Maharashtra Board Class 11 Maths Part 1 Chapter 8 Continuity Miscellaneous Solutions

Std 11 Maths 1 Miscellaneous Exercise 8 Solutions Commerce Maths

I. Discuss The Continuity Of The Following Functions At The Point(s) Or In The Interval Indicated Against Them.

Question 1. If \(f(x) = 2x^2 – 2x + 5\) for \(0 \le x < 2\) \( = \frac{1-3x-x^2}{1-x}\) for \(2 \le x < 4\) \( = \frac{7-x^2}{x-5}\) for \(4 \le x \le 7\) on its domain.
Answer:Solution: The domain of f is \([0, 5) \cup (5, 7]\) We observe that \(x = 5\) is not included in the domain as f is not defined at \(x = 5\)
(a) For \(0 \le x < 2\) \(f(x) = 2x^2 – 2x + 5\) It is a polynomial function and is continuous at all point in \([0, 2)\)
(b) For \(2 < x < 4\) \(f(x) = \frac{1-3x-x^2}{1-x}\) It is a rational function and is continuous everywhere except at points where its denominator becomes zero. Denominator becomes zero at \(x = 1\) But \(x = 1\) does not lie in the interval. \(f(x)\) is continuous at all points in \((2, 4)\)
(c) For \(4 < x \le 7, x \ne 5\) i.e. for \(x \in [4, 5) \cup (5, 7]\) \(\therefore f(x) = \frac{7-x^2}{x-5}\) It is a rational function and is continuous everywhere except possibly at points where its denominator becomes zero. Denominator becomes zero at \(x = 5\) But \(x = 5 \notin [4, 5) \cup (5, 7]\) \(\therefore\) f is continuous at all points in \((4, 7] - \{5\}\).
(d) Since the definition of function changes around \(x = 2, x = 4\) and \(x = 7\) \(\therefore\) there is disturbance in behaviour of the function. So we examine continuity at \(x = 2, 4, 7\) separately. Continuity at \(x = 2\): \(\lim_{x \to 2^-} f(x) = \lim_{x \to 2^-} (2x^2 – 2x + 5)\) \(= 2(2)^2 – 2(2) + 5\) \(= 8-4+5\) \(= 9\) \(\lim_{x \to 2^+} f(x) = \lim_{x \to 2^+} \frac{1-3x-x^2}{1-x}\) \(= \frac{1-3(2)-(2)^2}{1-2}\) \(= \frac{1-6-4}{-1}\) \(= \frac{-9}{-1}\) \(= 9\) Also, \(f(x) = \frac{1-3x-x^2}{1-x}\), at \(x = 2\) \(\therefore f(2) = \frac{1-3(2)-(2)^2}{1-2}\) \(f(2) = 9\) \(\therefore \lim_{x \to 2^-} f(x) = \lim_{x \to 2^+} f(x) = f(2)\) \(\therefore\) f is continuous at \(x = 2\)
(e) Continuity at \(x = 4\): \(\lim_{x \to 4^-} f(x) = \lim_{x \to 4^-} \frac{1-3x-x^2}{1-x}\) \(= \frac{1-3(4)-(4)^2}{1-4}\) \(= \frac{1-12-16}{-3}\) \(= \frac{-27}{-3}\) \(= 9\) \(\lim_{x \to 4^+} f(x) = \lim_{x \to 4^+} \frac{7-x^2}{x-5}\) \(= \frac{7-(4)^2}{4-5}\) \(= \frac{7-16}{-1}\) \(= \frac{-9}{-1}\) \(= 9\) \(f(x) = \frac{7-x^2}{x-5}\), at \(x = 4\) \(\therefore f(4) = \frac{7-(4)^2}{4-5}\) \(= 9\) \(\therefore \lim_{x \to 4^-} f(x) = \lim_{x \to 4^+} f(x) = f(4)\) \(\therefore\) f is continuous at \(x = 4\)In simple words: To check continuity, we examine the function in different intervals where it's defined and then check at the points where the function definition changes. If the left-hand limit, right-hand limit, and the function value are all equal at a point, the function is continuous there.

🎯 Exam Tip: Always evaluate continuity at the transition points of a piecewise function by comparing left-hand limit, right-hand limit, and function value at that point for full marks.

