} \)
Answer: हल : \( \lim_{z \to 1} \frac{z^{1/3} - 1}{z^{1/6} - 1} \)
माना \( z^{1/6} = y \)
तब \( z^{1/3} = (z^{1/6})^2 = y^2 \)
जैसे ही \( z \to 1 \), \( y \to 1 \)
\( = \lim_{y \to 1} \frac{y^2 - 1}{y - 1} \)
\( = \lim_{y \to 1} \frac{(y - 1)(y + 1)}{y - 1} \)
\( = \lim_{y \to 1} (y + 1) \)
\( = 1 + 1 = 2. \)
In simple words: This is an indeterminate form (0/0). Substitute \(z^{1/6} = y\) so the expression becomes \( (y^2-1)/(y-1) \). Factorize the numerator to \((y-1)(y+1)\), cancel \((y-1)\), and then substitute \(y=1\) into \((y+1)\) to get \(1+1=2\).

🎯 Exam Tip: When dealing with fractional powers, a substitution can often transform the expression into a more manageable polynomial form, allowing for easier factorization and simplification.

 

Question 11. lim \( \frac{ax^2 + bx + c}{cx^2 + bx + a} \), \(a + b + c \neq 0\).
\( \lim_{x \to 1} \frac{ax^2 + bx + c}{cx^2 + bx + a} \)
Answer: हल : \( \lim_{x \to 1} \frac{ax^2 + bx + c}{cx^2 + bx + a} = \frac{a(1)^2 + b(1) + c}{c(1)^2 + b(1) + a} \)
\( = \frac{a + b + c}{c + b + a} \)
\( = 1. \)
In simple words: Since \(a+b+c \neq 0\), direct substitution of \(x=1\) is valid. This yields \((a+b+c)/(c+b+a)\), which simplifies to 1.

🎯 Exam Tip: Always check the condition given (here, \(a+b+c \neq 0\)) as it often confirms that the denominator will not be zero upon direct substitution, making the limit calculation straightforward.

 

Question 12. lim \( \frac{\frac{1}{x} + \frac{1}{2}}{x + 2} \)
\( \lim_{x \to -2} \frac{\frac{1}{x} + \frac{1}{2}}{x + 2} \)
Answer: हल : \( \lim_{x \to -2} \frac{\frac{1}{x} + \frac{1}{2}}{x + 2} \)
\( = \lim_{x \to -2} \frac{\frac{2 + x}{2x}}{x + 2} \)
\( = \lim_{x \to -2} \frac{x + 2}{2x(x + 2)} \)
\( = \lim_{x \to -2} \frac{1}{2x} \)
\( = \frac{1}{2(-2)} \)
\( = - \frac{1}{4} \)
In simple words: This is an indeterminate form (0/0). Simplify the numerator by finding a common denominator, then cancel the common factor \((x+2)\) from the numerator and denominator. Finally, substitute \(x=-2\) into the simplified expression to get \(-1/4\).

🎯 Exam Tip: When fractions are present in the numerator or denominator, combine them first to simplify the expression, often revealing factors that can be cancelled.

 

Question 13. lim \( \frac{\sin ax}{bx} \)
\( \lim_{x \to 0} \frac{\sin ax}{bx} \)
Answer: हल : \( \lim_{x \to 0} \frac{\sin ax}{bx} \)
\( = \lim_{x \to 0} \frac{\sin ax}{ax} \times \frac{ax}{bx} \)
\( = \lim_{x \to 0} \frac{\sin ax}{ax} \times \frac{a}{b} \)
\( = 1 \times \frac{a}{b} \)
\( = \frac{a}{b} \)
[क्योंकि \( \lim_{x \to 0} \frac{\sin x}{x} = 1 \)]
In simple words: To evaluate this limit, we use the standard limit identity \( \lim_{x \to 0} \frac{\sin x}{x} = 1 \). Multiply and divide by \(ax\) in the numerator to transform the expression, allowing the application of this identity, and simplifying to \(a/b\).

🎯 Exam Tip: For trigonometric limits involving \(\sin x\) or \(\tan x\) as \(x \to 0\), manipulate the expression to fit the standard forms \(\lim_{x \to 0} \frac{\sin x}{x} = 1\) or \(\lim_{x \to 0} \frac{\tan x}{x} = 1\).

 

Question 14. lim \( \frac{\sin ax}{\sin bx} \), \(a, b \neq 0\).
\( \lim_{x \to 0} \frac{\sin ax}{\sin bx} \)
Answer: हल : \( \lim_{x \to 0} \frac{\sin ax}{\sin bx} \)
\( = \lim_{x \to 0} \frac{\frac{\sin ax}{ax} \times ax}{\frac{\sin bx}{bx} \times bx} \)
\( = \lim_{x \to 0} \frac{\frac{\sin ax}{ax}}{\frac{\sin bx}{bx}} \times \frac{ax}{bx} \)
\( = \frac{1}{1} \times \frac{a}{b} \)
\( = \frac{a}{b} \)
In simple words: This limit can be solved by manipulating the expression to use the identity \( \lim_{x \to 0} \frac{\sin x}{x} = 1 \). Multiply and divide the numerator by \(ax\) and the denominator by \(bx\), then rearrange and apply the identity to simplify the expression to \(a/b\).

🎯 Exam Tip: When both numerator and denominator contain \(\sin\) functions approaching 0, apply the standard limit identity separately to each, ensuring to adjust for the arguments \(ax\) and \(bx\).

 

Question 15. lim \( \frac{\sin(\pi - x)}{\pi (\pi - x)} \)
\( \lim_{x \to \pi} \frac{\sin(\pi - x)}{\pi (\pi - x)} \)
Answer: हल : \( \lim_{x \to \pi} \frac{\sin(\pi - x)}{\pi (\pi - x)} \)
माना \( \pi - x = \theta \) लीजिए, जब \( x \to \pi, \theta \to 0 \)
\( = \lim_{\theta \to 0} \frac{1}{\pi} \frac{\sin \theta}{\theta} \)
\( = \frac{1}{\pi} \times 1 \)
\( = \frac{1}{\pi} \)
In simple words: To evaluate this limit, let \( \theta = \pi - x \). As \(x\) approaches \(\pi\), \( \theta \) approaches 0. The expression then transforms into a known standard limit form \(\lim_{\theta \to 0} \frac{\sin \theta}{\theta} = 1\), resulting in \(1/\pi\).

🎯 Exam Tip: For limits where the variable approaches a non-zero value and the argument of a trigonometric function becomes 0, a substitution can often convert it into the standard \(\lim_{\theta \to 0} \frac{\sin \theta}{\theta} = 1\) form.

 

Question 16. lim \( \frac{\cos x}{\pi - x} \)
\( \lim_{x \to 0} \frac{\cos x}{\pi - x} \)
Answer: हल : \( \lim_{x \to 0} \frac{\cos x}{\pi - x} \)
\( = \frac{\cos 0}{\pi - 0} \)
\( = \frac{1}{\pi} \)
In simple words: Directly substitute \(x=0\) into the expression. The numerator becomes \(\cos 0 = 1\) and the denominator becomes \(\pi - 0 = \pi\), giving the limit as \(1/\pi\).

🎯 Exam Tip: If direct substitution yields a definite, non-indeterminate value, it is the limit. No further manipulation is needed.

 

Question 17. lim \( \frac{\cos 2x - 1}{\cos x - 1} \)
\( \lim_{x \to 0} \frac{\cos 2x - 1}{\cos x - 1} \)
Answer: हल : \( \lim_{x \to 0} \frac{\cos 2x - 1}{\cos x - 1} \)
\( = \lim_{x \to 0} \frac{(1 - 2\sin^2 x) - 1}{\cos x - 1} \)
\( = \lim_{x \to 0} \frac{-2\sin^2 x}{-(1 - \cos x)} \)
\( = \lim_{x \to 0} \frac{2\sin^2 x}{1 - \cos x} \)
\( = \lim_{x \to 0} \frac{2\sin^2 x}{(1 - \cos x)} \times \frac{(1 + \cos x)}{(1 + \cos x)} \)
\( = \lim_{x \to 0} \frac{2\sin^2 x (1 + \cos x)}{1 - \cos^2 x} \)
\( = \lim_{x \to 0} \frac{2\sin^2 x (1 + \cos x)}{\sin^2 x} \)
\( = \lim_{x \to 0} 2(1 + \cos x) \)
\( = 2(1 + \cos 0) \)
\( = 2(1 + 1) \)
\( = 2 \times 2 = 4. \)
In simple words: This is an indeterminate form (0/0). Use the double-angle identity \(\cos 2x = 1 - 2\sin^2 x\) in the numerator. Simplify the expression by cancelling the negative signs and multiplying by \((1+\cos x)/(1+\cos x)\) to use \(1-\cos^2 x = \sin^2 x\). After cancelling \(\sin^2 x\), substitute \(x=0\) to get \(2(1+1)=4\).

🎯 Exam Tip: For trigonometric limits involving \(\cos x - 1\) or \(\cos 2x - 1\), use identities like \(\cos 2x = 1 - 2\sin^2 x\) or multiply by conjugates like \((1+\cos x)\) to convert to \(\sin^2 x\), which can then be cancelled or simplified using \(\lim_{x \to 0} \frac{\sin x}{x} = 1\).

 

Question 18. lim \( \frac{ax + x \cos x}{b \sin x} \)
\( \lim_{x \to 0} \frac{ax + x \cos x}{b \sin x} \)
Answer: हल : \( \lim_{x \to 0} \frac{ax + x \cos x}{b \sin x} \)
\( = \lim_{x \to 0} \frac{x(a + \cos x)}{b \sin x} \)
\( = \lim_{x \to 0} \frac{a + \cos x}{b \frac{\sin x}{x}} \)
\( = \frac{a + \cos 0}{b \lim_{x \to 0} \frac{\sin x}{x}} \)
\( = \frac{a + 1}{b \times 1} \)
\( = \frac{a + 1}{b} \)
In simple words: This is an indeterminate form (0/0). Factor out \(x\) from the numerator. Rearrange the denominator to use the standard limit identity \( \lim_{x \to 0} \frac{\sin x}{x} = 1 \). Then substitute \(x=0\) to get \((a+1)/b\).

🎯 Exam Tip: When \(x\) is a factor in the numerator and \(\sin x\) in the denominator, always manipulate to form \(\frac{\sin x}{x}\) to use the fundamental trigonometric limit.

 

Question 19. lim \( x \sec x \).
\( \lim_{x \to 0} x \sec x \)
Answer: हल : \( \lim_{x \to 0} x \sec x \)
\( = \lim_{x \to 0} \frac{x}{\cos x} \)
\( = \frac{0}{\cos 0} \)
\( = \frac{0}{1} \)
\( = 0. \)
In simple words: Rewrite \(\sec x\) as \(1/\cos x\). Then substitute \(x=0\) directly. The numerator becomes 0 and the denominator becomes \(\cos 0 = 1\), resulting in a limit of 0.

🎯 Exam Tip: When evaluating limits of trigonometric functions, if direct substitution doesn't result in an indeterminate form, it is the limit. Remember \(\cos 0 = 1\) and \(\sin 0 = 0\).

 

Question 20. lim \( \frac{\sin ax + bx}{ax + \sin bx} \), \(a, b, a + b \neq 0\).
\( \lim_{x \to 0} \frac{\sin ax + bx}{ax + \sin bx} \)
Answer: हल : \( \lim_{x \to 0} \frac{\sin ax + bx}{ax + \sin bx} \)
अंश और हर को \(x\) से भाग करने पर,
\( = \lim_{x \to 0} \frac{\frac{\sin ax}{x} + b}{a + \frac{\sin bx}{x}} \)
\( = \frac{\lim_{x \to 0} (\frac{\sin ax}{x}) + b}{a + \lim_{x \to 0} (\frac{\sin bx}{x})} \)
\( = \frac{a + b}{a + b} \)
\( = 1 \), ( \(a + b \neq 0\) )
In simple words: This is an indeterminate form (0/0). Divide both the numerator and denominator by \(x\). Then apply the standard limit identity \( \lim_{x \to 0} \frac{\sin kx}{x} = k \) to both terms, which simplifies the expression to \((a+b)/(a+b)\), resulting in 1 (given \(a+b \neq 0\)).

🎯 Exam Tip: For expressions with sums of \(\sin x\) and linear terms, dividing by \(x\) (or the highest power of \(x\)) is a common strategy to use the standard trigonometric limits effectively.

 

Question 21. lim \( (\operatorname{cosec} x - \cot x) \).
\( \lim_{x \to 0} (\operatorname{cosec} x - \cot x) \)
Answer: हल : \( \lim_{x \to 0} (\operatorname{cosec} x - \cot x) \)
\( = \lim_{x \to 0} \left( \frac{1}{\sin x} - \frac{\cos x}{\sin x} \right) \)
\( = \lim_{x \to 0} \frac{1 - \cos x}{\sin x} \)
\( = \lim_{x \to 0} \frac{1 - \cos x}{\sin x} \times \frac{1 + \cos x}{1 + \cos x} \)
\( = \lim_{x \to 0} \frac{1 - \cos^2 x}{\sin x (1 + \cos x)} \)
\( = \lim_{x \to 0} \frac{\sin^2 x}{\sin x (1 + \cos x)} \)
\( = \lim_{x \to 0} \frac{\sin x}{1 + \cos x} \)
\( = \frac{\sin 0}{1 + \cos 0} \)
\( = \frac{0}{1 + 1} \)
\( = \frac{0}{2} \)
\( = 0. \)
In simple words: This is an indeterminate form (0/0). Convert \(\operatorname{cosec} x\) and \(\cot x\) to terms of \(\sin x\) and \(\cos x\). Combine them, then multiply by the conjugate \((1+\cos x)\) to transform \((1-\cos x)\) to \(\sin^2 x\). After simplification and cancellation, substitute \(x=0\) to get 0.

🎯 Exam Tip: For limits involving \(\operatorname{cosec} x\) and \(\cot x\), convert them to \(\sin x\) and \(\cos x\) first. Multiplying by the conjugate is often effective for expressions involving \((1 - \cos x)\) or \((1 + \cos x)\).

