CBSE Class 12 Mathematics HOTs Continuity And Differentiability Set 02

Very Short Answer Type Questions

Question. If \( y = x|x| \), find \( \frac{dy}{dx} \) for \( x < 0 \).
Answer: We have \( y = x|x| \).
When \( x < 0 \), then \( |x| = -x \). \[ y = x(-x) = -x^2 \] Differentiating both sides w.r.t. \( x \): \[ \frac{dy}{dx} = -2x \]

Question. Find \( \frac{dy}{dx} \), when \( 2x + 3y = \sin y \).
Answer: Given \( 2x + 3y = \sin y \).
Differentiating both sides w.r.t. \( x \), we get: \[ 2 + 3\frac{dy}{dx} = \cos y \left(\frac{dy}{dx}\right) \] Rearranging the terms: \[ \frac{dy}{dx} (\cos y - 3) = 2 \] \[ \frac{dy}{dx} = \frac{2}{\cos y - 3} \]

Question. Find the derivative of \( \sin(\tan^{-1} x) \) w.r.t. \( x \).
Answer: Let \( y = \sin(\tan^{-1} x) \).
On differentiating both sides w.r.t. \( x \), we get: \[ \frac{dy}{dx} = \cos(\tan^{-1} x) \cdot \frac{d}{dx}(\tan^{-1} x) \] \[ = \frac{\cos(\tan^{-1} x)}{1+x^2} \]

Question. If \( y\sqrt{1-x^2} + x\sqrt{1-y^2} = 1 \), then prove that \( \frac{dy}{dx} = -\sqrt{\frac{1-y^2}{1-x^2}} \).
Answer: Given, \( y\sqrt{1-x^2} + x\sqrt{1-y^2} = 1 \)      ...(i)
On putting \( x = \sin A \) and \( y = \sin B \) in Eq. (i), we get: \[ \sin B \cos A + \sin A \cos B = 1 \implies \sin(A+B) = 1 \] \[ A + B = \sin^{-1}(1) = \frac{\pi}{2} \] Substituting back \( A \) and \( B \): \[ \sin^{-1} x + \sin^{-1} y = \frac{\pi}{2} \]      ...(ii)
Now, differentiating both sides of Eq. (ii) w.r.t. \( x \), we get: \[ \frac{1}{\sqrt{1-x^2}} + \frac{1}{\sqrt{1-y^2}}\left(\frac{dy}{dx}\right) = 0 \] \[ \frac{dy}{dx} = -\frac{\sqrt{1-y^2}}{\sqrt{1-x^2}} = -\sqrt{\frac{1-y^2}{1-x^2}} \]

Question. Differentiate \( e^{\sqrt{x}} \) w.r.t. \( x \).
Answer: Let \( y = e^{\sqrt{x}} \).
On differentiating both sides w.r.t. \( x \), we get: \[ \frac{dy}{dx} = \frac{d}{dx}(e^{\sqrt{x}}) \] \[ = e^{\sqrt{x}} \frac{d}{dx}(\sqrt{x}) \] \[ = e^{\sqrt{x}} \cdot \frac{1}{2\sqrt{x}} = \frac{e^{\sqrt{x}}}{2\sqrt{x}} \]

Question. Differentiate \( e^{3x} \cdot \cos 2x \) w.r.t. \( x \).
Answer: Let \( y = e^{3x} \cdot \cos 2x \).
On differentiating both sides w.r.t. \( x \), we get: \[ \frac{dy}{dx} = \frac{d}{dx}(e^{3x} \cdot \cos 2x) \] Using the product rule: \[ \frac{dy}{dx} = e^{3x} \frac{d}{dx}(\cos 2x) + \cos 2x \frac{d}{dx}(e^{3x}) \] \[ = e^{3x} (-2 \sin 2x) + \cos 2x (3 e^{3x}) \] \[ = e^{3x} (3 \cos 2x - 2 \sin 2x) \]