Question 2. If \(f(x) = \frac{3^x + 3^{-x} - 2}{x^2}\) for \(x \ne 0\) \(= (\log 3)^2\) for \(x = 0\) at \(x = 0\)
Answer:Solution: \(f(0) = (\log 3)^2\) ...(given) \(\lim_{x \to 0} f(x) = \lim_{x \to 0} \frac{3^x + 3^{-x} - 2}{x^2}\) \(\therefore = \lim_{x \to 0} \frac{3^x + \frac{1}{3^x} - 2}{x^2}\) \(= \lim_{x \to 0} \frac{(3^x)^2 + 1 - 2(3^x)}{x^2(3^x)}\) \(= \lim_{x \to 0} \frac{(3^x - 1)^2}{x^2(3^x)}\) \(\ldots[\therefore a^2-2ab + b^2 = (a - b)^2]\) \(= \lim_{x \to 0} \left( \frac{3^x - 1}{x} \right)^2 \times \frac{1}{3^x}\) \(= \left( \lim_{x \to 0} \frac{3^x - 1}{x} \right)^2 \times \lim_{x \to 0} \frac{1}{3^x}\) \(= (\log 3)^2 \times \frac{1}{3^0}\) \(\ldots[\therefore \lim_{x \to 0} \frac{a^x - 1}{x} = \log a]\) \(= (\log 3)^2 \times \frac{1}{1}\) \(= (\log 3)^2\) \(\therefore \lim_{x \to 0} f(x) = f(0)\) \(\therefore\) f is continuous at \(x = 0\)In simple words: This problem involves checking the continuity of a function at a point using limits and a standard limit formula for exponential functions. Since the limit as x approaches 0 equals the function value at 0, the function is continuous.

🎯 Exam Tip: Remember the standard limit \(\lim_{x \to 0} \frac{a^x-1}{x} = \log a\). This is crucial for solving continuity problems involving exponential functions.

Question 3. If \(f(x) = \frac{5^x - e^x}{2x}\) for \(x \ne 0\) \(= \frac{1}{2} (\log 5 – 1)\) for \(x = 0\) at \(x = 0\)
Answer:Solution: \(f(0) = \frac{1}{2} (\log 5 – 1)\) ...(given) \(\lim_{x \to 0} f(x) = \lim_{x \to 0} \frac{5^x - e^x}{2x}\) \(= \lim_{x \to 0} \frac{5^x - 1 - e^x + 1}{2x}\) \(= \frac{1}{2} \lim_{x \to 0} \left[ \frac{(5^x - 1) - (e^x - 1)}{x} \right]\) \(= \frac{1}{2} \lim_{x \to 0} \left[ \frac{5^x - 1}{x} - \frac{e^x - 1}{x} \right]\) \(= \frac{1}{2} (\log 5 - \log e)\) \(\ldots[\therefore \lim_{x \to 0} \frac{a^x - 1}{x} = \log a]\) \(= \frac{1}{2} (\log 5 - 1)\) \(\ldots[\therefore \log e = 1]\) \(\therefore \lim_{x \to 0} f(x) = f(0)\) \(\therefore\) f is continuous at \(x = 0\)In simple words: The function's continuity at \(x=0\) is verified by evaluating the limit of the function as x approaches 0. By using the property of limits for exponential forms, the calculated limit matches the defined function value, confirming continuity.

🎯 Exam Tip: When dealing with limits involving multiple exponential terms, add and subtract '1' to create terms like \((a^x - 1)\) which are amenable to standard limit formulas.