 

Question 22. lim \( \frac{\tan 2x}{x - \frac{\pi}{2}} \)
\( \lim_{x \to \frac{\pi}{2}} \frac{\tan 2x}{x - \frac{\pi}{2}} \)
Answer: हल: \( \lim_{x \to \frac{\pi}{2}} \frac{\tan 2x}{x - \frac{\pi}{2}} \)
\( x = \frac{\pi}{2} + h \) रखने पर, जैसे ही \( x \to \frac{\pi}{2} \), \( h \to 0 \)
\( = \lim_{h \to 0} \frac{\tan 2(\frac{\pi}{2} + h)}{\frac{\pi}{2} + h - \frac{\pi}{2}} \)
\( = \lim_{h \to 0} \frac{\tan (\pi + 2h)}{h} \)
\( = \lim_{h \to 0} \frac{\tan 2h}{h} \)
\( = \lim_{h \to 0} \frac{\sin 2h}{h \cos 2h} \)
\( = \lim_{h \to 0} \frac{\sin 2h}{2h} \times \frac{2}{\cos 2h} \)
\( = 1 \times \frac{2}{\cos 0} \)
\( = \frac{2}{1} \)
\( = 2. \)
[क्योंकि \( \lim_{h \to 0} \frac{\sin 2h}{2h} = 1 \)]
In simple words: This is an indeterminate form (0/0). Substitute \(x = \frac{\pi}{2} + h\). As \(x \to \frac{\pi}{2}\), \(h \to 0\). Simplify the expression using trigonometric identities (\(\tan(\pi + \theta) = \tan \theta\)) and rearrange to apply the standard limit \(\lim_{k \to 0} \frac{\sin k}{k} = 1\). This results in 2.

🎯 Exam Tip: For limits where the variable approaches a value other than 0, a substitution that shifts the limit point to 0 (e.g., \(x = a+h\)) is often beneficial. Remember \(\tan(\pi + \theta) = \tan \theta\).

 

Question 23. lim f(x) और lim f(x) ज्ञात कीजिए, जहाँ
\( \lim_{x \to 0} f(x) \) और \( \lim_{x \to 1} f(x) \) ज्ञात कीजिए, जहाँ
ℹ️ चित्र व्याख्या (Diagram Explanation): फलन \(f(x)\) को एक piecewise function के रूप में परिभाषित किया गया है: यदि \(x \le 0\), \(f(x) = 2x+3\) यदि \(x > 0\), \(f(x) = 3(x+1)\)
Answer:हल : (i) जब \( x \to 0 \)
बाएँ हाथ की सीमा (LHL):
जब \(x < 0\), \(f(x) = 2x + 3\).
\( \lim_{x \to 0^-} f(x) = \lim_{x \to 0^-} (2x + 3) \)
\( = 2(0) + 3 = 3 \)
दाएँ हाथ की सीमा (RHL):
जब \(x > 0\), \(f(x) = 3(x + 1)\).
\( \lim_{x \to 0^+} f(x) = \lim_{x \to 0^+} 3(x + 1) \)
\( = 3(0 + 1) = 3 \)
चूँकि LHL = RHL = 3, अतः \( \lim_{x \to 0} f(x) = 3 \).

वैकल्पिक विधि (तालिकाओं का उपयोग करके):
जब \(x < 0\), \(f(x) = 2x + 3\)
\(x \to 0^-\) के लिए सारणी इस प्रकार है:

\( \implies \lim_{x \to 0^-} f(x) = 3 \)
जब \(x > 0\), \(f(x) = 3(x + 1)\)
\(x \to 0^+\) के लिए सारणी इस प्रकार है:

\( \implies \lim_{x \to 0^+} f(x) = 3 \)
\( \implies \lim_{x \to 0} f(x) = 3 \)
अतः \( \lim_{x \to 0} f(x) = 3 \).

(ii) जब \( x \to 1 \)
जब \(x < 1\), \(f(x) = 3(x + 1)\).
\( \lim_{x \to 1^-} f(x) = \lim_{x \to 1^-} 3(x + 1) \)
\( = 3(1 + 1) = 6 \)
जब \(x > 1\), \(f(x) = 3(x + 1)\).
\( \lim_{x \to 1^+} f(x) = \lim_{x \to 1^+} 3(x + 1) \)
\( = 3(1 + 1) = 6 \)
चूँकि LHL = RHL = 6, अतः \( \lim_{x \to 1} f(x) = 6 \).

वैकल्पिक विधि (तालिकाओं का उपयोग करके):
\( \lim_{x \to 1^-} f(x) \) ज्ञात करने के लिए \(f(x)\) में \(x\) का 1 के निकट और 1 से कम मान रखने पर

\( \lim_{x \to 1^-} 3(x + 1) = 6 \)
अब \(x\) का मान 1 के निकट और 1 से अधिक \(f(x)\) में रखने पर

\( \lim_{x \to 1^+} 3(x + 1) = 6 \)
\( \lim_{x \to 1} f(x) = 6 \)
वैकल्पिक विधि : \( \lim_{x \to 1} f(x) = 6 \).
अतः \( \lim_{x \to 1} f(x) = 6. \)
In simple words: For \( \lim_{x \to 0} f(x) \), evaluate the left-hand limit \((x<0 \Rightarrow 2x+3)\) and the right-hand limit \((x>0 \Rightarrow 3(x+1))\). Both are 3, so the limit exists and is 3. For \( \lim_{x \to 1} f(x) \), \(x=1\) falls into the \(x>0\) category, so \(f(x)=3(x+1)\) for both left and right limits. Both are 6, so the limit exists and is 6.

🎯 Exam Tip: For piecewise functions, evaluating the left-hand limit and right-hand limit at the point of interest is crucial. The limit exists only if both limits are equal.

 

Question 24. lim f(x) ज्ञात कीजिए, जहाँ
\( \lim_{x \to 1} f(x) \) ज्ञात कीजिए, जहाँ
ℹ️ चित्र व्याख्या (Diagram Explanation): फलन \(f(x)\) को एक piecewise function के रूप में परिभाषित किया गया है: यदि \(x \le 1\), \(f(x) = x^2-1\) यदि \(x > 1\), \(f(x) = -x^2-1\)
Answer:हल : जब \(x \to 1\)
बाएँ हाथ की सीमा (LHL):
जब \(x < 1\), \(f(x) = x^2 - 1\).
फलन में \(x\) का मान 1 से कम और 1 के निकट रखने पर,

\( \lim_{x \to 1^-} f(x) = 0 \)
दाएँ हाथ की सीमा (RHL):
जब \(x > 1\), \(f(x) = -x^2 - 1\).
फलन में \(x\) का मान 1 से अधिक और 1 के निकट रखने पर,

\( \lim_{x \to 1^+} f(x) = -2 \)
चूँकि \( \lim_{x \to 1^-} f(x) \neq \lim_{x \to 1^+} f(x) \) (अर्थात् \(0 \neq -2\))
अतः, \( \lim_{x \to 1} f(x) \) का अस्तित्व नहीं है।
समीकरण (i) और (ii) को जोड़ने पर, (This seems to be a leftover from a different problem)
\( 2b = 8 \) या \( b = 4 \)
समी (i) में \( b = 4 \) रखने पर,
\( 4 + a = 4 \) या \( a = 0 \)
अतः \( a = 0, b = 4. \)
In simple words: For \( \lim_{x \to 1} f(x) \), evaluate the left-hand limit \((x \le 1 \Rightarrow x^2-1)\) and the right-hand limit \((x > 1 \Rightarrow -x^2-1)\). The left-hand limit is \(1^2-1 = 0\). The right-hand limit is \(-(1^2)-1 = -2\). Since the left and right limits are not equal (\(0 \neq -2\)), the limit of \(f(x)\) as \(x\) approaches 1 does not exist. (Ignore the "समीकरण (i) और (ii)" part as it's unrelated to this question.)

🎯 Exam Tip: The existence of a limit at a point for a piecewise function strictly requires that the left-hand and right-hand limits at that point are equal.

 

Question 25. lim f(x) का मान प्राप्त कीजिए, जहाँ
\( \lim_{x \to 0} f(x) \) का मान प्राप्त कीजिए, जहाँ
ℹ️ चित्र व्याख्या (Diagram Explanation): फलन \(f(x)\) को एक piecewise function के रूप में परिभाषित किया गया है: यदि \(x \ne 0\), \(f(x) = \frac{|x|}{x}\) यदि \(x = 0\), \(f(x) = 0\)
Answer:हल : यदि \(x < 0\), तो \(|x| = -x\).
\( \lim_{x \to 0^-} f(x) = \lim_{x \to 0^-} \frac{|x|}{x} = \lim_{x \to 0^-} \frac{-x}{x} = -1 \)
और यदि \(x > 0\), तो \(|x| = x\).
\( \lim_{x \to 0^+} f(x) = \lim_{x \to 0^+} \frac{|x|}{x} = \lim_{x \to 0^+} \frac{x}{x} = 1 \)
चूँकि \( \lim_{x \to 0^-} f(x) \neq \lim_{x \to 0^+} f(x) \)
अतः \( x = 0 \) पर सीमा का अस्तित्व नहीं है।
In simple words: For \( \lim_{x \to 0} f(x) \), evaluate the left-hand limit and the right-hand limit. For \(x<0\), \(|x| = -x\), so \(f(x) = -x/x = -1\). For \(x>0\), \(|x| = x\), so \(f(x) = x/x = 1\). Since the left limit (\(-1\)) and right limit (\(1\)) are not equal, the limit does not exist at \(x=0\).

🎯 Exam Tip: For functions involving absolute values, always evaluate left and right limits by considering the definition of \(|x|\) (i.e., \(-x\) for \(x<0\) and \(x\) for \(x>0\)).

 

Question 26. lim f(x) ज्ञात कीजिए, जहाँ
\( \lim_{x \to 0} f(x) \) ज्ञात कीजिए, जहाँ
ℹ️ चित्र व्याख्या (Diagram Explanation): फलन \(f(x)\) को एक piecewise function के रूप में परिभाषित किया गया है: यदि \(x \ne 0\), \(f(x) = \frac{x}{|x|}\) यदि \(x = 0\), \(f(x) = 0\)
Answer:हल : यदि \(x < 0\), तो \(|x| = -x\).
\( \lim_{x \to 0^-} f(x) = \lim_{x \to 0^-} \frac{x}{|x|} = \lim_{x \to 0^-} \frac{x}{-x} = -1. \)
और यदि \(x > 0\), तो \(|x| = x\).
\( \lim_{x \to 0^+} f(x) = \lim_{x \to 0^+} \frac{x}{|x|} = \lim_{x \to 0^+} \frac{x}{x} = 1 \)
चूँकि \( \lim_{x \to 0^-} f(x) \neq \lim_{x \to 0^+} f(x) \)
अतः \( x = 0 \) पर सीमा का अस्तित्व नहीं है।
In simple words: For \( \lim_{x \to 0} f(x) \), evaluate the left-hand limit and the right-hand limit. For \(x<0\), \(|x| = -x\), so \(f(x) = x/(-x) = -1\). For \(x>0\), \(|x| = x\), so \(f(x) = x/x = 1\). Since the left limit (\(-1\)) and right limit (\(1\)) are not equal, the limit does not exist at \(x=0\).

🎯 Exam Tip: This question is identical to the previous one in its mathematical intent, highlighting the importance of correct interpretation of absolute value definitions for limits at zero.

 

Question 27. lim f(x) ज्ञात कीजिए, जहाँ f(x) = |x|- 5.
\( \lim_{x \to 5} f(x) \) ज्ञात कीजिए, जहाँ \(f(x) = |x| - 5\).
Answer: हल : \(f(x) = |x| - 5\)
यहाँ \(x \to 5\), जो कि \(x > 0\) है, इसलिए \(|x| = x\).
अतः \(f(x) = x - 5\).
बाएँ हाथ की सीमा (LHL):
\( \lim_{x \to 5^-} f(x) = \lim_{x \to 5^-} (|x| - 5) \)
माना \( x = 5 - h \), जैसे ही \( x \to 5 \), \( h \to 0 \)
\( = \lim_{h \to 0} (|5 - h| - 5) \)
चूंकि \( h \to 0 \), \( 5 - h \) धनात्मक होगा, इसलिए \( |5 - h| = 5 - h \).
\( = \lim_{h \to 0} (5 - h - 5) \)
\( = \lim_{h \to 0} (-h) = 0 \)
दाएँ हाथ की सीमा (RHL):
\( \lim_{x \to 5^+} f(x) = \lim_{x \to 5^+} (|x| - 5) \)
माना \( x = 5 + h \), जैसे ही \( x \to 5 \), \( h \to 0 \)
\( = \lim_{h \to 0} (|5 + h| - 5) \)
चूंकि \( h \to 0 \), \( 5 + h \) धनात्मक होगा, इसलिए \( |5 + h| = 5 + h \).
\( = \lim_{h \to 0} (5 + h - 5) \)
\( = \lim_{h \to 0} h = 0 \)
चूँकि \( \lim_{x \to 5^-} f(x) = \lim_{x \to 5^+} f(x) = 0 \)
अतः \( \lim_{x \to 5} f(x) = 0 \).
In simple words: For \( \lim_{x \to 5} f(x) \), since \(x\) approaches 5 (a positive value), \(|x|\) can be replaced with \(x\). So \(f(x) = x-5\). Then, direct substitution of \(x=5\) gives \(5-5=0\). Both left and right limits will be 0.

🎯 Exam Tip: When the limit point is clearly positive (or negative) and distinct from zero, the absolute value function can be directly replaced by \(x\) (or \(-x\)) before evaluating the limit.

 

Question 28. मान लीजिए f(x) =
ℹ️ चित्र व्याख्या (Diagram Explanation): फलन \(f(x)\) को एक piecewise function के रूप में परिभाषित किया गया है: यदि \(x < 1\), \(f(x) = a+bx\) यदि \(x = 1\), \(f(x) = 4\) यदि \(x > 1\), \(f(x) = b-ax\)
और यदि \( \lim_{x \to 1} f(x) = f(1) \), तो \(a\) और \(b\) के संभव मान क्या हैं?
Answer:हल : यदि \( \lim_{x \to 1} f(x) = f(1) \) है, तो सीमा का अस्तित्व होना चाहिए और \(f(1)\) के बराबर होना चाहिए।
हमें दिया गया है \(f(1) = 4\).
बाएँ हाथ की सीमा (LHL):
जब \(x < 1\), \(f(x) = a + bx\).
\( \lim_{x \to 1^-} f(x) = \lim_{x \to 1^-} (a + bx) = a + b(1) = a + b \)
दाएँ हाथ की सीमा (RHL):
जब \(x > 1\), \(f(x) = b - ax\).
\( \lim_{x \to 1^+} f(x) = \lim_{x \to 1^+} (b - ax) = b - a(1) = b - a \)
सीमा के अस्तित्व के लिए, LHL = RHL होना चाहिए।
\( a + b = b - a \)
\( a = -a \)
\( 2a = 0 \implies a = 0 \)
अब, चूँकि \( \lim_{x \to 1} f(x) = f(1) \), तो LHL = \(f(1)\) या RHL = \(f(1)\).
\( a + b = 4 \)
\( 0 + b = 4 \)
\( b = 4 \)
अतः \(a=0\) और \(b=4\) के लिए \( \lim_{x \to 1} f(x) = f(1) \) होगा।

वैकल्पिक विधि (तालिकाओं का उपयोग करके):
बाएँ पक्ष की सीमा ज्ञात हेतु, \(x\) का मान 1 से कम और 1 के निकट \(f(x)\) में रखने पर,

\( \lim_{x \to 1^-} f(x) = a + b \)
दाएं पक्ष की सीमा ज्ञात करने के लिए, \(f(x) = b - ax\), इसमें 1 से अधिक और 1 के निकट, \(x\) का मान रखने पर

\( \lim_{x \to 1^+} f(x) = b - a \)
यदि \(x = 1\) पर सीमा का अस्तित्व है तो
\( \lim_{x \to 1} f(x) = a + b = b + a = f(1) = 4 \)
\( b + a = 4 \) ...(i)
\( \lim_{x \to 1^+} f(x) = b - a = f(1) = 4 \)
\( b - a = 4 \) ...(ii)
समी (i) और (ii) से,
\( (b+a) + (b-a) = 4+4 \)
\( 2b = 8 \implies b = 4 \)
\( b+a = 4 \implies 4+a = 4 \implies a = 0 \)
(The line "अतः \(x = 1\) पर सीमा का अस्तित्व नहीं है।" contradicts the entire derivation and should be ignored, as the question asks to find \(a\) and \(b\) such that the limit exists and equals \(f(1)\). My interpretation clarifies this.)
In simple words: For the limit to exist and be equal to \(f(1)\), the left-hand limit \((a+bx)\), the right-hand limit \((b-ax)\), and \(f(1)\) must all be equal to 4. Setting \(a+b=4\) and \(b-a=4\), we solve these simultaneous equations to find \(a=0\) and \(b=4\).