Question. Find the derivative of \( \log(\sin x) \) w.r.t. \( x \).
Answer: Let \( y = \log(\sin x) \).
On differentiating both sides w.r.t. \( x \), we get: \[ \frac{dy}{dx} = \frac{d}{dx}(\log(\sin x)) \] Using the chain rule: \[ \frac{dy}{dx} = \frac{1}{\sin x} \cdot \frac{d}{dx}(\sin x) \] \[ = \frac{\cos x}{\sin x} = \cot x \]

Question. Find the second derivative of \( \log x \).
Answer: Let \( y = \log x \).
On differentiating both sides w.r.t. \( x \), we get: \[ \frac{dy}{dx} = \frac{1}{x} \] Again, differentiating both sides w.r.t. \( x \), we get: \[ \frac{d^2y}{dx^2} = \frac{d}{dx}\left(\frac{1}{x}\right) = -\frac{1}{x^2} \]

Short Answer Type Questions

Question. Discuss the continuity of secant.
Answer: Let \( f(x) = \sec x = \frac{1}{\cos x} \).
The domain of \( f(x) \) is all real numbers except those where \( \cos x = 0 \), i.e., \( x = (2n + 1)\frac{\pi}{2} \) for \( n \in \mathbb{Z} \).
Since \( \cos x \) is continuous everywhere, its reciprocal \( \sec x \) is continuous at all points in its domain. Thus, the secant function is continuous everywhere except at \( x = (2n+1)\frac{\pi}{2} \) where \( n \in \mathbb{Z} \).

Question. Examine the continuity of the function \[ f(x) = \begin{cases} \frac{|\sin x|}{x}, & \text{if } x \neq 0 \\ 1, & \text{if } x = 0 \end{cases} \] at \( x = 0 \).
Answer: Let us find the LHL and RHL at \( x = 0 \): \[ \text{LHL} = \lim_{x \to 0^-} f(x) = \lim_{x \to 0^-} \frac{-\sin x}{x} = -1 \] \[ \text{RHL} = \lim_{x \to 0^+} f(x) = \lim_{x \to 0^+} \frac{\sin x}{x} = 1 \] Since \( \text{LHL} \neq \text{RHL} \), the limit of \( f(x) \) as \( x \to 0 \) does not exist.
Therefore, \( f(x) \) is not continuous at \( x = 0 \).

Question. If the function \( f(x) = \begin{cases} 3ax + b, & \text{if } x > 1 \\ 11, & \text{if } x = 1 \\ 5ax - 2b, & \text{if } x < 1 \end{cases} \) is continuous at \( x = 1 \), then find the values of \( a \) and \( b \).
Answer: Since \( f(x) \) is continuous at \( x = 1 \), we have: \[ \text{LHL} = \text{RHL} = f(1) \] Evaluate the limits: \[ \text{LHL} = \lim_{x \to 1^-} f(x) = \lim_{x \to 1^-} (5ax - 2b) = 5a - 2b \] \[ \text{RHL} = \lim_{x \to 1^+} f(x) = \lim_{x \to 1^+} (3ax + b) = 3a + b \] Also, \( f(1) = 11 \).
Thus, we set up the system of equations: \[ 5a - 2b = 11 \]      ...(i) \[ 3a + b = 11 \]      ...(ii) Multiplying Eq. (ii) by 2 and adding Eq. (i): \[ (5a - 2b) + 2(3a + b) = 11 + 22 \] \[ 11a = 33 \implies a = 3 \] Substituting \( a = 3 \) in Eq. (ii): \[ 3(3) + b = 11 \implies b = 2 \] Hence, \( a = 3 \) and \( b = 2 \).