Question 4. If \(f(x) = \frac{\sqrt{x+3}-2}{x^3-1}\) for \(x \ne 1\) \(= 2\) for \(x = 1\), at \(x = 1\)
Answer:Solution: \(f(1) = 2\) ...(given) \(\lim_{x \to 1} f(x) = \lim_{x \to 1} \frac{\sqrt{x+3}-2}{x^3-1}\) \(= \lim_{x \to 1} \frac{\sqrt{x+3}-2}{x^3-1} \times \frac{\sqrt{x+3}+2}{\sqrt{x+3}+2}\) \(= \lim_{x \to 1} \frac{x+3-4}{(x^3-1)(\sqrt{x+3}+2)}\) \(= \lim_{x \to 1} \frac{x-1}{(x-1)(x^2+x+1)(\sqrt{x+3}+2)}\) \(= \lim_{x \to 1} \frac{1}{(x^2+x+1)(\sqrt{x+3}+2)}\) \(\ldots[\text{As } x \to 1, x \ne 1 \implies x-1 \ne 0]\) \(= \frac{1}{\lim_{x \to 1} (x^2+x+1) \times \lim_{x \to 1} (\sqrt{x+3}+2)}\) \(= \frac{1}{(1^2+1+1) \times (\sqrt{1+3}+2)}\) \(= \frac{1}{3(2+2)}\) \(= \frac{1}{12}\) \(\therefore \lim_{x \to 1} f(x) \ne f(1)\) \(\therefore\) f is discontinuous at \(x = 1\)In simple words: To check continuity, we calculated the limit of the function as x approaches 1 and compared it with the function's value at \(x=1\). Since the limit (1/12) is not equal to the function's defined value (2), the function is discontinuous at \(x=1\).

🎯 Exam Tip: When evaluating limits involving square roots, rationalize the numerator or denominator by multiplying by the conjugate to simplify the expression and avoid indeterminate forms.

Question 5. If \(f(x) = \frac{\log x-\log 3}{x-3}\) for \(x \ne 3\) \(= 3\) for \(x = 3\), at \(x = 3\)
Answer:Solution: \(f(3) = 3\) ...(given) \(\lim_{x \to 3} f(x) = \lim_{x \to 3} \frac{\log x - \log 3}{x-3}\) Substitute \(x-3=h\)
\(\implies x = 3+h\). as \(x \to 3, h \to 0\) \(\lim_{h \to 0} f(x) = \lim_{h \to 0} \frac{\log(h+3)-\log 3}{3+h-3}\) \(= \lim_{h \to 0} \frac{\log(\frac{h+3}{3})}{h}\) \(= \lim_{h \to 0} \frac{\log(1+\frac{h}{3})}{h}\) \(= \frac{1}{3} \lim_{h \to 0} \frac{\log(1+\frac{h}{3})}{\frac{h}{3}}\) \(= \frac{1}{3}(1)\) \(\ldots[\therefore \lim_{x \to 0} \frac{\log(1+x)}{x} = 1]\) \(= \frac{1}{3}\) \(\therefore \lim_{x \to 3} f(x) \ne f(3)\) \(\therefore\) f is discontinuous at \(x = 3\)In simple words: This problem involves checking the continuity of a logarithmic function. After substituting \(x-3=h\) to change the limit point to 0, and using a standard logarithmic limit, the calculated limit (1/3) does not match the function's defined value (3) at \(x=3\), indicating discontinuity.

🎯 Exam Tip: For limits involving logarithms, especially near a point where the denominator goes to zero, use substitution like \(x-a=h\) to transform it into the standard form \(\lim_{x \to 0} \frac{\log(1+x)}{x} = 1\).

