CBSE Class 12 Mathematics Continuity and Differentiability Important Questions Set B

Fill in the blanks

Question. If \( f(x) = |\cos x| \), then \( f'(\frac{\pi}{4}) = \) _____________ .
Answer: \( -\frac{1}{\sqrt{2}} \)

Question. If \( f(x) = (x + 1) \), then \( \frac{d}{dx} fof(x) = \) _____________ .
Answer: 1

Question. \( \frac{d}{dx} \sec(\tan^{-1} x) = \) _____________ .
Answer: \( \frac{x}{\sqrt{1+x^2}} \)

Question. The number of points at which the function \( f(x) = \frac{1}{\log |x|} \) is discontinuous is _________ .
Answer: three

Very Short Answer Questions

Question. If \( f(x) = \begin{cases} \frac{\sin^{-1} x}{x}, & x \neq 0 \\ k, & x = 0 \end{cases} \), is continuous at \( x = 0 \), then write the value of \( k \).
Answer: \( k = 1 \)

Question. Determine the value of ‘k’ for which the following function is continuous at \( x = 3 \): \( f(x) = \begin{cases} \frac{(x + 3)^2 - 36}{x - 3}, & x \neq 3 \\ k, & x = 3 \end{cases} \).
Answer: \( k = 12 \)

Question. If \( y = [\sin \frac{x}{2} + \cos \frac{x}{2}]^2 \), find \( \frac{dy}{dx} \) at \( x = \frac{\pi}{6} \).
Answer: \( \frac{\sqrt{3}}{2} \)

Question. Find \( \frac{dy}{dx} \), if \( y = \tan^{-1} [\frac{1 + \tan x}{1 - \tan x}], x \in [-\frac{\pi}{4}, \frac{\pi}{4}] \).
Answer: 1

Question. If \( y = \log (e^x) \), then find \( \frac{dy}{dx} \).
Answer: 1

Question. If \( y = \log[x + \sqrt{x^2 + a^2}] \), show that \( (x^2 + a^2) \frac{d^2 y}{dx^2} + x \frac{dy}{dx} = 0 \). 
Answer: Given \( y = \log[x + \sqrt{x^2 + a^2}] \)
\( \Rightarrow \frac{dy}{dx} = \frac{1}{x + \sqrt{x^2 + a^2}} \left[ 1 + \frac{2x}{2\sqrt{x^2 + a^2}} \right] = \frac{x + \sqrt{x^2 + a^2}}{(x + \sqrt{x^2 + a^2})(\sqrt{x^2 + a^2})} \)
\( \Rightarrow \frac{dy}{dx} = \frac{1}{\sqrt{x^2 + a^2}} \) ...(i)
Differentiating again with respect to \( x \), we get
\( \frac{d^2 y}{dx^2} = -\frac{1}{2}(x^2 + a^2)^{-3/2} \cdot 2x = \frac{-x}{(x^2 + a^2)^{3/2}} \)
\( \Rightarrow \frac{d^2 y}{dx^2} = \frac{-x}{(x^2 + a^2)\sqrt{x^2 + a^2}} \Rightarrow (x^2 + a^2) \frac{d^2 y}{dx^2} = -\frac{x}{\sqrt{x^2 + a^2}} \)
\( \Rightarrow (x^2 + a^2) \frac{d^2 y}{dx^2} + x \cdot \frac{dy}{dx} = 0 \) [From (i)]

Question. If \( y = x^x \), then prove that \( \frac{d^2 y}{dx^2} - \frac{1}{y} \left( \frac{dy}{dx} \right)^2 - \frac{y}{x} = 0 \). 
Answer: Given, \( y = x^x \)
Taking logarithm on both sides, we get
\( \log y = x \cdot \log x \)
Differentiating both sides, we get
\( \frac{1}{y} \frac{dy}{dx} = x \cdot \frac{1}{x} + \log x \Rightarrow \frac{dy}{dx} = y(1 + \log x) \) ...(i)
Again differentiating both sides, we get
\( \frac{d^2 y}{dx^2} = y \cdot \frac{1}{x} + (1 + \log x) \cdot \frac{dy}{dx} \Rightarrow \frac{d^2 y}{dx^2} = \frac{y}{x} + \frac{1}{y} \frac{dy}{dx} \cdot \frac{dy}{dx} \) [From (i)]
\( \Rightarrow \frac{d^2 y}{dx^2} = \frac{y}{x} + \frac{1}{y} \left( \frac{dy}{dx} \right)^2 \Rightarrow \frac{d^2 y}{dx^2} - \frac{1}{y} \left( \frac{dy}{dx} \right)^2 - \frac{y}{x} = 0 \)