🎯 Exam Tip: For problems involving piecewise functions and continuity (where the limit equals the function value), set the left-hand limit, right-hand limit, and function value equal to each other to form equations and solve for unknown constants.

 

Question 29. मान लीजिए \(a_1, a_2, \ldots, a_n\) अचर वास्तविक संख्याएँ हैं और एक फलन \(f(x) = (x - a_1)(x - a_2)\ldots(x - a_n)\) से परिभाषित है। \( \lim_{x \to a_1} f(x) \) क्या है? किसी \(a \neq a_1, a_2, \ldots, a_n\) के लिए \( \lim_{x \to a} f(x) \) का परिकलन कीजिए।
Answer:हल :
(i) \( \lim_{x \to a_1} f(x) \) के लिए:
हमें दिया गया है \( f(x) = (x - a_1)(x - a_2)\ldots(x - a_n) \).
\( \lim_{x \to a_1} f(x) = \lim_{x \to a_1} [(x - a_1)(x - a_2)\ldots(x - a_n)] \)
चूँकि यह एक बहुपद फलन है, हम सीधे प्रतिस्थापन कर सकते हैं।
\( = (a_1 - a_1)(a_1 - a_2)\ldots(a_1 - a_n) \)
\( = 0 \times (a_1 - a_2)(a_1 - a_3)\ldots(a_1 - a_n) \)
\( = 0 \)

(ii) किसी \( a \neq a_1, a_2, \ldots, a_n \) के लिए \( \lim_{x \to a} f(x) \) का परिकलन:
\( \lim_{x \to a} f(x) = \lim_{x \to a} [(x - a_1)(x - a_2)\ldots(x - a_n)] \)
चूँकि \(a\) दिए गए \(a_i\) में से कोई भी नहीं है, और \(f(x)\) एक बहुपद फलन है, हम सीधे \(x=a\) को प्रतिस्थापित कर सकते हैं।
\( = (a - a_1)(a - a_2)\ldots(a - a_n) \)
In simple words: For \( \lim_{x \to a_1} f(x) \), substitute \(x=a_1\) into the polynomial. Since \((x-a_1)\) is a factor, the expression becomes \((a_1-a_1)\) multiplied by other terms, resulting in 0. For \( \lim_{x \to a} f(x) \) where \(a\) is not any of the \(a_i\), directly substitute \(x=a\) into the polynomial expression, which yields \((a-a_1)(a-a_2)\ldots(a-a_n)\).

🎯 Exam Tip: For polynomial functions, limits at any point can be found by direct substitution. If the point is a root of a factor, the limit will be zero for that factor.

 

Question 30. यदि f(x) = \( \begin{cases} |x|+1, & x<0 \\ 0, & x=0 \\ |x|-1, & x>0 \end{cases} \) तो a के किन मानों के लिए \( \lim_{x \to a} f(x) \) का अस्तित्व है?
Answer: हल : दिया गया फलन :
\( f(x) = \begin{cases} |x|+1, & x<0 \\ 0, & x=0 \\ |x|-1, & x>0 \end{cases} \)
जब \( x<0 \) तब \( |x| = -x \). अत: \( f(x) = -x+1 = 1-x \).
जब \( x>0 \) तब \( |x| = x \). अत: \( f(x) = x-1 \).

(i) x = 0 पर,
बायां हाथ सीमा: \( \lim_{x \to 0^-} f(x) = \lim_{x \to 0^-} (1-x) = 1-0 = 1 \)
दायां हाथ सीमा: \( \lim_{x \to 0^+} f(x) = \lim_{x \to 0^+} (x-1) = 0-1 = -1 \)
\( \implies \lim_{x \to 0^-} f(x) \neq \lim_{x \to 0^+} f(x) \)
अतः x = 0 पर \( \lim_{x \to 0} f(x) \) का अस्तित्व नहीं है।
(ii) जब a < 0
बायां हाथ सीमा: \( \lim_{x \to a^-} f(x) = \lim_{x \to a^-} (1-x) = 1-a \)
दायां हाथ सीमा: \( \lim_{x \to a^+} f(x) = \lim_{x \to a^+} (1-x) = 1-a \)

\( \implies \lim_{x \to a^-} f(x) = \lim_{x \to a^+} f(x) \)
अर्थात \( \lim_{x \to a} f(x) = 1-a \). अतः \( a < 0 \) के लिए सीमा का अस्तित्व है।
(iii) जब a > 0
बायां हाथ सीमा: \( \lim_{x \to a^-} f(x) = \lim_{x \to a^-} (x-1) = a-1 \)
दायां हाथ सीमा: \( \lim_{x \to a^+} f(x) = \lim_{x \to a^+} (x-1) = a-1 \)

\( \implies \lim_{x \to a^-} f(x) = \lim_{x \to a^+} f(x) \)
अतः \( \lim_{x \to a} f(x) = a-1 \). अतः \( a > 0 \) के लिए सीमा का अस्तित्व है।
अतः सभी a, \( a \neq 0 \) के लिए \( \lim_{x \to a} f(x) \) का अस्तित्व है।
In simple words: The limit of a piecewise function exists if the left-hand limit equals the right-hand limit at a specific point. For \( x=0 \), the left and right limits are different, so the limit doesn't exist. For any other value of \( a \) (i.e., \( a \neq 0 \)), the function is continuous, and thus the limit exists.

🎯 Exam Tip: When evaluating limits of piecewise functions, always check the left-hand and right-hand limits at the points where the function definition changes. For other points, apply the standard limit rules based on the specific function definition for that interval.

 

Question 31. यदि फलन f(x), \( \lim_{x \to 1} \frac{f(x)-2}{x^2-1} = \pi \) को संतुष्ट करता है, तो \( \lim_{x \to 1} f(x) \) का मान ज्ञात कीजिए।
Answer: हल : जैसे ही \( x \to 1 \), फलन \( \frac{f(x)-2}{x^2-1} \to \pi \) (दिया है)
चूंकि हर \( x^2-1 \to 0 \) जैसे ही \( x \to 1 \), और सीमा एक परिमित मान \( \pi \) है, तो अंश को भी शून्य होना चाहिए।
\( \implies \) जैसे ही \( x \to 1 \), \( f(x)-2 \to 0 \)
\( \implies \lim_{x \to 1} (f(x)-2) = 0 \)
\( \implies \lim_{x \to 1} f(x) - \lim_{x \to 1} 2 = 0 \)
\( \implies \lim_{x \to 1} f(x) - 2 = 0 \)
\( \implies \lim_{x \to 1} f(x) = 2 \).
In simple words: For a fraction's limit to be a finite non-zero number when the denominator approaches zero, the numerator must also approach zero. By setting the numerator's limit to zero, we can find the value of \( \lim_{x \to 1} f(x) \).

🎯 Exam Tip: This question tests a fundamental property of limits: if \( \lim_{x \to a} \frac{g(x)}{h(x)} = L \) (where L is a finite non-zero number) and \( \lim_{x \to a} h(x) = 0 \), then it must be that \( \lim_{x \to a} g(x) = 0 \).

 

Question 32. किन पूर्णांकों m और n के लिए \( \lim_{x \to 0} f(x) \) और, \( \lim_{x \to 1} f(x) \) दोनों का अस्तित्व है। यदि \( f(x) = \begin{cases} mx^2 + n, & x < 0 \\ nx + m, & 0 \le x \le 1 \\ nx^3+m, & x > 1 \end{cases} \)
Answer: हल:
(i) x = 0 पर, सीमा के अस्तित्व के लिए, बायां हाथ सीमा = दायां हाथ सीमा होनी चाहिए।
बायां हाथ सीमा: \( \lim_{x \to 0^-} f(x) = \lim_{x \to 0^-} (mx^2 + n) = m(0)^2 + n = n \)
दायां हाथ सीमा: \( \lim_{x \to 0^+} f(x) = \lim_{x \to 0^+} (nx + m) = n(0) + m = m \)
\( \implies \) सीमा के अस्तित्व के लिए, \( m=n \) होना चाहिए।
(ii) x = 1 पर, सीमा के अस्तित्व के लिए, बायां हाथ सीमा = दायां हाथ सीमा होनी चाहिए।
बायां हाथ सीमा: \( \lim_{x \to 1^-} f(x) = \lim_{x \to 1^-} (nx + m) = n(1) + m = n+m \)
दायां हाथ सीमा: \( \lim_{x \to 1^+} f(x) = \lim_{x \to 1^+} (nx^3 + m) = n(1)^3 + m = n+m \)
चूंकि बायां हाथ सीमा और दायां हाथ सीमा दोनों \( n+m \) के बराबर हैं, तो \( \lim_{x \to 1} f(x) \) हमेशा अस्तित्व में रहता है, बशर्ते \( n \) और \( m \) कोई भी पूर्णांक हों।

अतः \( \lim_{x \to 0} f(x) \) और \( \lim_{x \to 1} f(x) \) के अस्तित्व हेतु, \( m = n \) अनिवार्य रूप से होना चाहिए; \( m \) तथा \( n \) के किसी भी पूर्णांक मान के लिए \( \lim_{x \to 1} f(x) \) का अस्तित्व है।
In simple words: For a limit to exist at a point, the function's value approached from the left must be equal to the value approached from the right. By applying this condition at \( x=0 \) and \( x=1 \), we find that \( m \) must be equal to \( n \) for both limits to exist simultaneously.

🎯 Exam Tip: Always analyze the continuity points for piecewise functions. For limits to exist at transition points, the sub-function limits must match. At other points, just evaluate the relevant sub-function.

 

Exercise 13(B)

Avkalan Class 11 प्रश्न 1. x = 10 पर \( x^2 – 2 \) का अवकलज ज्ञात कीजिए।
Answer: हल : \( x = a \) पर \( f(x) \) का अवकलज
\( f'(a) = \lim_{h \to 0} \frac{f(a+h)-f(a)}{h} \)

\( x = 10 \) पर \( f(x) = x^2 - 2 \) का अवकलज:
\( f'(10) = \lim_{h \to 0} \frac{[(10+h)^2-2]-(10^2-2)}{h} \)
\( = \lim_{h \to 0} \frac{(100+20h+h^2-2)-(100-2)}{h} \)
\( = \lim_{h \to 0} \frac{100+20h+h^2-2-100+2}{h} \)
\( = \lim_{h \to 0} \frac{20h+h^2}{h} \)
\( = \lim_{h \to 0} (20+h) \)
\( = 20+0 = 20 \).
In simple words: Using the first principle of derivatives, we substitute \( x \) with \( 10+h \) and \( 10 \) in the function \( x^2-2 \), then simplify the expression and take the limit as \( h \) approaches zero to find the derivative at \( x=10 \).

🎯 Exam Tip: Master the first principle definition of a derivative. Expanding terms like \( (a+h)^n \) correctly and simplifying the numerator to factor out \( h \) are crucial steps for accuracy.

 

सीमा और अवकलज कक्षा 11 प्रश्न 2. x = 100 पर 99x का अवकलज ज्ञात कीजिए।
Answer: हल :
\( f(x) = 99x \)
\( f'(a) = \lim_{h \to 0} \frac{f(a+h)-f(a)}{h} \)
\( f'(100) = \lim_{h \to 0} \frac{99(100+h) - 99 \times 100}{h} \)
\( = \lim_{h \to 0} \frac{9900+99h-9900}{h} \)
\( = \lim_{h \to 0} \frac{99h}{h} \)
\( = 99 \).
In simple words: The derivative of \( 99x \) at \( x=100 \) is found using the first principle, which involves calculating the limit of the difference quotient. After simplification, the limit as \( h \) approaches zero yields \( 99 \).

🎯 Exam Tip: For linear functions like \( f(x) = kx \), the derivative is simply the constant \( k \). Applying the first principle here confirms this general rule. Always simplify algebraic expressions before taking the limit.

 

सीमा और डेरिवेटिव वर्ग 11 प्रश्न 3. x = 1 पर x का अवकलज ज्ञात कीजिए ।
Answer: हल :
\( f(x) = x \)
\( f'(a) = \lim_{h \to 0} \frac{f(a+h)-f(a)}{h} \)
\( f'(1) = \lim_{h \to 0} \frac{(1+h)-1}{h} \)
\( = \lim_{h \to 0} \frac{h}{h} \)
\( = \lim_{h \to 0} 1 = 1 \).
In simple words: The derivative of \( x \) at \( x=1 \) is calculated using the first principle. After substituting \( f(1+h) \) and \( f(1) \), the expression simplifies to \( h/h \), which is \( 1 \), yielding the derivative as \( 1 \).

🎯 Exam Tip: The derivative of \( x \) with respect to \( x \) is \( 1 \). This result is consistent whether found by the first principle or standard differentiation rules. It's a foundational derivative to remember.

 

सीमा और अवकलज प्रश्न 4. प्रथम सिद्धांत से निम्नलिखित फलनों का अवकलज ज्ञात कीजिए:
(i) \( x^3 - 27 \).
Answer: हल : दिया है.
माना \( f(x) = x^3 - 27 \)
तब \( f(x+h) = (x+h)^3 - 27 \)
\( = (x^3 + 3x^2h + 3xh^2 + h^3) - 27 \)
अब, \( f(x+h) - f(x) = (x^3 + 3x^2h + 3xh^2 + h^3 - 27) - (x^3 - 27) \)
\( = 3x^2h + 3xh^2 + h^3 \)
\( = h(3x^2 + 3xh + h^2) \)
प्रथम सिद्धांत से,
\( f'(x) = \lim_{h \to 0} \frac{f(x+h)-f(x)}{h} \)
\( = \lim_{h \to 0} \frac{h(3x^2 + 3xh + h^2)}{h} \)
\( = \lim_{h \to 0} (3x^2 + 3xh + h^2) \)
\( = 3x^2 + 3x(0) + (0)^2 \)
\( = 3x^2 \).
In simple words: To find the derivative of \( x^3 - 27 \) using the first principle, we compute the difference \( f(x+h) - f(x) \), factor out \( h \), and then take the limit as \( h \) approaches zero. This process yields \( 3x^2 \).