II. Find K If Following Functions Are Continuous At The Points Indicated Against Them.

Question 1. If \(f(x) = \left( \frac{5x-8}{8-3x} \right)^{\frac{3}{2x-4}}\) for \(x \ne 2\) \(= k\) for \(x = 2\) at \(x = 2\)
Answer:Solution: f is continuous at \(x = 2\) \(\therefore f(2) = \lim_{x \to 2} f(x)\)
\(\therefore k = \lim_{x \to 2} \left( \frac{5x-8}{8-3x} \right)^{\frac{3}{2x-4}}\) Substitute \(x-2=h\)
\(\implies x = 2+h\). As \(x \to 2, h \to 0\)
\(\therefore k = \lim_{h \to 0} \left[ \frac{5(2+h)-8}{8-3(2+h)} \right]^{\frac{3}{2(2+h)-4}}\) \(= \lim_{h \to 0} \left[ \frac{10+5h-8}{8-6-3h} \right]^{\frac{3}{2h}}\) \(= \lim_{h \to 0} \left[ \frac{2+5h}{2-3h} \right]^{\frac{3}{2h}}\) \(= \lim_{h \to 0} \left[ \frac{2(1+\frac{5h}{2})}{2(1-\frac{3h}{2})} \right]^{\frac{3}{2h}}\) \(= \lim_{h \to 0} \left[ \frac{1+\frac{5h}{2}}{1-\frac{3h}{2}} \right]^{\frac{3}{2h}}\) \(= \frac{\lim_{h \to 0} \left(1+\frac{5h}{2}\right)^{\frac{3}{2h}}}{\lim_{h \to 0} \left(1-\frac{3h}{2}\right)^{\frac{3}{2h}}}\) \(= \frac{\lim_{h \to 0} \left[ \left(1+\frac{5h}{2}\right)^{\frac{2}{5h}} \right]^{\frac{5h}{2} \cdot \frac{3}{2h}}}{\lim_{h \to 0} \left[ \left(1-\frac{3h}{2}\right)^{-\frac{2}{3h}} \right]^{-\frac{3h}{2} \cdot \frac{3}{2h}}}\) \(= \frac{e^{\frac{15}{4}}}{e^{-\frac{9}{4}}}\)
\(\ldots[\because h \to 0, \frac{5h}{2} \to 0, \frac{-3h}{2} \to 0]\) \(\ldots[\text{and } \lim_{x \to 0} (1+x)^{\frac{1}{x}} = e]\) \(= e^{\frac{15}{4} - (-\frac{9}{4})}\) \(= e^{\frac{15}{4} + \frac{9}{4}}\) \(= e^{\frac{24}{4}}\) \(\therefore k = e^6\)In simple words: Since the function is continuous at \(x=2\), the value of k must be equal to the limit of the function as x approaches 2. By using the special limit form \(\lim_{x \to 0} (1+ax)^{1/x} = e^a\), we manipulate the expression to find that \(k = e^6\).

🎯 Exam Tip: For limits of the form \(1^\infty\), always try to convert the expression into \((1+f(x))^{1/f(x)}\) structure to utilize the \(e\) limit, which is key for finding the value of 'k'.

Question 2. If \(f(x) = \frac{45^x - 9^x - 5^x + 1}{(k^x-1)(3^x-1)}\) for \(x \ne 0\) \(= \frac{2}{3}\) for \(x = 0\), at \(x = 0\)
Answer:Solution: f is continuous at \(x = 0\) \(\therefore \lim_{x \to 0} f(x) = f(0)\)
\(\therefore \lim_{x \to 0} \frac{45^x - 9^x - 5^x + 1}{(k^x-1)(3^x-1)} = \frac{2}{3}\)
\(\therefore \lim_{x \to 0} \frac{9^x \cdot 5^x - 9^x - 5^x + 1}{(k^x-1)(3^x-1)} = \frac{2}{3}\)
\(\therefore \lim_{x \to 0} \frac{9^x(5^x-1)-1(5^x-1)}{(k^x-1)(3^x-1)} = \frac{2}{3}\)
\(\therefore \lim_{x \to 0} \frac{(5^x-1)(9^x-1)}{(k^x-1)(3^x-1)} = \frac{2}{3}\)
\(\therefore \lim_{x \to 0} \frac{\frac{(5^x-1)(9^x-1)}{x^2}}{\frac{(k^x-1)(3^x-1)}{x^2}} = \frac{2}{3}\) \(\ldots[\because x \to 0, x \ne 0 \therefore x^2 \ne 0]\) \(\ldots[\text{Divide Numerator and Denominator by } x^2]\)
\(\therefore \frac{\lim_{x \to 0} \frac{5^x-1}{x} \cdot \lim_{x \to 0} \frac{9^x-1}{x}}{\lim_{x \to 0} \frac{k^x-1}{x} \cdot \lim_{x \to 0} \frac{3^x-1}{x}} = \frac{2}{3}\)
\(\ldots[\therefore \lim_{x \to 0} \frac{a^x-1}{x} = \log a]\) \(\therefore \frac{\log 5 \cdot \log 9}{\log k \cdot \log 3} = \frac{2}{3}\) \(\therefore \frac{\log 5 \cdot \log (3)^2}{\log k \cdot \log 3} = \frac{2}{3}\)
\(\therefore \frac{\log 5 \cdot 2 \log 3}{\log k \cdot \log 3} = \frac{2}{3}\)
\(\therefore \frac{2 \log 5}{\log k} = \frac{2}{3}\)
\(\therefore 3 \log 5 = \log k\)
\(\therefore \log(5)^3 = \log k\)
\(\therefore (5)^3 = k\)
\(\therefore k = 125\)In simple words: For the function to be continuous at \(x=0\), its limit as x approaches 0 must equal its defined value at \(x=0\). By factoring the numerator and dividing both numerator and denominator by \(x^2\), we apply the standard limit \(\lim_{x \to 0} \frac{a^x-1}{x} = \log a\) to find that k equals 125.