Question. Find the value of \( k \), if the function \[ f(x) = \begin{cases} \frac{1-\sin x}{(\pi - 2x)^2}, & \text{if } x \neq \frac{\pi}{2} \\ k, & \text{if } x = \frac{\pi}{2} \end{cases} \] is continuous at \( x = \frac{\pi}{2} \).
Answer: Since \( f(x) \) is continuous at \( x = \frac{\pi}{2} \), we have: \[ \lim_{x \to \frac{\pi}{2}} f(x) = f\left(\frac{\pi}{2}\right) = k \] Let \( x = \frac{\pi}{2} + h \). As \( x \to \frac{\pi}{2} \), \( h \to 0 \). \[ k = \lim_{h \to 0} \frac{1-\sin\left(\frac{\pi}{2}+h\right)}{\left(\pi - 2\left(\frac{\pi}{2}+h\right)\right)^2} \] \[ = \lim_{h \to 0} \frac{1-\cos h}{(-2h)^2} \] \[ = \lim_{h \to 0} \frac{2\sin^2\left(\frac{h}{2}\right)}{4h^2} \] \[ = \frac{2}{4} \lim_{h \to 0} \left( \frac{\sin\left(\frac{h}{2}\right)}{2 \cdot \frac{h}{2}} \right)^2 \] \[ = \frac{1}{2} \cdot \lim_{h \to 0} \left( \frac{\sin\left(\frac{h}{2}\right)}{\frac{h}{2}} \cdot \frac{1}{2} \right)^2 \] \[ = \frac{1}{2} \cdot \left( 1 \cdot \frac{1}{2} \right)^2 = \frac{1}{2} \cdot \frac{1}{4} = \frac{1}{8} \] Hence, \( k = \frac{1}{8} \).

Question. Find the value of \( k \) so that the function \( f \) is continuous at \( x = \frac{\pi}{2} \), where \[ f(x) = \begin{cases} \frac{k \cos x}{\pi - 2x}, & \text{if } x \neq \frac{\pi}{2} \\ 3, & \text{if } x = \frac{\pi}{2} \end{cases} \]
Answer: Since \( f(x) \) is continuous at \( x = \frac{\pi}{2} \), we have: \[ \lim_{x \to \frac{\pi}{2}} f(x) = f\left(\frac{\pi}{2}\right) = 3 \] Let \( x = \frac{\pi}{2} + h \). As \( x \to \frac{\pi}{2} \), \( h \to 0 \). \[ \lim_{h \to 0} \frac{k \cos\left(\frac{\pi}{2}+h\right)}{\pi - 2\left(\frac{\pi}{2}+h\right)} = 3 \] \[ \lim_{h \to 0} \frac{k(-\sin h)}{-2h} = 3 \] \[ \frac{k}{2} \lim_{h \to 0} \frac{\sin h}{h} = 3 \] Since \( \lim_{h \to 0} \frac{\sin h}{h} = 1 \): \[ \frac{k}{2}(1) = 3 \implies k = 6 \]

Question. The function \( f(x) = \begin{cases} x^2, & \text{if } 0 \leq x < 1 \\ a, & \text{if } 1 \leq x < \sqrt{2} \\ \frac{2b^2-4b}{a}, & \text{if } \sqrt{2} \leq x < \infty \end{cases} \) is continuous on \( [0, \infty) \). Find the most suitable values of \( a \) and \( b \).
Answer: Since \( f(x) \) is continuous on \( [0, \infty) \), it must be continuous at the transition points \( x = 1 \) and \( x = \sqrt{2} \).
For continuity at \( x = 1 \): \[ \lim_{x \to 1^-} f(x) = \lim_{x \to 1^+} f(x) = f(1) \] \[ \lim_{x \to 1^-} x^2 = a \implies 1 = a \implies a = 1 \] For continuity at \( x = \sqrt{2} \): \[ \lim_{x \to \sqrt{2}^-} f(x) = \lim_{x \to \sqrt{2}^+} f(x) = f(\sqrt{2}) \] \[ a = \frac{2b^2-4b}{a} \] Substituting \( a = 1 \): \[ 1 = \frac{2b^2-4b}{1} \implies 2b^2 - 4b - 1 = 0 \] Solving the quadratic equation for \( b \): \[ b = \frac{-(-4) \pm \sqrt{(-4)^2 - 4(2)(-1)}}{2(2)} \] \[ = \frac{4 \pm \sqrt{16+8}}{4} = \frac{4 \pm 2\sqrt{6}}{4} = 1 \pm \frac{\sqrt{6}}{2} \] Thus, the most suitable values are \( a = 1 \) and \( b = 1 \pm \frac{\sqrt{6}}{2} \). (If \( a = -1 \) is considered, \( b = 1 \pm \frac{\sqrt{6}}{2} \) is also obtained. In some interpretations, \( a = -1, b = 1 \) or \( a = 1, b = 1 \pm \sqrt{2} \) are given depending on alternative expressions of the boundary condition).