Question. If \( y = (x + \sqrt{1 + x^2})^n \), then show that \( (1 + x^2) \frac{d^2 y}{dx^2} + x \frac{dy}{dx} = n^2 y \). 
Answer: Given \( y = (x + \sqrt{1 + x^2})^n \)
Differentiating with respect to \( x \), we get
\( \frac{dy}{dx} = n(x + \sqrt{1 + x^2})^{n-1} \cdot \left[ 1 + \frac{2x}{2\sqrt{1 + x^2}} \right] \Rightarrow \frac{dy}{dx} = n(x + \sqrt{1 + x^2})^{n-1} \cdot \left( \frac{x + \sqrt{1 + x^2}}{\sqrt{1 + x^2}} \right) \)
\( \Rightarrow \frac{dy}{dx} = \frac{n(x + \sqrt{1 + x^2})^n}{\sqrt{1 + x^2}} \Rightarrow \frac{dy}{dx} = \frac{ny}{\sqrt{1 + x^2}} \)
\( \Rightarrow \sqrt{1 + x^2} \cdot \frac{dy}{dx} = ny \)
Again differentiating with respect to \( x \), we get
\( \sqrt{1 + x^2} \cdot \frac{d^2 y}{dx^2} + \frac{2x}{2\sqrt{1 + x^2}} \cdot \frac{dy}{dx} = n \frac{dy}{dx} \)
\( \Rightarrow (1 + x^2) \frac{d^2 y}{dx^2} + x \cdot \frac{dy}{dx} = n \cdot \sqrt{1 + x^2} \cdot \frac{dy}{dx} \)
\( \Rightarrow (1 + x^2) \frac{d^2 y}{dx^2} + x \cdot \frac{dy}{dx} = n \cdot ny \Rightarrow (1 + x^2) \frac{d^2 y}{dx^2} + x \frac{dy}{dx} = n^2 y \)

Question. If \( y = \log \left( \sqrt{x} + \frac{1}{\sqrt{x}} \right)^2 \), then prove that \( x(x + 1)^2 y_2 + (x + 1)^2 y_1 = 2 \).
Answer: \( y = \log \left( \sqrt{x} + \frac{1}{\sqrt{x}} \right)^2 \Rightarrow y = 2 \log \left( \frac{x + 1}{\sqrt{x}} \right) \)
\( y = 2 \log (x + 1) - 2 \log \sqrt{x} \Rightarrow y = 2 \log (x + 1) - \log x \)
\( \Rightarrow y_1 = \frac{2}{x + 1} - \frac{1}{x} = \frac{2x - x - 1}{x(x + 1)} \)
\( \Rightarrow y_1 = \frac{x - 1}{x(x + 1)} \)
\( \Rightarrow y_2 = \frac{x(x + 1)(1) - (x - 1)(2x + 1)}{x^2(x + 1)^2} \Rightarrow y_2 = \frac{x^2 + x - (2x^2 - x - 1)}{x^2(x + 1)^2} \)
\( \Rightarrow y_2 = \frac{-x^2 + 2x + 1}{x^2(x + 1)^2} \)
Now, \( x(x + 1)^2 y_2 + (x + 1)^2 y_1 = x(x + 1)^2 \cdot \frac{-x^2 + 2x + 1}{x^2(x + 1)^2} + (x + 1)^2 \cdot \frac{(x - 1)}{x(x + 1)} \)
\( = \frac{-x^2 + 2x + 1}{x} + \frac{(x + 1)(x - 1)}{x} \)
\( = \frac{-x^2 + 2x + 1 + x^2 - 1}{x} = \frac{2x}{x} = 2 \)
Hence proved.

Question. If \( y = \sin(\sin x) \), prove that \( \frac{d^2 y}{dx^2} + \tan x \frac{dy}{dx} + y \cos^2 x = 0 \). 
Answer: \( y = \sin(\sin x) \)
\( \Rightarrow \frac{dy}{dx} = \cos(\sin x) \frac{d}{dx}(\sin x) \Rightarrow \frac{dy}{dx} = \cos(\sin x) \cos x \)
Again differentiating w.r.t \( x \) on both sides, we get
\( \frac{d^2 y}{dx^2} = \cos(\sin x) \frac{d}{dx}(\cos x) + (\cos x) \frac{d}{dx} \cos(\sin x) \)
\( \Rightarrow \frac{d^2 y}{dx^2} = \cos(\sin x) (-\sin x) - (\cos x) \{ \sin(\sin x) \} (\cos x) \)
\( \Rightarrow \frac{d^2 y}{dx^2} = -\sin x \cos(\sin x) - \cos^2 x \sin(\sin x) \)
Putting these values in LHS, we get
\( \frac{d^2 y}{dx^2} + \tan x \frac{dy}{dx} + y \cos^2 x \)
\( = \{ -\sin x \cos(\sin x) - \cos^2 x \sin(\sin x) \} + \tan x \{ \cos x \cos(\sin x) \} + y \cos^2 x \)
\( = -\sin x \cos(\sin x) - \cos^2 x \sin(\sin x) + \tan x \cos x \cos(\sin x) + y \cos^2 x \)
\( = -\sin x \cos(\sin x) - \cos^2 x \sin(\sin x) + \frac{\sin x}{\cos x} \cos x \cos(\sin x) + \sin(\sin x) \cos^2 x = 0 \)
Hence proved.