🎯 Exam Tip: Factorization of exponential terms like \(45^x - 9^x - 5^x + 1\) into \((5^x-1)(9^x-1)\) is a common strategy. Remember to divide by \(x\) or \(x^2\) in both numerator and denominator when \(x \to 0\) to use the standard limit formulas effectively.

Question 3. If \(f(x) = (1 + kx)^{\frac{1}{x}}\) for \(x \ne 0\) \(= e^{\frac{3}{2}}\) for \(x = 0\), at \(x = 0\)
Answer:Solution: f is continuous at \(x = 0\) \(\therefore \lim_{x \to 0} f(x) = f(0)\)
\(\therefore \lim_{x \to 0} (1 + kx)^{\frac{1}{x}} = e^{\frac{3}{2}}\)
\(\therefore \lim_{x \to 0} \left[ (1 + kx)^{\frac{1}{kx}} \right]^k = e^{\frac{3}{2}}\)
\(\therefore e^k = e^{\frac{3}{2}}\) \(\ldots[\therefore \lim_{x \to 0} (1+x)^{\frac{1}{x}} = e]\)
\(\therefore k = \frac{3}{2}\)In simple words: Since the function is continuous at \(x=0\), we set the limit of \(f(x)\) as \(x \to 0\) equal to \(f(0)\). By using the special limit \(\lim_{x \to 0} (1+ax)^{1/x} = e^a\), we found that the value of \(k\) is \(\frac{3}{2}\).

🎯 Exam Tip: For limits involving \((1+ \text{something})^{\frac{1}{\text{something}}}\) as something approaches zero, aim to transform the expression into the form \((1+f(x))^{1/f(x)}\) so you can directly substitute \(e\).

III. Find A And B If Following Functions Are Continuous At The Point Indicated Against Them.

Question 1. If \(f(x) = x^2 + a\), for \(x \ge 0\) \(= 2\sqrt{x^2 + 1} + b\), for \(x < 0\) and \(f(1) = 2\), is continuous at \(x = 0\)
Answer:Solution: Since, \(f(x) = x^2 + a\), \(x \ge 0\)
\(\therefore f(1) = (1)^2 + a\)
\(\therefore 2 = 1 + a\) \(\ldots[\therefore f(1) = 2]\)
\(\therefore a = 1\) Also f is continuous at \(x = 0\)
\(\therefore \lim_{x \to 0^-} f(x) = \lim_{x \to 0^+} f(x)\)
\(\therefore \lim_{x \to 0^-} (2\sqrt{x^2 + 1} + b) = \lim_{x \to 0^+} (x^2 + a)\)
\(\therefore 2\sqrt{0^2 + 1} + b = 0^2 + a\)
\(\therefore 2(1) + b = a\)
\(\therefore 2 + b = a\) Substitute \(a=1\), we get
\(\therefore 2 + b = 1\)
\(\therefore b = -1\)
\(\therefore a = 1\) and \(b = -1\)In simple words: We first use the given condition \(f(1)=2\) to find the value of 'a'. Then, because the function is continuous at \(x=0\), we equate the left-hand limit and the right-hand limit at \(x=0\) to find 'b'. The consistent values are \(a=1\) and \(b=-1\).

🎯 Exam Tip: For piecewise functions continuous at a point, always ensure to: (1) use any given function value in a relevant piece to find a variable, and (2) equate the left-hand limit, right-hand limit, and function value at the continuity point to find the remaining variables.