🎯 Exam Tip: Remember the expansion of \( (a+b)^3 \) as \( a^3 + 3a^2b + 3ab^2 + b^3 \). Careful algebraic manipulation and simplification are key to accurately deriving the derivative using the first principle.

 

(ii) \( (x - 1)(x – 2) \).
Answer: हल :
माना \( f(x) = (x - 1)(x - 2) \)
\( = x^2 - 2x - x + 2 = x^2 - 3x + 2 \)
तब \( f(x+h) = (x+h)^2 - 3(x+h) + 2 \)
\( = (x^2 + 2xh + h^2) - (3x + 3h) + 2 \)
अब, \( f(x+h) - f(x) = [(x^2 + 2xh + h^2) - (3x + 3h) + 2] - [x^2 - 3x + 2] \)
\( = x^2 + 2xh + h^2 - 3x - 3h + 2 - x^2 + 3x - 2 \)
\( = 2xh + h^2 - 3h \)
\( = h(2x + h - 3) \)
प्रथम सिद्धांत से,
\( f'(x) = \lim_{h \to 0} \frac{f(x+h)-f(x)}{h} \)
\( = \lim_{h \to 0} \frac{h(2x+h-3)}{h} \)
\( = \lim_{h \to 0} (2x+h-3) \)
\( = 2x+0-3 \)
\( = 2x-3 \).
In simple words: To differentiate \( (x-1)(x-2) \) using first principles, expand the function to \( x^2-3x+2 \), find \( f(x+h)-f(x) \), simplify to factor out \( h \), and then evaluate the limit as \( h \) tends to zero.

🎯 Exam Tip: Always simplify the original function first, if possible (e.g., expand products), before applying the first principle of differentiation. This often makes the algebraic steps much clearer and less prone to error.

 

(iii) \( \frac{1}{x^2} \).
Answer: हल :
माना \( f(x) = \frac{1}{x^2} \)
तब \( f(x+h) = \frac{1}{(x+h)^2} \)
अब, \( f(x+h) - f(x) = \frac{1}{(x+h)^2} - \frac{1}{x^2} \)
\( = \frac{x^2 - (x+h)^2}{x^2(x+h)^2} \)
\( = \frac{x^2 - (x^2+2xh+h^2)}{x^2(x+h)^2} \)
\( = \frac{x^2 - x^2 - 2xh - h^2}{x^2(x+h)^2} \)
\( = \frac{-2xh - h^2}{x^2(x+h)^2} \)
\( = \frac{-h(2x + h)}{x^2(x+h)^2} \)
प्रथम सिद्धांत से,
\( f'(x) = \lim_{h \to 0} \frac{f(x+h)-f(x)}{h} \)
\( = \lim_{h \to 0} \frac{\frac{-h(2x + h)}{x^2(x+h)^2}}{h} \)
\( = \lim_{h \to 0} \frac{-(2x+h)}{x^2(x+h)^2} \)
\( = \frac{-(2x+0)}{x^2(x+0)^2} \)
\( = \frac{-2x}{x^2 \cdot x^2} = \frac{-2x}{x^4} = -\frac{2}{x^3} \).
In simple words: To find the derivative of \( 1/x^2 \) using the first principle, we calculate \( f(x+h)-f(x) \), combine the fractions, factor out \( h \) from the numerator, and then evaluate the limit as \( h \) approaches zero.

🎯 Exam Tip: When dealing with rational functions in first principle differentiation, finding a common denominator and carefully expanding and simplifying the numerator are essential steps before factoring out \( h \).

 

(iv) \( \frac{x+1}{x-1} \).
Answer: हल :
माना \( f(x) = \frac{x+1}{x-1} \)
तब \( f(x+h) = \frac{x+h+1}{x+h-1} \)
अब, \( f(x+h) - f(x) = \frac{x+h+1}{x+h-1} - \frac{x+1}{x-1} \)
\( = \frac{((x+1)+h)(x-1) - (x+1)((x-1)+h)}{(x+h-1)(x-1)} \)
\( = \frac{(x+1)(x-1) + h(x-1) - (x+1)(x-1) - h(x+1)}{(x+h-1)(x-1)} \)
\( = \frac{h(x-1) - h(x+1)}{(x+h-1)(x-1)} \)
\( = \frac{h(x-1-x-1)}{(x+h-1)(x-1)} \)
\( = \frac{h(-2)}{(x+h-1)(x-1)} \)
प्रथम सिद्धांत से,
\( f'(x) = \lim_{h \to 0} \frac{f(x+h)-f(x)}{h} \)
\( = \lim_{h \to 0} \frac{\frac{-2h}{(x+h-1)(x-1)}}{h} \)
\( = \lim_{h \to 0} \frac{-2}{(x+h-1)(x-1)} \)
\( = \frac{-2}{(x+0-1)(x-1)} \)
\( = \frac{-2}{(x-1)(x-1)} = \frac{-2}{(x-1)^2} \).
In simple words: To find the derivative of \( \frac{x+1}{x-1} \) by the first principle, we subtract \( f(x) \) from \( f(x+h) \), simplify the numerator by finding a common denominator, factor out \( h \), and then evaluate the limit as \( h \) approaches zero.

🎯 Exam Tip: For rational functions, pay close attention to algebraic simplification of the numerator after finding a common denominator. Mistakes in combining terms or factoring are common pitfalls.

 

Question 5. फलन \( f(x) = \frac{x^{100}}{100} + \frac{x^{99}}{99} +.....+\frac{x^2}{2} + x + 1 \) के लिए सिद्ध कीजिए कि \( f '(1) = 100 f'(0) \).
Answer: हल : हम जानते हैं कि \( \frac{d}{dx} x^n = nx^{n-1} \).
दिए गए फलन \( f(x) = \frac{x^{100}}{100} + \frac{x^{99}}{99} +.....+\frac{x^2}{2} + x + 1 \)
इसका अवकलन करने पर,
\( f'(x) = \frac{1}{100} (100x^{99}) + \frac{1}{99} (99x^{98}) +.....+ \frac{1}{2} (2x) + 1 + 0 \)
\( = x^{99} + x^{98} +....+ x + 1 \).

x = 1 पर, \( f'(1) = 1^{99} + 1^{98} +....+ 1 + 1 \)
\( = 1 + 1 +....+ 1 \) (100 बार)
\( = 100 \).

x = 0 पर, \( f'(0) = (0)^{99} + (0)^{98} +....+ (0) + 1 \)
\( = 0 + 0 +....+ 0 + 1 \)
\( = 1 \).

बायाँ पक्ष: \( f'(1) = 100 \).
दायाँ पक्ष: \( 100 f'(0) = 100 \times 1 = 100 \).

अत : बायाँ पक्ष = दायाँ पक्ष। यह सिद्ध हुआ।
In simple words: We first find the derivative of the given polynomial function term by term. Then, we evaluate this derivative at \( x=1 \) and \( x=0 \). Finally, we verify that the value of \( f'(1) \) is indeed 100 times the value of \( f'(0) \).

🎯 Exam Tip: When differentiating a sum of terms, differentiate each term separately. Also, recognize that \( \frac{x^n}{n} \) differentiates to \( x^{n-1} \), leading to a simplified derivative for this type of series.

 

Question 6. किसी अचर वास्तविक संख्या a के लिए \( x^n + ax^{n-1} + a^2x^{n-2} +....+ a^{n-1}x + a^n \) का अवकलज ज्ञात कीजिएं।
Answer: हल : हम जानते हैं कि \( \frac{d}{dx} [f(x)] = f' (x) \).
और \( \frac{d}{dx} x^n = nx^{n-1} \).

माना \( f(x) = x^n + ax^{n-1} + a^2x^{n-2} + ..... + a^{n-1}x + a^n \).
इसका अवकलन करने पर,
\( f'(x) = nx^{n-1} + a .(n - 1)x^{n-2} + a^2 (n-2)x^{n-3} +.....+ a^{n-1} .1 + 0 \)
\( = nx^{n-1} + (n - 1)ax^{n-2} + (n-2) a^2x^{n-3} + .....+ a^{n-1} \).
In simple words: To find the derivative of this polynomial, we apply the power rule \( \frac{d}{dx}(x^k) = kx^{k-1} \) to each term. The constants \( a, a^2, \ldots, a^{n-1} \) are treated as coefficients, and the derivative of the constant term \( a^n \) is zero.

🎯 Exam Tip: Treat constant coefficients carefully when differentiating. The power rule is fundamental here. Remember that the derivative of a constant term is always zero.

 

Seema Aur Avkalan प्रश्न 7. किन्हीं अचरों a और b के लिए

 

(i) \( (x - a) (x - b) \) का अवकलज ज्ञात कीजिए।
Answer: हल : माना \( f(x) = (x-a)(x-b) \)
सबसे पहले इसे विस्तारित करें: \( f(x) = x^2 - bx - ax + ab = x^2 - (a+b)x + ab \).
इसका अवकलन करने पर,
\( f'(x) = \frac{d}{dx}(x^2) - \frac{d}{dx}((a+b)x) + \frac{d}{dx}(ab) \)
\( = 2x^{2-1} - (a+b) \cdot 1 + 0 \)
\( = 2x - (a+b) \).
In simple words: We first expand the given product into a polynomial form, then differentiate each term using the power rule. Constants like \( a \) and \( b \) are treated as coefficients or parts of constant terms.

🎯 Exam Tip: For products of simple linear terms, expanding first and then differentiating term-by-term is often easier than using the product rule. Always simplify the function before differentiating if it leads to a simpler form.

 

(ii) \( (ax^2 + b)^2 \) का अवकलज ज्ञात कीजिए।
Answer: हल : माना \( f(x) = (ax^2 + b)^2 \)
सबसे पहले इसे विस्तारित करें: \( f(x) = a^2x^4 + 2abx^2 + b^2 \).
इसका अवकलन करने पर,
\( f'(x) = \frac{d}{dx}(a^2x^4) + \frac{d}{dx}(2abx^2) + \frac{d}{dx}(b^2) \)
\( = a^2 \cdot 4x^{4-1} + 2ab \cdot 2x^{2-1} + 0 \)
\( = 4a^2x^3 + 4abx \)
\( = 4ax(ax^2 + b) \).
In simple words: We expand the squared term into a polynomial, then apply the power rule to each term to find the derivative. Constant coefficients \( a^2 \) and \( 2ab \) are multiplied by the derivatives of the \( x \) terms, and the constant \( b^2 \) differentiates to zero.

🎯 Exam Tip: Recognizing that constants like \( a, b \) are just numerical values helps in applying differentiation rules. For expressions like \( (ax^2+b)^2 \), expanding before differentiating is a valid approach, though the chain rule would also be applicable.

 

(iii) \( \frac{x-a}{x-b} \) का अवकलज ज्ञात कीजिए।
Answer: हल : माना \( f(x) = \frac{x-a}{x-b} \).
यह \( \frac{u}{v} \) के रूप का है, जहाँ \( u = x-a \) और \( v = x-b \).
हम जानते हैं कि \( \frac{d}{dx} \left( \frac{u}{v} \right) = \frac{u'v - uv'}{v^2} \).
यहां \( u' = \frac{d}{dx}(x-a) = 1 \) और \( v' = \frac{d}{dx}(x-b) = 1 \).
अतः,
\( f'(x) = \frac{(1)(x-b) - (x-a)(1)}{(x-b)^2} \)
\( = \frac{x-b-x+a}{(x-b)^2} \)
\( = \frac{a-b}{(x-b)^2} \).
In simple words: To find the derivative of a rational function like \( \frac{x-a}{x-b} \), we use the quotient rule: \( \frac{(derivative\ of\ numerator \times denominator) - (numerator \times derivative\ of\ denominator)}{(denominator)^2} \).

🎯 Exam Tip: The quotient rule is essential for differentiating rational functions. Clearly identify \( u \) and \( v \), find their derivatives, and carefully substitute them into the formula. Pay attention to signs during subtraction in the numerator.

 

Question 8. किसी अचर a के लिए \( \frac{x^n-a^n}{x-a} \) का अवकलज ज्ञात कीजिए।
Answer: हल : माना \( f(x) = \frac{x^n-a^n}{x-a} \).
यह \( \frac{u}{v} \) के रूप का है, जहाँ \( u = x^n-a^n \) और \( v = x-a \).
हम जानते हैं कि \( \frac{d}{dx} \left( \frac{u}{v} \right) = \frac{u'v - uv'}{v^2} \).
यहां \( u' = \frac{d}{dx}(x^n-a^n) = nx^{n-1} \) (क्योंकि \( a^n \) एक अचर है, उसका अवकलज 0 है) और \( v' = \frac{d}{dx}(x-a) = 1 \).
अतः,
\( f'(x) = \frac{(nx^{n-1})(x-a) - (x^n-a^n)(1)}{(x-a)^2} \)
\( = \frac{nx^{n-1} \cdot x - nx^{n-1} \cdot a - x^n + a^n}{(x-a)^2} \)
\( = \frac{nx^n - nax^{n-1} - x^n + a^n}{(x-a)^2} \)
\( = \frac{(n-1)x^n - nax^{n-1} + a^n}{(x-a)^2} \).
In simple words: To find the derivative of \( \frac{x^n-a^n}{x-a} \), we use the quotient rule. We differentiate the numerator and denominator separately, then substitute these into the quotient rule formula and simplify the expression.

🎯 Exam Tip: When \( a \) is a constant, its derivative is zero. Be careful with algebraic simplification in the numerator, especially when combining terms with powers of \( x \).

 

Limits And Derivatives Class 11 प्रश्न 9. निम्नलिखित के अवकलज ज्ञात कीजिए:

 

(i) \( 2x - \frac{3}{4} \).
Answer: हल :
माना \( f(x) = 2x - \frac{3}{4} \).
इसका अवकलन करने पर,
\( f'(x) = \frac{d}{dx}(2x) - \frac{d}{dx}\left(\frac{3}{4}\right) \)
\( = 2 \cdot 1 - 0 \)
\( = 2 \).
In simple words: The derivative of \( 2x - \frac{3}{4} \) is found by differentiating each term. The derivative of \( 2x \) is \( 2 \), and the derivative of the constant \( \frac{3}{4} \) is \( 0 \).

🎯 Exam Tip: Remember that the derivative of \( cx \) is \( c \) and the derivative of any constant is \( 0 \). Apply these basic rules when differentiating linear expressions.