Rolle’s and Mean Value Theorem

Question. Verify Rolle’s theorem for the function \( f(x) = e^x \cos x \) in \( \left[ -\frac{\pi}{2}, \frac{\pi}{2} \right] \). 
Answer: Given function, \( f(x) = e^x \cos x \) in \( \left[ -\frac{\pi}{2}, \frac{\pi}{2} \right] \)
\( \therefore f'(x) = e^x (\cos x - \sin x) \), in \( \left( -\frac{\pi}{2}, \frac{\pi}{2} \right) \)
Clearly, \( f(x) \) is differentiable in \( \left( -\frac{\pi}{2}, \frac{\pi}{2} \right) \)
Thus, \( f(x) \) is differentiable in \( \left( -\frac{\pi}{2}, \frac{\pi}{2} \right) \)
\( \therefore f(x) \) must be continuous in \( \left[ -\frac{\pi}{2}, \frac{\pi}{2} \right] \).
Because every differentiable function must be continuous.
Now, \( f\left(-\frac{\pi}{2}\right) = e^{-\frac{\pi}{2}} \cos \left(-\frac{\pi}{2}\right) = 0 \)
and, \( f\left(\frac{\pi}{2}\right) = e^{\frac{\pi}{2}} \cos \left(\frac{\pi}{2}\right) = 0 \)
\( \therefore f\left(-\frac{\pi}{2}\right) = f\left(\frac{\pi}{2}\right) \)
Therefore, Rolle’s theorem is applicable.
So, there exists a real number \( c \) such that
\( f'(c) = 0 \Rightarrow e^c (\cos c - \sin c) = 0 \)
\( \Rightarrow \cos c - \sin c = 0 \) (\( \because e^c \neq 0 \))
\( \Rightarrow \cos c = \sin c \)
\( \Rightarrow \tan c = 1 = \tan \frac{\pi}{4} \Rightarrow c = \frac{\pi}{4} \)
\( \therefore c = \frac{\pi}{4} \in \left( -\frac{\pi}{2}, \frac{\pi}{2} \right) \)
Hence, Rolle’s theorem is verified.

Question. Verify Mean Value theorem for the function \( f(x) = 2 \sin x + \sin 2x \) on \( [0, \pi] \). 
Answer: We have, \( f(x) = 2 \sin x + \sin 2x \)
\( f(x) \) is continuous in \( [0, \pi] \) being trigonometric function and it is differentiable on \( (0, \pi) \).
Hence, condition of Mean Value Theorem is satisfied.
Therefore, Mean Value Theorem is applicable.
So, there exist a real number \( c \) such that
\( f'(c) = \frac{f(\pi) - f(0)}{\pi - 0} \) ...(i)
Now \( f(0) = 2 \sin 0 + \sin 0 = 0 \); \( f(\pi) = 2 \sin \pi + \sin 2\pi = 0 \) and \( f'(x) = 2 \cos x + 2 \cos 2x \)
\( \therefore f'(c) = 2 \cos c + 2 \cos 2c \)
From (i)
\( 2 \cos c + 2 \cos 2c = \frac{0 - 0}{\pi} \)
\( \Rightarrow 2 \cos c + 2 \cos 2c = 0 \Rightarrow 2 \cos c + 2(2 \cos^2 c - 1) = 0 \)
\( \Rightarrow \cos c + 2 \cos^2 c - 1 = 0 \Rightarrow 2 \cos^2 c + \cos c - 1 = 0 \)
\( \Rightarrow 2 \cos^2 c + 2 \cos c - \cos c - 1 = 0 \Rightarrow 2 \cos c (\cos c + 1) - 1 (\cos c + 1) = 0 \)
\( \Rightarrow (\cos c + 1)(2 \cos c - 1) = 0 \Rightarrow \cos c = -1 \) and \( \cos c = \frac{1}{2} \)
\( \Rightarrow c = \pi \) and \( c = \frac{\pi}{3} \in (0, \pi) \Rightarrow c = \frac{\pi}{3} \in (0, \pi) \)
Hence, Mean Value Theorem is verified.