Question. If the function \( f(x) \) defined by \[ f(x) = \begin{cases} \frac{\log(1+ax) - \log(1-bx)}{x}, & \text{if } x \neq 0 \\ k, & \text{if } x = 0 \end{cases} \] is continuous at \( x = 0 \), then find the value of \( k \).
Answer: Since the function \( f(x) \) is continuous at \( x = 0 \), we have: \[ \lim_{x \to 0} f(x) = f(0) = k \] \[ \lim_{x \to 0} \frac{\log(1+ax) - \log(1-bx)}{x} = k \] \[ \lim_{x \to 0} \left[ \frac{\log(1+ax)}{x} - \frac{\log(1-bx)}{x} \right] = k \] \[ a \lim_{x \to 0} \frac{\log(1+ax)}{ax} - (-b) \lim_{x \to 0} \frac{\log(1-bx)}{-bx} = k \] Using the standard limit \( \lim_{t \to 0} \frac{\log(1+t)}{t} = 1 \), we get: \[ a(1) - (-b)(1) = k \implies k = a + b \]

Long Answer Type Questions

Question. Find the value of \( a \) for which the function \( f \) is defined as \[ f(x) = \begin{cases} a \sin \frac{\pi}{2}(x+1), & x \leq 0 \\ \frac{\tan x - \sin x}{x^3}, & x > 0 \end{cases} \] is continuous at \( x = 0 \).
Answer: Since \( f(x) \) is continuous at \( x = 0 \), we have: \[ \text{LHL} = \text{RHL} = f(0) \] Let's evaluate LHL: \[ \text{LHL} = \lim_{x \to 0^-} a \sin \frac{\pi}{2}(x+1) = a \sin \frac{\pi}{2} = a \] Let's evaluate RHL: \[ \text{RHL} = \lim_{x \to 0^+} \frac{\tan x - \sin x}{x^3} \] \[ = \lim_{x \to 0^+} \frac{\frac{\sin x}{\cos x} - \sin x}{x^3} \] \[ = \lim_{x \to 0^+} \frac{\sin x(1-\cos x)}{x^3 \cos x} \] \[ = \lim_{x \to 0^+} \left( \frac{\sin x}{x} \cdot \frac{1-\cos x}{x^2} \cdot \frac{1}{\cos x} \right) \] Using the limits \( \lim_{x \to 0} \frac{\sin x}{x} = 1 \), \( \lim_{x \to 0} \frac{1-\cos x}{x^2} = \frac{1}{2} \), and \( \lim_{x \to 0} \cos x = 1 \): \[ \text{RHL} = 1 \cdot \frac{1}{2} \cdot \frac{1}{1} = \frac{1}{2} \] Equating LHL and RHL: \[ a = \frac{1}{2} \]

Question. Find the value of \( k \) so that the function \( f \) defined by \( f(x) = \begin{cases} kx + 1, & \text{if } x \le \pi \\ \cos x, & \text{if } x > \pi \end{cases} \) is continuous at \( x = \pi \).
Answer: Given, \( f(x) = \begin{cases} kx + 1, & \text{if } x \le \pi \\ \cos x, & \text{if } x > \pi \end{cases} \) is continuous at \( x = \pi \).
Then, \( \text{LHL} = \text{RHL} = f(\pi) \) ...(i)
Now, \( \text{LHL} = \lim_{x \to \pi^-} f(x) = \lim_{x \to \pi^-} (kx + 1) \)
\( = \lim_{h \to 0} [k(\pi - h) + 1] \) [since \( x = \pi - h \); when \( x \to \pi^- \), then \( h \to 0 \)]
\( = \lim_{h \to 0} (k\pi - kh + 1) = k\pi + 1 \)
and \( \text{RHL} = \lim_{x \to \pi^+} f(x) = \lim_{x \to \pi^+} \cos x \)
\( = \lim_{h \to 0} \cos(\pi + h) \) [since \( x = \pi + h \); when \( x \to \pi^+ \), then \( h \to 0 \)]
\( = \cos \pi = -1 \) [since \( \cos \pi = -1 \)]
Now, from Eq. (i), we get
\( \text{LHL} = \text{RHL} \Rightarrow k\pi + 1 = -1 \Rightarrow k\pi = -2 \Rightarrow k = -\frac{2}{\pi} \)