Question 2. If \(f(x) = \frac{x^2-9}{x-3} + a\), for \(x > 3\) \(= 5\), for \(x = 3\) \(= 2x^2 + 3x + b\), for \(x < 3\) is continuous at \(x = 3\)
Answer:Solution: f is continuous at \(x = 3\)
\(\therefore f(3) = \lim_{x \to 3^-} f(x)\)
\(\therefore 5 = \lim_{x \to 3^-} (2x^2 + 3x + b)\)
\(\therefore 5 = 2(3)^2 + 3(3) + b\)
\(\therefore 5 = 18 + 9 + b\)
\(\therefore 5 = 27 + b\)
\(\therefore b = 5 - 27\)
\(\therefore b = -22\) Also, \(f(3) = \lim_{x \to 3^+} f(x)\)
\(\therefore 5 = \lim_{x \to 3^+} \left( \frac{x^2-9}{x-3} + a \right)\) \(= \lim_{x \to 3^+} \left( \frac{(x+3)(x-3)}{x-3} + a \right)\) \(\ldots[\because x \to 3; x \ne 3 \therefore x-3 \ne 0]\) \(= \lim_{x \to 3^+} (x+3) + a\) \(= (3 + 3) + a\) \(= 6 + a\)
\(\therefore 5 = 6 + a\)
\(\therefore a = 5 - 6\)
\(\therefore a = -1\)
\(\therefore a = -1\), \(b = -22\)In simple words: Since the function is continuous at \(x=3\), we equate \(f(3)\) with both the left-hand limit and the right-hand limit. We use \(f(3) = \lim_{x \to 3^-} f(x)\) to find 'b', and \(f(3) = \lim_{x \to 3^+} f(x)\) to find 'a', resulting in \(a=-1\) and \(b=-22\).

🎯 Exam Tip: When simplifying rational expressions like \(\frac{x^2-9}{x-3}\), remember to factorize the numerator \((x^2-9) = (x-3)(x+3)\) to cancel common terms, which is crucial for evaluating the limit at the point of discontinuity in the rational part.

Question 3. If \(f(x) = \frac{32^x-1}{8^x-1} + a\), for \(x > 0\) \(= 2\), for \(x = 0\) \(= x + 5 – 2b\), for \(x < 0\) is continuous at \(x = 0\)
Answer:Solution: f is continuous at \(x = 0\)
\(\therefore \lim_{x \to 0^-} f(x) = f(0)\)
\(\therefore \lim_{x \to 0^-} (x + 5 – 2b) = 2\)
\(\therefore 0 + 5 – 2b = 2\)
\(\therefore 5 – 2 = 2b\)
\(\therefore 3 = 2b\)
\(\therefore b = \frac{3}{2}\) Also \(\lim_{x \to 0^+} f(x) = f(0)\)
\(\therefore \lim_{x \to 0^+} \left( \frac{32^x-1}{8^x-1} + a \right) = 2\)
\(\therefore \lim_{x \to 0^+} \left( \frac{\frac{32^x-1}{x}}{\frac{8^x-1}{x}} + a \right) = 2\)
\(\ldots[\therefore \lim_{x \to 0} \frac{a^x-1}{x} = \log a]\)
\(\therefore \frac{\log 32}{\log 8} + a = 2\)
\(\therefore \frac{\log (2)^5}{\log (2)^3} + a = 2\)
\(\therefore \frac{5 \log 2}{3 \log 2} + a = 2\)
\(\therefore \frac{5}{3} + a = 2\)
\(\therefore a = 2 - \frac{5}{3}\)
\(\therefore a = \frac{6-5}{3}\)
\(\therefore a = \frac{1}{3}\)
\(\therefore a = \frac{1}{3}\) and \(b = \frac{3}{2}\)In simple words: For continuity at \(x=0\), the left-hand limit, right-hand limit, and \(f(0)\) must all be equal. We use \(\lim_{x \to 0^-} f(x) = f(0)\) to find 'b' and \(\lim_{x \to 0^+} f(x) = f(0)\) along with logarithmic properties to find 'a', giving \(a=\frac{1}{3}\) and \(b=\frac{3}{2}\).

🎯 Exam Tip: When evaluating limits of exponential functions divided by each other, divide both numerator and denominator by 'x' (or a suitable power of x) to leverage the standard limit \(\lim_{x \to 0} \frac{a^x-1}{x} = \log a\).