 

(ii) \( (5x^3 + 3x - 1) (x - 1) \).
Answer: हल :
माना \( f(x) = (5x^3 + 3x - 1) (x - 1) \).
यह \( u \cdot v \) के रूप का है, जहाँ \( u = 5x^3 + 3x - 1 \) और \( v = x - 1 \).
हम जानते हैं कि \( \frac{d}{dx} (uv) = u'v + uv' \).
यहां \( u' = \frac{d}{dx}(5x^3 + 3x - 1) = 15x^2 + 3 \) और \( v' = \frac{d}{dx}(x - 1) = 1 \).
अतः,
\( f'(x) = (15x^2 + 3)(x - 1) + (5x^3 + 3x - 1)(1) \)
\( = (15x^3 - 15x^2 + 3x - 3) + (5x^3 + 3x - 1) \)
\( = 15x^3 - 15x^2 + 3x - 3 + 5x^3 + 3x - 1 \)
\( = (15x^3 + 5x^3) - 15x^2 + (3x + 3x) - (3 + 1) \)
\( = 20x^3 - 15x^2 + 6x - 4 \).
In simple words: To differentiate the product of two functions, we apply the product rule: \( (derivative\ of\ first \times second) + (first \times derivative\ of\ second) \). Then, simplify the resulting polynomial.

🎯 Exam Tip: The product rule is crucial for differentiating functions that are products of two or more expressions. Make sure to correctly find the derivatives of both \( u \) and \( v \) before applying the formula and simplifying.

 

(iii) \( x^{-3}(5 + 3x) \).
Answer: हल :
माना \( f(x) = x^{-3}(5 + 3x) \).
सबसे पहले इसे विस्तारित करें: \( f(x) = 5x^{-3} + 3x^{-3}x^1 = 5x^{-3} + 3x^{-2} \).
अवकलन करने पर,
\( f'(x) = \frac{d}{dx}(5x^{-3}) + \frac{d}{dx}(3x^{-2}) \)
\( = 5(-3)x^{-3-1} + 3(-2)x^{-2-1} \)
\( = -15x^{-4} - 6x^{-3} \)
\( = -\frac{15}{x^4} - \frac{6}{x^3} \)
\( = -\frac{15}{x^4} - \frac{6x}{x^4} = -\frac{15+6x}{x^4} \).
In simple words: Expand the function first by multiplying \( x^{-3} \) into the bracket, simplifying the terms. Then, differentiate each term using the power rule \( \frac{d}{dx}(x^n) = nx^{n-1} \) and combine the results.

🎯 Exam Tip: When dealing with negative exponents, be careful with sign changes during differentiation and common denominator simplification. Expanding the function simplifies it into a sum of terms, making differentiation easier.

 

(iv) \( x^5(3 - 6x^{-9}) \).
Answer: हल :
माना \( f(x) = x^5(3 - 6x^{-9}) \).
सबसे पहले इसे विस्तारित करें: \( f(x) = 3x^5 - 6x^5x^{-9} = 3x^5 - 6x^{-4} \).
अवकलन करने पर,
\( f'(x) = \frac{d}{dx}(3x^5) - \frac{d}{dx}(6x^{-4}) \)
\( = 3(5)x^{5-1} - 6(-4)x^{-4-1} \)
\( = 15x^4 + 24x^{-5} \)
\( = 15x^4 + \frac{24}{x^5} \).
In simple words: First, distribute \( x^5 \) across the terms inside the parenthesis and simplify the exponents. Then, differentiate each resulting term using the power rule, paying attention to signs for negative exponents.

🎯 Exam Tip: Always simplify the expression before differentiating. Distributing terms with exponents means adding the exponents (\( x^a \cdot x^b = x^{a+b} \)). This converts complex products into simpler sums for differentiation.

 

(v) \( x^4 (3 – 4x^{-5}) \).
Answer: हल :
माना \( f(x) = x^4 (3 - 4x^{-5}) \).
सबसे पहले इसे विस्तारित करें: \( f(x) = 3x^4 - 4x^4x^{-5} = 3x^4 - 4x^{-1} \).
अवकलन करने पर,
\( f'(x) = \frac{d}{dx}(3x^4) - \frac{d}{dx}(4x^{-1}) \)
\( = 3(4)x^{4-1} - 4(-1)x^{-1-1} \)
\( = 12x^3 + 4x^{-2} \)
\( = 12x^3 + \frac{4}{x^2} \).
In simple words: Distribute \( x^4 \) into the parenthesis to simplify the function into a polynomial form with simpler exponents. Then, differentiate each term using the power rule, correctly handling the negative exponent.

🎯 Exam Tip: Remember that \( x^a \cdot x^b = x^{a+b} \). Simplifying the function to a sum of terms before differentiating often prevents errors. Pay careful attention to the signs when differentiating terms with negative exponents.

 

(vi) \( \frac{2}{x+1} - \frac{x^2}{3x-1} \).
Answer: हल :
माना \( f(x) = \frac{2}{x+1} - \frac{x^2}{3x-1} \).
यह \( \frac{u}{v} \) के रूप का है, इसलिए हम भागफल नियम \( \frac{d}{dx} \left( \frac{u}{v} \right) = \frac{u'v - uv'}{v^2} \) का उपयोग करेंगे।

पहले पद \( \frac{2}{x+1} \) के लिए:
\( u=2, v=x+1 \implies u'=0, v'=1 \).
\( \frac{d}{dx}\left(\frac{2}{x+1}\right) = \frac{0(x+1) - 2(1)}{(x+1)^2} = \frac{-2}{(x+1)^2} \).

दूसरे पद \( \frac{x^2}{3x-1} \) के लिए:
\( u=x^2, v=3x-1 \implies u'=2x, v'=3 \).
\( \frac{d}{dx}\left(\frac{x^2}{3x-1}\right) = \frac{2x(3x-1) - x^2(3)}{(3x-1)^2} \)
\( = \frac{6x^2-2x-3x^2}{(3x-1)^2} = \frac{3x^2-2x}{(3x-1)^2} \).

अतः, \( f'(x) = \frac{d}{dx}\left(\frac{2}{x+1}\right) - \frac{d}{dx}\left(\frac{x^2}{3x-1}\right) \)
\( = \frac{-2}{(x+1)^2} - \frac{3x^2-2x}{(3x-1)^2} \).
In simple words: We differentiate each rational term separately using the quotient rule, then combine the results by subtracting the derivative of the second term from the derivative of the first term.

🎯 Exam Tip: When differentiating a sum or difference of rational functions, apply the quotient rule to each term independently. Be careful with the minus sign between the terms when combining the final derivatives.

 

Seema Aur Avkalaj प्रश्न 10. प्रथम सिद्धांत से cos x का अवकलज ज्ञात कीजिए।
Answer: हल :
माना \( f(x) = \cos x \).
तब \( f(x+h) = \cos(x+h) \).
प्रथम सिद्धांत से,
\( f'(x) = \lim_{h \to 0} \frac{f(x+h)-f(x)}{h} \)
\( = \lim_{h \to 0} \frac{\cos(x+h) - \cos x}{h} \)
हम जानते हैं कि \( \cos C - \cos D = -2\sin\left(\frac{C+D}{2}\right)\sin\left(\frac{C-D}{2}\right) \).
\( = \lim_{h \to 0} \frac{-2\sin\left(\frac{x+h+x}{2}\right)\sin\left(\frac{x+h-x}{2}\right)}{h} \)
\( = \lim_{h \to 0} \frac{-2\sin\left(x+\frac{h}{2}\right)\sin\left(\frac{h}{2}\right)}{h} \)
\( = \lim_{h \to 0} \left[ -\sin\left(x+\frac{h}{2}\right) \cdot \frac{\sin\left(\frac{h}{2}\right)}{\frac{h}{2}} \right] \)
\( = -\lim_{h \to 0} \sin\left(x+\frac{h}{2}\right) \cdot \lim_{h \to 0} \frac{\sin\left(\frac{h}{2}\right)}{\frac{h}{2}} \)
चूंकि \( \lim_{\theta \to 0} \frac{\sin\theta}{\theta} = 1 \).
\( = -\sin(x+0) \cdot 1 \)
\( = -\sin x \).
अतः \( \frac{d}{dx} \cos x = -\sin x \).
In simple words: The derivative of \( \cos x \) from the first principle involves using the trigonometric sum-to-product formula for \( \cos(x+h) - \cos x \), then manipulating the expression to use the standard limit \( \lim_{\theta \to 0} \frac{\sin\theta}{\theta} = 1 \).

🎯 Exam Tip: Memorize the trigonometric sum-to-product formulas, especially for \( \cos C - \cos D \). This is a common step in first principle differentiation for trigonometric functions. Ensure correct handling of the \( h/2 \) term in the limit evaluation.

 

Chapter 13 Class 11 Maths प्रश्न 11. निम्नलिखित फलनों के अवकलज ज्ञात कीजिए:

 

(i) \( \sin x \cos x \).
Answer: हल :
माना \( f(x) = \sin x \cos x \).
यह \( u \cdot v \) के रूप का है, जहाँ \( u = \sin x \) और \( v = \cos x \).
हम जानते हैं कि \( \frac{d}{dx} (uv) = u'v + uv' \).
यहां \( u' = \frac{d}{dx}(\sin x) = \cos x \) और \( v' = \frac{d}{dx}(\cos x) = -\sin x \).
अतः,
\( f'(x) = (\cos x)(\cos x) + (\sin x)(-\sin x) \)
\( = \cos^2 x - \sin^2 x \).
हम जानते हैं कि \( \cos^2 x - \sin^2 x = \cos 2x \).
अतः, \( f'(x) = \cos 2x \).
In simple words: To differentiate \( \sin x \cos x \), we apply the product rule: \( (derivative\ of\ \sin x \times \cos x) + (\sin x \times derivative\ of\ \cos x) \). The result can then be simplified using the double angle identity for cosine.

🎯 Exam Tip: The product rule is key for differentiating products of functions. Recognizing and applying trigonometric identities (like \( \cos^2 x - \sin^2 x = \cos 2x \)) can simplify your final answer and is often expected.

 

(ii) \( \sec x \).
Answer: हल :
माना \( f(x) = \sec x \).
तब \( f(x+h) = \sec(x+h) \).
प्रथम सिद्धांत से,
\( f'(x) = \lim_{h \to 0} \frac{f(x+h)-f(x)}{h} \)
\( = \lim_{h \to 0} \frac{\sec(x+h) - \sec x}{h} \)
\( = \lim_{h \to 0} \frac{\frac{1}{\cos(x+h)} - \frac{1}{\cos x}}{h} \)
\( = \lim_{h \to 0} \frac{\cos x - \cos(x+h)}{h \cos(x+h)\cos x} \)
हम जानते हैं कि \( \cos C - \cos D = 2\sin\left(\frac{C+D}{2}\right)\sin\left(\frac{D-C}{2}\right) \).
\( = \lim_{h \to 0} \frac{2\sin\left(\frac{x+h+x}{2}\right)\sin\left(\frac{x+h-x}{2}\right)}{h \cos(x+h)\cos x} \)
\( = \lim_{h \to 0} \frac{2\sin\left(x+\frac{h}{2}\right)\sin\left(\frac{h}{2}\right)}{h \cos(x+h)\cos x} \)
\( = \lim_{h \to 0} \left[ \frac{\sin\left(x+\frac{h}{2}\right)}{\cos(x+h)\cos x} \cdot \frac{\sin\left(\frac{h}{2}\right)}{\frac{h}{2}} \right] \)
\( = \left( \lim_{h \to 0} \frac{\sin\left(x+\frac{h}{2}\right)}{\cos(x+h)\cos x} \right) \cdot \left( \lim_{h \to 0} \frac{\sin\left(\frac{h}{2}\right)}{\frac{h}{2}} \right) \)
\( = \frac{\sin(x+0)}{\cos(x+0)\cos x} \cdot 1 \)
\( = \frac{\sin x}{\cos x \cos x} = \frac{\sin x}{\cos x} \cdot \frac{1}{\cos x} \)
\( = \tan x \sec x \).
In simple words: To find the derivative of \( \sec x \) using the first principle, convert it to \( 1/\cos x \), find a common denominator, apply the sum-to-product trigonometric identity, and then use the fundamental limit \( \lim_{\theta \to 0} \frac{\sin\theta}{\theta} = 1 \).

🎯 Exam Tip: When differentiating reciprocal trigonometric functions by the first principle, convert them to their sine/cosine forms. Careful application of sum-to-product identities and limit properties is essential for correct derivation.

 

(iii) \( 5 \sec x + 4 \cos x \).
Answer: हल :
माना \( f(x) = 5 \sec x + 4 \cos x \).
योग नियम का उपयोग करके,
\( f'(x) = \frac{d}{dx}(5 \sec x) + \frac{d}{dx}(4 \cos x) \)
\( = 5 \frac{d}{dx}(\sec x) + 4 \frac{d}{dx}(\cos x) \).
हमने पिछले भागों (ii) और प्रश्न 10 से \( \frac{d}{dx}(\sec x) = \sec x \tan x \) और \( \frac{d}{dx}(\cos x) = -\sin x \) ज्ञात किया है।
अतः,
\( f'(x) = 5 (\sec x \tan x) + 4 (-\sin x) \)
\( = 5 \sec x \tan x - 4 \sin x \).
In simple words: To differentiate a sum of functions, differentiate each term separately. The constants \( 5 \) and \( 4 \) act as coefficients. We use known derivative formulas for \( \sec x \) and \( \cos x \) to find the overall derivative.

🎯 Exam Tip: The sum/difference rule states that the derivative of a sum/difference is the sum/difference of the derivatives. Always apply scalar multiples correctly. Have the basic derivatives of trigonometric functions memorized.

 

(ii) sec x.

Answer:हल : माना \( f(x) = \sec x \) \( f(x + h) = \sec (x + h) \) \( f(x + h) - f(x) = \sec (x + h) - \sec x \)
\( = \frac{1}{\cos(x + h)} - \frac{1}{\cos x} \)
\( = \frac{\cos x - \cos(x + h)}{\cos(x + h)\cos x} \)
\( = \frac{2\sin\left(x + \frac{h}{2}\right)\sin\left(\frac{h}{2}\right)}{\cos(x + h)\cos x} \) \( f'(x) = \lim_{h \to 0} \frac{f(x + h) - f(x)}{h} \)
\( = \lim_{h \to 0} \frac{2\sin\left(x + \frac{h}{2}\right)\sin\left(\frac{h}{2}\right)}{h\cos(x + h)\cos x} \)
\( = \lim_{h \to 0} \frac{\sin\left(x + \frac{h}{2}\right)}{\cos(x + h)\cos x} \cdot \lim_{h \to 0} \frac{\sin\left(\frac{h}{2}\right)}{\frac{h}{2}} \)
\( = \frac{\sin x}{\cos x \cdot \cos x} \cdot 1 \)
\( = \frac{1}{\cos x} \cdot \frac{\sin x}{\cos x} = \sec x \tan x \)In simple words: The derivative of sec x is found using the first principle by setting up the limit of the difference quotient and simplifying it using trigonometric identities and standard limit formulas. The result is sec x tan x.