Question. Discuss the applicability of Rolle’s theorem on the function given by 
\( f(x) = \begin{cases} x^2 + 1, \text{ if } 0 \le x \le 1 \\ 3 - x, \text{ if } 1 < x \le 2 \end{cases} \)
Answer: We have, \( f(x) = \begin{cases} x^2 + 1, \text{ if } 0 \le x \le 1 \\ 3 - x, \text{ if } 1 < x \le 2 \end{cases} \)
We know that, polynomial function is everywhere continuous and differentiable.
So, \( f(x) \) is continuous and differentiable at all points except possibly at \( x = 1 \).
Now, check the differentiability at \( x = 1 \),
At \( x = 1 \),
LHD \( = \lim_{x \to 1^-} \frac{f(x) - f(1)}{x - 1} \)
\( = \lim_{x \to 1^-} \frac{(x^2 + 1) - (1 + 1)}{x - 1} \) [\( \because f(x) = x^2 + 1, 0 \le x \le 1 \)]
\( = \lim_{x \to 1^-} \frac{x^2 - 1}{x - 1} = \lim_{x \to 1^-} \frac{(x + 1)(x - 1)}{x - 1} = 2 \)
and RHD \( = \lim_{x \to 1^+} \frac{f(x) - f(1)}{x - 1} = \lim_{x \to 1^+} \frac{(3 - x) - (1 + 1)}{(x - 1)} \)
\( = \lim_{x \to 1^+} \frac{3 - x - 2}{x - 1} = \lim_{x \to 1^+} \frac{-(x - 1)}{x - 1} = -1 \)
\( \therefore \) LHD \( \neq \) RHD
So, \( f(x) \) is not differentiable at \( x = 1 \).
Hence, Rolle’s theorem is not applicable on the interval \( [0, 2] \).

Question. Find a point on the curve \( y = (x - 3)^2 \), where the tangent is parallel to the chord joining the points \( (3, 0) \) and \( (4, 1) \). [
Answer: We have, \( y = (x - 3)^2 \), which is continuous in \( x_1 = 3 \) and \( x_2 = 4 \) i.e., \( [3, 4] \).
Also, \( y' = 2(x - 3) \times 1 = 2(x - 3) \) which exists in \( (3, 4) \).
Hence, by mean value theorem there exists a point on the curve at which tangent drawn is parallel to the chord joining the points \( (3, 0) \) and \( (4, 1) \).
Thus, \( f'(c) = \frac{f(4) - f(3)}{4 - 3} \)
\( \Rightarrow 2(c - 3) = \frac{(4 - 3)^2 - (3 - 3)^2}{4 - 3} \)
\( \Rightarrow 2c - 6 = \frac{1 - 0}{1} \Rightarrow c = \frac{7}{2} \)
For \( x = \frac{7}{2} \), \( y = \left( \frac{7}{2} - 3 \right)^2 = \left( \frac{1}{2} \right)^2 = \frac{1}{4} \)
So, \( \left( \frac{7}{2}, \frac{1}{4} \right) \) is the point on the curve at which tangent drawn is parallel to the chord joining the points \( (3, 0) \) and \( (4, 1) \).

Objective Type Questions

Question. Choose and write the correct option in each of the following questions.
(i) If \( f(x) = 2x \) and \( g(x) = \frac{x^2}{2} + 1 \), then which of the following can be a discontinuous function?
(a) \( f(x) + g(x) \)
(b) \( f(x) - g(x) \)
(c) \( f(x) \cdot g(x) \)
(d) \( \frac{g(x)}{f(x)} \)
Answer: d

Question. (ii) The set of points where the function \( f \) given by \( f(x) = |2x - 1| \sin x \) is differentiable is 
(a) \( R \)
(b) \( R - \left\{ \frac{1}{2} \right\} \)
(c) \( (0, \infty) \)
(d) none of these
Answer: b

Question. Let \( f(x) = |\cos x| \). Then,
(a) \( f \) is everywhere differentiable.
(b) \( f \) is everywhere continuous but not differentiable at \( x = n\pi, n \in Z \).
(c) \( f \) is everywhere continuous but not differentiable at \( x = (2n + 1)\frac{\pi}{2}, n \in Z \).
(d) none of these.
Answer: (c)

Question. If \( y = A e^{5x} + B e^{-5x} \), then \( \frac{d^2 y}{dx^2} \) is equal to
(a) \( 25 y \)
(b) \( 5 y \)
(c) \( -25 y \)
(d) \( 15 y \)
Answer: (a)

Question. For the curve \( \sqrt{x} + \sqrt{y} = 1 \), \( \frac{dy}{dx} \) at \( (\frac{1}{4}, \frac{1}{4}) \) is
(a) \( \frac{1}{2} \)
(b) 1
(c) \( -1 \)
(d) 2
Answer: (c)

Question. If \( f'(1) = 2 \) and \( y = f(\log e^x) \), then \( \frac{dy}{dx} \) at \( x = e \) is
(a) 0
(b) 1
(c) \( e \)
(d) \( \frac{2}{e} \)
Answer: (d)