🎯 Exam Tip: Remember the trigonometric identity for \( \cos A - \cos B \) and the fundamental limit \( \lim_{\theta \to 0} \frac{\sin \theta}{\theta} = 1 \). These are crucial for first principle derivative problems.

 

(iii) 5 sec x + 4 cos x

Answer:हल : माना \( f(x) = 5\sec x + 4\cos x \) हम जानते हैं कि \( \frac{d}{dx}(\sec x) = \sec x \tan x \) और \( \frac{d}{dx}(\cos x) = -\sin x \) अतः \( f'(x) = 5\frac{d}{dx}(\sec x) + 4\frac{d}{dx}(\cos x) \)
\( = 5(\sec x \tan x) + 4(-\sin x) \)
\( = 5\sec x \tan x - 4\sin x \)In simple words: To find the derivative of a sum of functions, we take the derivative of each function separately and then add them. We use the known derivatives of sec x and cos x.

🎯 Exam Tip: Familiarize yourself with the derivatives of common trigonometric functions. Remember that the derivative of a constant times a function is the constant times the derivative of the function.

 

(iv) cosec x.

Answer:हल : माना \( f(x) = \operatorname{cosec} x \) और \( f(x + h) = \operatorname{cosec} (x + h) \) \( f(x + h) - f(x) = \operatorname{cosec} (x + h) - \operatorname{cosec} x \)
\( = \frac{1}{\sin(x + h)} - \frac{1}{\sin x} \)
\( = \frac{\sin x - \sin(x + h)}{\sin(x + h)\sin x} \)
\( = \frac{-2\cos\left(x + \frac{h}{2}\right)\sin\left(\frac{h}{2}\right)}{\sin(x + h)\sin x} \) \( f'(x) = \lim_{h \to 0} \frac{f(x + h) - f(x)}{h} \)
\( = \lim_{h \to 0} \frac{1}{h} \left[ \frac{-2\cos\left(x + \frac{h}{2}\right)\sin\left(\frac{h}{2}\right)}{\sin(x + h)\sin x} \right] \)
\( = \lim_{h \to 0} \frac{-\cos\left(x + \frac{h}{2}\right)}{\sin(x + h)\sin x} \cdot \lim_{h \to 0} \frac{\sin\left(\frac{h}{2}\right)}{\frac{h}{2}} \)
\( = \frac{-\cos x}{\sin x \cdot \sin x} \cdot 1 \)
\( = -\frac{1}{\sin x} \cdot \frac{\cos x}{\sin x} = -\operatorname{cosec} x \cot x \)In simple words: The derivative of cosec x is found using the first principle by applying the limit definition and using trigonometric sum-to-product identities. The result is -cosec x cot x.

🎯 Exam Tip: Pay close attention to negative signs that arise from trigonometric identities. Ensure the \( \frac{\sin \theta}{\theta} \) limit is applied correctly by adjusting the denominator.

 

(v) 3 cot x + 5 cosec x

Answer:हल : माना \( f(x) = 3\cot x + 5\operatorname{cosec} x \) हम जानते हैं कि \( \frac{d}{dx}(\cot x) = -\operatorname{cosec}^2 x \) और \( \frac{d}{dx}(\operatorname{cosec} x) = -\operatorname{cosec} x \cot x \) अब \( f'(x) = 3\frac{d}{dx}(\cot x) + 5\frac{d}{dx}(\operatorname{cosec} x) \)
\( = 3(-\operatorname{cosec}^2 x) + 5(-\operatorname{cosec} x \cot x) \)
\( = -3\operatorname{cosec}^2 x - 5\operatorname{cosec} x \cot x \)In simple words: The derivative of a sum of functions is the sum of their individual derivatives. We use the known derivatives of cot x and cosec x to find the total derivative.

🎯 Exam Tip: Be careful with the signs when differentiating trigonometric functions, especially for cot x and cosec x. Memorizing the basic derivative formulas is essential.

 

(vi) 5 sin x - 6 cos x + 7.

Answer:हल : माना \( f(x) = 5\sin x - 6\cos x + 7 \) \( f'(x) = 5\frac{d}{dx}(\sin x) - 6\frac{d}{dx}(\cos x) + \frac{d}{dx}(7) \)
\( = 5(\cos x) - 6(-\sin x) + 0 \)
\( = 5\cos x + 6\sin x \)In simple words: We differentiate each term in the function. The derivative of a constant (7) is zero, and we apply the known derivatives for sin x and cos x, respecting the coefficients.

🎯 Exam Tip: Remember that the derivative of any constant is zero. Also, be mindful of the negative sign in the derivative of cos x, which changes the subtraction to addition in this case.

 

(vii) 2 tan x - 7 sec x.

Answer:हल : माना \( f(x) = 2\tan x - 7\sec x \) हम जानते हैं कि \( \frac{d}{dx}(\tan x) = \sec^2 x \) और \( \frac{d}{dx}(\sec x) = \sec x \tan x \) \( f'(x) = 2\frac{d}{dx}(\tan x) - 7\frac{d}{dx}(\sec x) \)
\( = 2(\sec^2 x) - 7(\sec x \tan x) \)
\( = 2\sec^2 x - 7\sec x \tan x \)In simple words: We find the derivative of each term separately. The derivative of tan x is sec² x, and the derivative of sec x is sec x tan x. We then combine these with their respective constant multipliers.

🎯 Exam Tip: Make sure to recall the derivatives of tan x and sec x accurately. These are frequently used in calculus problems and need to be memorized for quick application.

 

Chapter 13 Maths Class 11 प्रश्न 1.
प्रथम सिद्धांत से निम्नलिखित फलनों का अवकलज ज्ञात कीजिए:
(i) -x.

Answer:हल : मान लीजिए \( f(x) = -x \) \( f(x + h) = -(x + h) = -x - h \) \( \frac{d}{dx}(-x) = \lim_{h \to 0} \frac{f(x + h) - f(x)}{h} \)
\( = \lim_{h \to 0} \frac{(-x - h) - (-x)}{h} \)
\( = \lim_{h \to 0} \frac{-x - h + x}{h} \)
\( = \lim_{h \to 0} \frac{-h}{h} \)
\( = \lim_{h \to 0} (-1) = -1 \)In simple words: We use the definition of the derivative (first principle) to find the slope of the tangent line. By plugging the function into the limit formula and simplifying, we find the derivative is -1.

🎯 Exam Tip: The first principle (limit definition of derivative) is fundamental. Practice simplifying the difference quotient \( \frac{f(x+h)-f(x)}{h} \) thoroughly before taking the limit as \( h \to 0 \).

 

(ii) (-x)-1.

Answer:हल : मान लीजिए \( f(x) = (-x)^{-1} = -\frac{1}{x} \) \( f(x + h) = -\frac{1}{x + h} \) \( f'(x) = \lim_{h \to 0} \frac{f(x + h) - f(x)}{h} \)
\( = \lim_{h \to 0} \frac{-\frac{1}{x + h} - \left(-\frac{1}{x}\right)}{h} \)
\( = \lim_{h \to 0} \frac{-\frac{1}{x + h} + \frac{1}{x}}{h} \)
\( = \lim_{h \to 0} \frac{\frac{-x + (x + h)}{x(x + h)}}{h} \)
\( = \lim_{h \to 0} \frac{\frac{h}{x(x + h)}}{h} \)
\( = \lim_{h \to 0} \frac{1}{x(x + h)} \)
\( = \frac{1}{x(x + 0)} = \frac{1}{x^2} \)In simple words: Using the first principle, we find the derivative of -1/x. After substituting into the limit definition and simplifying the complex fraction, we find the derivative to be 1/x².

🎯 Exam Tip: When dealing with rational functions in the first principle, the common denominator technique is crucial for simplifying the numerator before canceling the 'h' term. Watch out for algebraic simplification errors.

 

(iii) sin (x + 1).

Answer:हल : माना \( f(x) = \sin(x + 1) \) \( f(x + h) = \sin(x + h + 1) \) \( f'(x) = \lim_{h \to 0} \frac{f(x + h) - f(x)}{h} \)
\( = \lim_{h \to 0} \frac{\sin(x + h + 1) - \sin(x + 1)}{h} \) Using the identity \( \sin A - \sin B = 2\cos\left(\frac{A + B}{2}\right)\sin\left(\frac{A - B}{2}\right) \): Let \( A = x + h + 1 \) and \( B = x + 1 \). \( A + B = 2x + h + 2 \implies \frac{A + B}{2} = x + 1 + \frac{h}{2} \) \( A - B = h \implies \frac{A - B}{2} = \frac{h}{2} \) So, \( \sin(x + h + 1) - \sin(x + 1) = 2\cos\left(x + 1 + \frac{h}{2}\right)\sin\left(\frac{h}{2}\right) \)
\( f'(x) = \lim_{h \to 0} \frac{2\cos\left(x + 1 + \frac{h}{2}\right)\sin\left(\frac{h}{2}\right)}{h} \)
\( = \lim_{h \to 0} \left[ \cos\left(x + 1 + \frac{h}{2}\right) \cdot \frac{\sin\left(\frac{h}{2}\right)}{\frac{h}{2}} \right] \)
\( = \cos(x + 1 + 0) \cdot 1 \)
\( = \cos(x + 1) \)In simple words: The derivative of sin(x + 1) is found using the first principle by applying the sum-to-product trigonometric identity for sin A - sin B and then evaluating the limit. The result is cos(x + 1).

🎯 Exam Tip: Mastering trigonometric identities, especially sum-to-product and product-to-sum, is crucial for finding derivatives of trigonometric functions using the first principle. Remember to group terms to apply \( \lim_{\theta \to 0} \frac{\sin \theta}{\theta} = 1 \).

 

(iv) cos\( \left(x - \frac{\pi}{8}\right) \).

Answer:हल : माना \( f(x) = \cos\left(x - \frac{\pi}{8}\right) \) \( f(x + h) = \cos\left(x + h - \frac{\pi}{8}\right) \) \( f'(x) = \lim_{h \to 0} \frac{f(x + h) - f(x)}{h} \)
\( = \lim_{h \to 0} \frac{\cos\left(x + h - \frac{\pi}{8}\right) - \cos\left(x - \frac{\pi}{8}\right)}{h} \) Using the identity \( \cos A - \cos B = -2\sin\left(\frac{A + B}{2}\right)\sin\left(\frac{A - B}{2}\right) \): Let \( A = x + h - \frac{\pi}{8} \) and \( B = x - \frac{\pi}{8} \). \( A + B = 2x + h - \frac{2\pi}{8} = 2x + h - \frac{\pi}{4} \implies \frac{A + B}{2} = x - \frac{\pi}{8} + \frac{h}{2} \) \( A - B = h \implies \frac{A - B}{2} = \frac{h}{2} \) So, \( \cos\left(x + h - \frac{\pi}{8}\right) - \cos\left(x - \frac{\pi}{8}\right) = -2\sin\left(x - \frac{\pi}{8} + \frac{h}{2}\right)\sin\left(\frac{h}{2}\right) \)
\( f'(x) = \lim_{h \to 0} \frac{-2\sin\left(x - \frac{\pi}{8} + \frac{h}{2}\right)\sin\left(\frac{h}{2}\right)}{h} \)
\( = \lim_{h \to 0} \left[ -\sin\left(x - \frac{\pi}{8} + \frac{h}{2}\right) \cdot \frac{\sin\left(\frac{h}{2}\right)}{\frac{h}{2}} \right] \)
\( = -\sin\left(x - \frac{\pi}{8} + 0\right) \cdot 1 \)
\( = -\sin\left(x - \frac{\pi}{8}\right) \)In simple words: To differentiate \( \cos(x - \frac{\pi}{8}) \) using the first principle, we apply the cosine difference formula, simplify the expression, and then take the limit as h approaches zero. The derivative is \( -\sin(x - \frac{\pi}{8}) \).

🎯 Exam Tip: Ensure proper handling of the constant term (like \( -\frac{\pi}{8} \)) when using sum/difference identities. The negative sign from the \( \cos A - \cos B \) identity is critical for the correct final answer.

 

प्रश्न 2.
(x + a)

Answer:हल : \( \frac{d}{dx}(x + a) = \frac{d}{dx}(x) + \frac{d}{dx}(a) = 1 + 0 = 1. \)In simple words: The derivative of \( x + a \) is found by differentiating each term. The derivative of \( x \) is 1, and the derivative of a constant \( a \) is 0.

🎯 Exam Tip: Remember that the derivative of x with respect to x is 1, and the derivative of any constant (like 'a') is always 0.

 

प्रश्न 3.
\( (px + q)\left(\frac{r}{x} + s\right) \)

Answer:हल : माना \( f(x) = (px + q)\left(\frac{r}{x} + s\right) \) गुणनफल नियम (uv)' = u'v + uv' का उपयोग करते हुए: Let \( u = px + q \) and \( v = rx^{-1} + s \). Then \( u' = p \) and \( v' = -rx^{-2} \). \( f'(x) = p\left(\frac{r}{x} + s\right) + (px + q)\left(-\frac{r}{x^2}\right) \)
\( = \frac{pr}{x} + ps - \frac{prx}{x^2} - \frac{qr}{x^2} \)
\( = \frac{pr}{x} + ps - \frac{pr}{x} - \frac{qr}{x^2} \)
\( = ps - \frac{qr}{x^2} \)In simple words: To find the derivative of this product, we apply the product rule. We differentiate each part of the product and combine them according to the rule, then simplify the resulting expression.

🎯 Exam Tip: Always identify the 'u' and 'v' parts clearly before applying the product rule. Simplify the result carefully, especially when dealing with negative exponents.

 

प्रश्न 4.
\( (ax + b)(cx + d)^2 \)

Answer:हल : माना, \( f(x) = (ax + b)(cx + d)^2 \) गुणनफल नियम (uv)' = u'v + uv' का उपयोग करते हुए: Let \( u = ax + b \) and \( v = (cx + d)^2 \). Then \( u' = a \). And \( v' = 2(cx + d) \cdot c = 2c(cx + d) \) (using chain rule for \( (cx + d)^2 \)). \( f'(x) = \frac{d}{dx}[(ax + b)(cx + d)^2] \)
\( = a(cx + d)^2 + (ax + b)[2c(cx + d)] \)
\( = (cx + d)[a(cx + d) + 2c(ax + b)] \)
\( = (cx + d)[acx + ad + 2acx + 2bc] \)
\( = (cx + d)[3acx + ad + 2bc] \)In simple words: We apply the product rule, where one part is \( ax + b \) and the other is \( (cx + d)^2 \). For \( (cx + d)^2 \), we also use the chain rule for its derivative. Then, we combine terms and simplify.

🎯 Exam Tip: This problem requires both the product rule and the chain rule. Remember to differentiate \( (cx+d)^2 \) as \( 2(cx+d) \cdot c \). Factoring out common terms like \( (cx+d) \) in the final step can simplify the expression.

 

प्रश्न 5.
\( \frac{ax + b}{cx + d} \)

Answer:हल : माना \( f(x) = \frac{ax + b}{cx + d} \) भागफल नियम \( \left(\frac{u}{v}\right)' = \frac{u'v - uv'}{v^2} \) का उपयोग करते हुए: Let \( u = ax + b \) and \( v = cx + d \). Then \( u' = a \) and \( v' = c \). \( f'(x) = \frac{a(cx + d) - (ax + b)c}{(cx + d)^2} \)
\( = \frac{acx + ad - acx - bc}{(cx + d)^2} \)
\( = \frac{ad - bc}{(cx + d)^2} \)In simple words: We use the quotient rule to differentiate this rational function. We find the derivatives of the numerator and the denominator, then substitute them into the quotient rule formula and simplify.

🎯 Exam Tip: The order of terms in the numerator of the quotient rule \( (u'v - uv') \) is critical. A common mistake is swapping them, which results in a sign error. Be meticulous with algebraic simplification.

 

प्रश्न 6.
\( \frac{1 + \frac{1}{x}}{1 - \frac{1}{x}} \)

Answer:हल : माना \( f(x) = \frac{1 + \frac{1}{x}}{1 - \frac{1}{x}} \) पहले फलन को सरल कीजिए: \( f(x) = \frac{\frac{x + 1}{x}}{\frac{x - 1}{x}} = \frac{x + 1}{x - 1} \) अब भागफल नियम \( \left(\frac{u}{v}\right)' = \frac{u'v - uv'}{v^2} \) का उपयोग करते हुए: Let \( u = x + 1 \) and \( v = x - 1 \). Then \( u' = 1 \) and \( v' = 1 \). \( f'(x) = \frac{1(x - 1) - (x + 1)1}{(x - 1)^2} \)
\( = \frac{x - 1 - x - 1}{(x - 1)^2} \)
\( = \frac{-2}{(x - 1)^2} \)In simple words: First, simplify the given complex fraction into a simpler rational function. Then, apply the quotient rule to find its derivative.

🎯 Exam Tip: Always simplify the function algebraically *before* differentiating if possible. This can significantly reduce the complexity of applying differentiation rules and prevent errors.

 

प्रश्न 7.
\( \frac{1}{ax^2 + bx + c} \)

Answer:हल : माना \( f(x) = \frac{1}{ax^2 + bx + c} \) भागफल नियम \( \left(\frac{u}{v}\right)' = \frac{u'v - uv'}{v^2} \) का उपयोग करते हुए: Let \( u = 1 \) and \( v = ax^2 + bx + c \). Then \( u' = 0 \) and \( v' = 2ax + b \). \( f'(x) = \frac{0(ax^2 + bx + c) - 1(2ax + b)}{(ax^2 + bx + c)^2} \)
\( = \frac{-(2ax + b)}{(ax^2 + bx + c)^2} \)
\( = -\frac{2ax + b}{(ax^2 + bx + c)^2} \)In simple words: We treat the function as a quotient, where the numerator is a constant. Applying the quotient rule, the derivative of the constant numerator is zero, simplifying the expression significantly.

🎯 Exam Tip: For functions of the form \( \frac{1}{g(x)} \), you can also use the chain rule by writing it as \( [g(x)]^{-1} \). This would give \( -1[g(x)]^{-2} \cdot g'(x) \), which is \( -\frac{g'(x)}{[g(x)]^2} \). Both methods yield the same result.

 

प्रश्न 8.
\( \frac{ax + b}{px^2 + qx + r} \)

Answer:हल : माना \( f(x) = \frac{ax + b}{px^2 + qx + r} \) भागफल नियम \( \left(\frac{u}{v}\right)' = \frac{u'v - uv'}{v^2} \) का उपयोग करते हुए: Let \( u = ax + b \) and \( v = px^2 + qx + r \). Then \( u' = a \) and \( v' = 2px + q \). \( f'(x) = \frac{a(px^2 + qx + r) - (ax + b)(2px + q)}{(px^2 + qx + r)^2} \)
\( = \frac{apx^2 + aqx + ar - (2apx^2 + aqx + 2bpx + bq)}{(px^2 + qx + r)^2} \)
\( = \frac{apx^2 + aqx + ar - 2apx^2 - aqx - 2bpx - bq}{(px^2 + qx + r)^2} \)
\( = \frac{-apx^2 + (aq - aq)x + (ar - 2bpx) - bq}{(px^2 + qx + r)^2} \)
\( = \frac{-apx^2 - 2bpx + ar - bq}{(px^2 + qx + r)^2} \)In simple words: We apply the quotient rule directly to this rational function. We differentiate the numerator and denominator, substitute into the formula, and then expand and simplify the terms in the numerator.

🎯 Exam Tip: Be careful with the distribution of the negative sign across the second term in the numerator \( -(ax+b)(2px+q) \). This is a common source of error. Group and cancel terms systematically.

 

प्रश्न 9.
\( \frac{px^2 + qx + r}{ax + b} \)

Answer:हल : माना \( f(x) = \frac{px^2 + qx + r}{ax + b} \) भागफल नियम \( \left(\frac{u}{v}\right)' = \frac{u'v - uv'}{v^2} \) का उपयोग करते हुए: Let \( u = px^2 + qx + r \) and \( v = ax + b \). Then \( u' = 2px + q \) and \( v' = a \). \( f'(x) = \frac{(2px + q)(ax + b) - (px^2 + qx + r)a}{(ax + b)^2} \)
\( = \frac{(2apx^2 + 2bpx + aqx + bq) - (apx^2 + aqx + ar)}{(ax + b)^2} \)
\( = \frac{2apx^2 + 2bpx + aqx + bq - apx^2 - aqx - ar}{(ax + b)^2} \)
\( = \frac{apx^2 + 2bpx + bq - ar}{(ax + b)^2} \)In simple words: We apply the quotient rule. We differentiate the quadratic numerator and the linear denominator, then substitute into the quotient rule formula, expand, and simplify the terms.

🎯 Exam Tip: Algebraic expansion and simplification are crucial here. Ensure that each term is multiplied correctly and that like terms are combined accurately in the numerator.

 

प्रश्न 10.
\( \frac{a}{x^4} - \frac{b}{x^2} + \cos x. \)

Answer:हल : माना \( f(x) = \frac{a}{x^4} - \frac{b}{x^2} + \cos x \) \( = ax^{-4} - bx^{-2} + \cos x \) \( f'(x) = \frac{d}{dx}(ax^{-4}) - \frac{d}{dx}(bx^{-2}) + \frac{d}{dx}(\cos x) \)
\( = a(-4)x^{-4-1} - b(-2)x^{-2-1} + (-\sin x) \)
\( = -4ax^{-5} + 2bx^{-3} - \sin x \)
\( = -\frac{4a}{x^5} + \frac{2b}{x^3} - \sin x \)In simple words: We rewrite terms with negative exponents and then apply the power rule for differentiation for each algebraic term and the standard derivative for cos x.

🎯 Exam Tip: Convert fractions with \( x \) in the denominator to negative exponents before applying the power rule. Remember the derivative of \( \cos x \) is \( -\sin x \).

 

प्रश्न 11.
\( 4\sqrt{x} - 2. \)

Answer:हल : \( \frac{d}{dx}(4\sqrt{x} - 2) = \frac{d}{dx}(4x^{1/2} - 2) \)
\( = 4 \cdot \frac{1}{2} x^{\frac{1}{2}-1} - 0 \)
\( = 2x^{-1/2} \)
\( = \frac{2}{\sqrt{x}} \)In simple words: We rewrite the square root as a fractional exponent. Then, we apply the power rule to differentiate the term with x and remember that the derivative of a constant is zero.

🎯 Exam Tip: Always remember that \( \sqrt{x} = x^{1/2} \) when applying the power rule. The power rule \( \frac{d}{dx}(x^n) = nx^{n-1} \) is fundamental here.

 

प्रश्न 12.
\( (ax + b)^n \)

Answer:हल : माना \( f(x) = (ax + b)^n \) x के सापेक्ष अवकलन करने पर \( f'(x) = n(ax + b)^{n-1} \cdot \frac{d}{dx}(ax + b) \)
\( = n(ax + b)^{n-1} \cdot a \)
\( = na(ax + b)^{n-1} \)In simple words: We use the chain rule to differentiate this function. We differentiate the outer power function first, then multiply by the derivative of the inner function \( ax + b \).

🎯 Exam Tip: The chain rule is essential for composite functions. Differentiate "outside-in": first the power, then the expression inside the parenthesis. Don't forget to multiply by the derivative of the inner function.

 

प्रश्न 13.
\( (ax + b)^n (cx + d)^m \)

Answer:हल : माना, \( f(x) = (ax + b)^n (cx + d)^m \) गुणनफल नियम (uv)' = u'v + uv' का उपयोग करते हुए: Let \( u = (ax + b)^n \) and \( v = (cx + d)^m \). Then \( u' = n(ax + b)^{n-1} \cdot a = na(ax + b)^{n-1} \) (using chain rule). And \( v' = m(cx + d)^{m-1} \cdot c = mc(cx + d)^{m-1} \) (using chain rule). \( f'(x) = na(ax + b)^{n-1}(cx + d)^m + (ax + b)^n mc(cx + d)^{m-1} \)
\( = (ax + b)^{n-1}(cx + d)^{m-1} [na(cx + d) + mc(ax + b)] \)In simple words: This problem requires both the product rule and the chain rule for each term. We differentiate each part using the chain rule, then combine them using the product rule, and factor out common terms for simplification.

🎯 Exam Tip: This is a multi-step differentiation problem. Be very systematic in applying the chain rule to each factor first, then combining them with the product rule. Factoring out the lowest powers of the common bases is a good way to simplify the final answer.

 

प्रश्न 14.
sin (x + a).

Answer:हल : माना \( f(x) = \sin(x + a) \) x के सापेक्ष अवकलन करने पर \( f'(x) = \frac{d}{dx}[\sin(x + a)] \) Let \( u = x + a \). Then \( \frac{du}{dx} = 1 \). So, \( f'(x) = \frac{d}{du}(\sin u) \cdot \frac{du}{dx} \)
\( = \cos u \cdot 1 \)
\( = \cos(x + a) \)In simple words: We use the chain rule to find the derivative of sin(x + a). We differentiate the sine function (outer function) and then multiply by the derivative of the argument (inner function), x + a.

🎯 Exam Tip: The chain rule is crucial for composite functions. Differentiate the outer function (sin) with respect to its argument, then multiply by the derivative of the inner argument (x + a).

 

प्रश्न 15.
cosec x cot x.

Answer:हल : माना \( f(x) = \operatorname{cosec} x \cot x \) गुणनफल नियम (uv)' = u'v + uv' का उपयोग करते हुए: Let \( u = \operatorname{cosec} x \) and \( v = \cot x \). Then \( u' = -\operatorname{cosec} x \cot x \) and \( v' = -\operatorname{cosec}^2 x \). \( f'(x) = (-\operatorname{cosec} x \cot x)(\cot x) + (\operatorname{cosec} x)(-\operatorname{cosec}^2 x) \)
\( = -\operatorname{cosec} x \cot^2 x - \operatorname{cosec}^3 x \)
\( = -\operatorname{cosec} x (\cot^2 x + \operatorname{cosec}^2 x) \) Using the identity \( \cot^2 x + 1 = \operatorname{cosec}^2 x \), so \( \cot^2 x = \operatorname{cosec}^2 x - 1 \). \( = -\operatorname{cosec} x (\operatorname{cosec}^2 x - 1 + \operatorname{cosec}^2 x) \)
\( = -\operatorname{cosec} x (2\operatorname{cosec}^2 x - 1) \)
\( = -\operatorname{cosec} x + 2\operatorname{cosec}^3 x \)In simple words: We apply the product rule, differentiating cosec x and cot x separately, and then combining the results according to the rule. Further simplification is done using a trigonometric identity.

🎯 Exam Tip: Remember the derivatives of cosec x and cot x. Be careful with the negative signs and simplify the final expression using trigonometric identities if possible, as it shows a deeper understanding.

 

Question 16. \( \frac{\cos x}{1 + \sin x} \)
Answer: हल : माना \( f(x) = \frac{\cos x}{1 + \sin x} \) \[ \frac{u}{v} = \frac{u'v - uv'}{v^2} \]
\( f '(x) = \frac{\frac{d}{dx}(\cos x) \times (1 + \sin x) - \cos x \times \frac{d}{dx}(1 + \sin x)}{(1 + \sin x)^2} \)
\( = \frac{- \sin x (1 + \sin x) - \cos x \times \cos x}{(1 + \sin x)^2} \)
\( = \frac{- \sin x - \sin^2 x - \cos^2 x}{(1 + \sin x)^2} \)
\( = \frac{- \sin x - (\sin^2 x + \cos^2 x)}{(1 + \sin x)^2} \)
\( = \frac{- \sin x - 1}{(1 + \sin x)^2} \)
\( = \frac{- (1 + \sin x)}{(1 + \sin x)^2} \)
\( = \frac{- 1}{1 + \sin x} \)In simple words: To find the derivative of this fraction, we apply the quotient rule. We differentiate the numerator and multiply by the denominator, then subtract the numerator multiplied by the derivative of the denominator, all divided by the square of the denominator.

🎯 Exam Tip: Mastering the quotient rule for derivatives is crucial for questions involving rational functions. Remember \( \frac{d}{dx} \left( \frac{u}{v} \right) = \frac{u'v - uv'}{v^2} \).

 

Question 17. \( \frac{\sin x + \cos x}{\sin x - \cos x} \)
Answer: हल : माना, \( f(x) = \frac{\sin x + \cos x}{\sin x - \cos x} \) \[ \left( \frac{u}{v} \right)' = \frac{u'v - uv'}{v^2} \]
\( f '(x) = \frac{\frac{d}{dx}(\sin x + \cos x) (\sin x - \cos x) - (\sin x + \cos x) \frac{d}{dx}(\sin x - \cos x)}{(\sin x - \cos x)^2} \)
\( = \frac{(\cos x - \sin x)(\sin x - \cos x) - (\sin x + \cos x)(\cos x + \sin x)}{(\sin x - \cos x)^2} \)
\( = \frac{- (\sin x - \cos x)^2 - (\sin x + \cos x)^2}{(\sin x - \cos x)^2} \)
\( = \frac{- [(\sin^2 x + \cos^2 x - 2 \sin x \cos x) + (\sin^2 x + \cos^2 x + 2 \sin x \cos x)]}{(\sin x - \cos x)^2} \)
\( = \frac{- [ (1 - 2 \sin x \cos x) + (1 + 2 \sin x \cos x) ]}{(\sin x - \cos x)^2} \)
\( = \frac{- [1 - 2 \sin x \cos x + 1 + 2 \sin x \cos x]}{(\sin x - \cos x)^2} \)
\( = \frac{- 2}{(\sin x - \cos x)^2} \)In simple words: This problem uses the quotient rule. We differentiate the numerator and denominator, then apply the formula, simplifying trigonometric identities like \( \sin^2 x + \cos^2 x = 1 \) along the way.

🎯 Exam Tip: Be cautious with negative signs and trigonometric identities during simplification. Double-check each step to avoid algebraic errors.

 

Question 18. \( \frac{\sec x - 1}{\sec x + 1} \)
Answer: हल : माना \( f(x) = \frac{\sec x - 1}{\sec x + 1} \) \[ \frac{u}{v} = \frac{u'v - uv'}{v^2} \]
\( f '(x) = \frac{\frac{d}{dx}(\sec x - 1) (\sec x + 1) - (\sec x - 1) \frac{d}{dx}(\sec x + 1)}{(\sec x + 1)^2} \)
\( = \frac{(\sec x \tan x)(\sec x + 1) - (\sec x - 1)(\sec x \tan x)}{(\sec x + 1)^2} \)
\( = \frac{\sec^2 x \tan x + \sec x \tan x - (\sec^2 x \tan x - \sec x \tan x)}{(\sec x + 1)^2} \)
\( = \frac{\sec^2 x \tan x + \sec x \tan x - \sec^2 x \tan x + \sec x \tan x}{(\sec x + 1)^2} \)
\( = \frac{2 \sec x \tan x}{(\sec x + 1)^2} \)In simple words: We find the derivative using the quotient rule, remembering the derivative of \( \sec x \) is \( \sec x \tan x \). The common terms cancel out, leading to the simplified expression.

🎯 Exam Tip: Knowing standard trigonometric derivatives and applying the quotient rule correctly are key. Pay attention to distributing terms and combining like terms.

 

Question 19. \( \sin^n x. \)
Answer: हल : माना \( f(x) = \sin^n x \) sin x को u रखने पर \( f(x) = u^n \) x के सापेक्ष अवकलन करने पर
\( f'(x) = \frac{d}{dx} u^n = \frac{d}{du} u^n \times \frac{du}{dx} \)
\( = n u^{n-1} \frac{du}{dx} \)
\( = n \sin^{n-1} x \frac{d}{dx} \sin x = n \sin^{n-1} x \cdot \cos x \)
\( = n \cos x \sin^{n-1} x. \)In simple words: To differentiate \( \sin^n x \), we use the chain rule. We treat \( \sin x \) as an inner function, differentiating \( u^n \) with respect to \( u \) and then multiplying by the derivative of \( \sin x \) with respect to \( x \).

🎯 Exam Tip: The chain rule is fundamental for differentiating composite functions. Clearly identify the outer and inner functions for accurate application.

 

Question 20. \( \frac{a + b \sin x}{c + d \cos x} \)
Answer: हल: माना \( f(x) = \frac{a + b \sin x}{c + d \cos x} \) \[ \left( \frac{u}{v} \right)' = \frac{u'v - uv'}{v^2} \]
\( f'(x) = \frac{\frac{d}{dx}(a + b \sin x) (c + d \cos x) - (a + b \sin x) \frac{d}{dx}(c + d \cos x)}{(c + d \cos x)^2} \)
\( = \frac{b \cos x (c + d \cos x) - (a + b \sin x)(- d \sin x)}{(c + d \cos x)^2} \)
\( = \frac{bc \cos x + bd \cos^2 x + ad \sin x + bd \sin^2 x}{(c + d \cos x)^2} \)
\( = \frac{bc \cos x + ad \sin x + bd(\sin^2 x + \cos^2 x)}{(c + d \cos x)^2} \)
\( = \frac{bc \cos x + ad \sin x + bd}{(c + d \cos x)^2} \)In simple words: We apply the quotient rule to differentiate this function. After differentiating the numerator and denominator, we substitute them into the formula and use the identity \( \sin^2 x + \cos^2 x = 1 \) to simplify.

🎯 Exam Tip: Simplify the numerator carefully after applying the derivative rules. Look for common trigonometric identities to reach the most compact form.

 

Question 21. \( \frac{\sin(x+a)}{\cos x} \)
Answer: हल : माना \( f(x) = \frac{\sin (x+a)}{\cos x} \) \[ \left( \frac{u}{v} \right)' = \frac{u'v - uv'}{v^2} \]
\( f'(x) = \frac{\frac{d}{dx}(\sin(x+a)) \cos x - \sin(x+a) \frac{d}{dx}(\cos x)}{\cos^2 x} \)
\( = \frac{\cos(x+a) \cos x - \sin(x+a)(-\sin x)}{\cos^2 x} \)
\( = \frac{\cos(x+a) \cos x + \sin(x+a) \sin x}{\cos^2 x} \)
\( = \frac{\cos(x+a-x)}{\cos^2 x} \)
\( = \frac{\cos a}{\cos^2 x} \)In simple words: This derivative is found using the quotient rule. The key simplification involves recognizing the trigonometric identity \( \cos A \cos B + \sin A \sin B = \cos(A-B) \) in the numerator.

🎯 Exam Tip: Familiarity with sum/difference trigonometric identities (like \( \cos(A-B) \)) can significantly simplify complex derivatives after applying the quotient rule.

 

Question 22. \( x^4 (5 \sin x - 3 \cos x). \)
Answer: हल : \( (uv)' = u'v + uv' \)
\( \frac{d}{dx} [x^4 (5 \sin x - 3 \cos x)] = \left(\frac{d}{dx}x^4\right) (5 \sin x - 3 \cos x) + x^4 \left(\frac{d}{dx}(5 \sin x - 3 \cos x)\right) \)
\( = 4x^3 (5 \sin x - 3 \cos x) + x^4(5 \cos x + 3 \sin x) \)
\( = x^3 (20 \sin x - 12 \cos x + 5x \cos x + 3x \sin x). \)In simple words: We apply the product rule to find the derivative of this function. We differentiate \( x^4 \) and multiply by the second term, then add \( x^4 \) multiplied by the derivative of the second term, which involves basic trigonometric derivatives.

🎯 Exam Tip: When using the product rule, differentiate each part carefully, especially trigonometric functions. Factoring out common terms like \( x^3 \) in the final step is good practice.

 

Question 23. \( (x^2 + 1) \cos x. \)
Answer: हल : \( \frac{d}{dx} [(x^2 + 1) \cos x] = \left[\frac{d}{dx}(x^2+1)\right] \cos x + (x^2 + 1) \frac{d}{dx} \cos x \)
\( = 2x \cos x + (x^2 + 1) (- \sin x) \)
\( = 2x \cos x - (x^2 + 1) \sin x \)
\( = - x^2 \sin x - \sin x + 2x \cos x. \)In simple words: We use the product rule to differentiate this function. We differentiate the first term \( (x^2 + 1) \) and multiply by \( \cos x \), then add the first term multiplied by the derivative of \( \cos x \).

🎯 Exam Tip: Remember the derivatives of basic power functions and trigonometric functions. Apply the product rule systematically to avoid errors.

 

Question 24. \( (ax^2 + \sin x) (p + q \cos x). \)
Answer: हल : \( (uv)' = u'v + uv' \)
\( \frac{d}{dx} [(ax^2 + \sin x)(p + q \cos x)] = \left[\frac{d}{dx}(ax^2 + \sin x)\right] (p + q \cos x) + (ax^2 + \sin x) \frac{d}{dx} (p + q \cos x) \)
\( = (2ax + \cos x) (p + q \cos x) + (ax^2 + \sin x) (- q \sin x) \)
\( = 2ax(p + q \cos x) + \cos x(p + q \cos x) - q \sin x (ax^2 + \sin x) \)In simple words: This problem requires the product rule. We differentiate each factor, \( (ax^2 + \sin x) \) and \( (p + q \cos x) \), and combine them according to the product rule formula.

🎯 Exam Tip: Be careful with signs when differentiating trigonometric functions, especially \( \frac{d}{dx}(\cos x) = -\sin x \). Ensure all terms are correctly expanded and combined.

 

Question 25. \( (x + \cos x)(x - \tan x) \)
Answer: हल : \( (uv)' = u'v + uv' \)
\( \frac{d}{dx} [(x + \cos x) (x - \tan x)] = \left[\frac{d}{dx}(x + \cos x)\right] (x - \tan x) + (x + \cos x) \left[\frac{d}{dx}(x - \tan x)\right] \)
\( = (1 - \sin x)(x - \tan x) + (x + \cos x)(1 - \sec^2 x) \)
\( = (1 - \sin x)(x - \tan x) + (x + \cos x) (-\tan^2 x). \)In simple words: We apply the product rule here. Differentiate \( (x + \cos x) \) and \( (x - \tan x) \) separately, then substitute into the product rule formula, remembering that \( \frac{d}{dx}(\tan x) = \sec^2 x \).

🎯 Exam Tip: Keep track of the derivatives of all trigonometric functions. The identity \( 1 - \sec^2 x = -\tan^2 x \) is useful for simplifying the expression.

 

Question 26. \( \frac{4x + 5 \sin x}{3x + 7 \cos x} \)
Answer: हल : \[ \left( \frac{u}{v} \right)' = \frac{u'v - uv'}{v^2} \]
\( \frac{d}{dx} \left( \frac{4x + 5 \sin x}{3x + 7 \cos x} \right) = \frac{\frac{d}{dx}(4x + 5 \sin x) (3x + 7 \cos x) - (4x + 5 \sin x) \frac{d}{dx}(3x + 7 \cos x)}{(3x + 7 \cos x)^2} \)
\( = \frac{(4 + 5 \cos x)(3x + 7 \cos x) - (4x + 5 \sin x)(3 - 7 \sin x)}{(3x + 7 \cos x)^2} \)
\( = \frac{(12x + 28 \cos x + 15x \cos x + 35 \cos^2 x) - (12x - 28x \sin x + 15 \sin x - 35 \sin^2 x)}{(3x + 7 \cos x)^2} \)
\( = \frac{28 \cos x + 15x \cos x + 35 \cos^2 x + 28x \sin x - 15 \sin x + 35 \sin^2 x}{(3x + 7 \cos x)^2} \)
\( = \frac{28 \cos x + 15x \cos x + 28x \sin x - 15 \sin x + 35(\cos^2 x + \sin^2 x)}{(3x + 7 \cos x)^2} \)
\( = \frac{35 + 15x \cos x + 28 \cos x + 28x \sin x - 15 \sin x}{(3x + 7 \cos x)^2} \)In simple words: We apply the quotient rule and expand the terms in the numerator. After differentiation, we combine like terms and use the identity \( \sin^2 x + \cos^2 x = 1 \) to simplify the expression further.

🎯 Exam Tip: Be meticulous with algebraic expansion and consolidation of terms in the numerator. Errors in signs or distribution are common pitfalls.

 

Question 27. \( \frac{x^2 \cos (\frac{\pi}{4})}{\sin x} \)
Answer: हल : माना \( f(x) = \frac{x^2 \cos \frac{\pi}{4}}{\sin x} = \frac{x^2 \frac{1}{\sqrt{2}}}{\sin x} \) \[ \left( \frac{u}{v} \right)' = \frac{u'v - uv'}{v^2} \]
\( f'(x) = \frac{1}{\sqrt{2}} \frac{\frac{d}{dx}(x^2) \sin x - x^2 \frac{d}{dx}(\sin x)}{\sin^2 x} \)
\( = \frac{1}{\sqrt{2}} \frac{2x \sin x - x^2 \cos x}{\sin^2 x} \) या
\( = \frac{x(2 \sin x - x \cos x)}{\sqrt{2} \sin^2 x} \)In simple words: First, we substitute the value of \( \cos(\frac{\pi}{4}) \) and then apply the quotient rule to differentiate the function. The constant factor \( \frac{1}{\sqrt{2}} \) is carried through the differentiation process.

🎯 Exam Tip: Recognize and substitute constant trigonometric values (like \( \cos(\frac{\pi}{4}) \)) early to simplify the derivative process. The constant multiplier rule applies here.

 

Question 28. \( \frac{x}{1+\tan x} \)
Answer: हल : \[ \left( \frac{u}{v} \right)' = \frac{u'v - uv'}{v^2} \]
\( \frac{d}{dx} \left( \frac{x}{1+\tan x} \right) = \frac{\left(\frac{d}{dx}x\right)(1+\tan x) - x\left(\frac{d}{dx}(1+\tan x)\right)}{(1+\tan x)^2} \)
\( = \frac{1 \cdot (1 + \tan x) - x \sec^2 x}{(1 + \tan x)^2} \)
\( = \frac{1 + \tan x - x \sec^2 x}{(1 + \tan x)^2} \)In simple words: We apply the quotient rule for differentiation. The derivative of \( x \) is 1, and the derivative of \( (1 + \tan x) \) is \( \sec^2 x \). Substitute these into the formula to get the result.

🎯 Exam Tip: Standard derivatives of basic functions like \( x \) and \( \tan x \) are essential. Pay attention to the structure of the quotient rule and substitute accurately.

 

Question 29. \( (x+ \sec x) (x - \tan x). \)
Answer: हल : \( (uv)' = u'v + uv' \)
\( \frac{d}{dx} [(x + \sec x) (x - \tan x)] = \left[\frac{d}{dx}(x + \sec x)\right] (x - \tan x) + (x + \sec x) \left[\frac{d}{dx}(x - \tan x)\right] \)
\( = (1 + \sec x \tan x)(x - \tan x) + (x + \sec x) (1 - \sec^2 x). \)In simple words: We use the product rule to find the derivative. We differentiate the first factor \( (x + \sec x) \) and the second factor \( (x - \tan x) \) and then combine them using the product rule formula.

🎯 Exam Tip: Remember the derivatives of \( \sec x \) and \( \tan x \). The identity \( 1 - \sec^2 x = -\tan^2 x \) can sometimes offer further simplification, but the given form is also acceptable.

 

Question 30. \( \frac{x}{\sin^n x} \)
Answer: हल : माना \( f(x) = \frac{x}{\sin^n x} \) \[ \left( \frac{u}{v} \right)' = \frac{u'v - uv'}{v^2} \]
\( f'(x) = \frac{\left(\frac{d}{dx}x\right) \sin^n x - x \left(\frac{d}{dx}(\sin^n x)\right)}{(\sin^n x)^2} \)
\( = \frac{1 \cdot \sin^n x - x \cdot n \cos x \sin^{n-1} x}{\sin^{2n} x} \)
\( = \frac{\sin^{n-1} x(\sin x - nx \cos x)}{\sin^{2n} x} \)
\( = \frac{\sin x - nx \cos x}{\sin^{n+1} x} \)In simple words: This derivative uses the quotient rule. We differentiate the numerator and denominator, applying the chain rule for \( \sin^n x \). Then, we simplify by factoring out common terms in the numerator and reducing the powers of \( \sin x \).

🎯 Exam Tip: Pay close attention to the power of \( n \) and apply the chain rule correctly when differentiating \( \sin^n x \). Careful simplification of exponential terms is also crucial.