Question. For what values of \( \lambda \), is the function \( f(x) = \begin{cases} \lambda(x^2 - 2x), & \text{if } x \le 0 \\ 4x + 1, & \text{if } x > 0 \end{cases} \) continuous at \( x = 0 \)?
Answer: Let \( f(x) = \begin{cases} \lambda(x^2 - 2x), & \text{if } x \le 0 \\ 4x + 1, & \text{if } x > 0 \end{cases} \) is continuous at \( x = 0 \).
Then, \( (\text{LHL})_{x=0} = (\text{RHL})_{x=0} = f(0) \) ...(i)
Now, \( f(0) = \lambda[0 - 0] = 0 \),
\( \text{LHL} = \lim_{x \to 0^-} f(x) = \lim_{h \to 0} f(0 - h) \)
\( = \lambda \lim_{h \to 0} [(0 - h)^2 - 2(0 - h)] \)
\( = \lambda \times 0 = 0 \)
and \( \text{RHL} = \lim_{x \to 0^+} f(x) = \lim_{h \to 0} f(0 + h) \)
\( = \lim_{h \to 0} [4(0 + h) + 1] = 1 \)
\( \because \text{LHL} \neq \text{RHL} \), which is a contradiction to Eq. (i).
\( \therefore \) There is no value of \( \lambda \) for which \( f(x) \) is continuous at \( x = 0 \).

Question. Find the value of \( a \), if the function \( f(x) \) defined by \( f(x) = \begin{cases} 2x - 1, & x < 2 \\ a, & x = 2 \\ x + 1, & x > 2 \end{cases} \) is continuous at \( x = 2 \). Also, discuss the continuity of \( f(x) \) at \( x = 3 \).
Answer: Given, \( f(x) = \begin{cases} 2x - 1, & x < 2 \\ a, & x = 2 \\ x + 1, & x > 2 \end{cases} \) is continuous at \( x = 2 \).
\( \therefore (\text{LHL})_{x=2} = (\text{RHL})_{x=2} = f(2) \) ...(i)
Now, \( f(2) = a \)
and \( \text{LHL} = \lim_{x \to 2^-} f(x) = \lim_{x \to 2^-} (2x - 1) \)
\( = \lim_{h \to 0} [2(2 - h) - 1] = 3 \) [since \( x = 2 - h \); when \( x \to 2^- \), then \( h \to 0 \)]
From Eq. (i), we get
\( \text{LHL} = f(2) \Rightarrow a = 3 \)
Now, let us check the continuity at \( x = 3 \).
Consider, \( \lim_{x \to 3} f(x) = \lim_{x \to 3} (x + 1) \) [since \( f(x) = x + 1 \) for \( x > 2 \)]
\( = 4 = f(3) \) [since \( f(3) = 3 + 1 = 4 \)]
\( \therefore f(x) \) is continuous at \( x = 3 \).

Question. If a function \( f \) is differentiable at a point \( c \), then prove that it is also continuous at that point.
Answer: Given, \( f \) is differentiable at \( c \).
\( \therefore \lim_{x \to c} \frac{f(x) - f(c)}{x - c} = f'(c) \)
Now, \( \lim_{x \to c} [f(x) - f(c)] = \lim_{x \to c} \left[ \frac{f(x) - f(c)}{x - c} \cdot (x - c) \right] \)
\( = \lim_{x \to c} \frac{f(x) - f(c)}{x - c} \cdot \lim_{x \to c} (x - c) = f'(c) \cdot 0 = 0 \)
\( \Rightarrow \lim_{x \to c} f(x) = f(c) \)
\( \therefore f(x) \) is continuous at \( x = c \).

Question. Show that the function \( f(x) = |x - 3|, x \in \mathbb{R} \) is continuous but not differentiable at \( x = 3 \).
Answer: Given, \( f(x) = |x - 3| \)
\( \Rightarrow f(x) = \begin{cases} -(x - 3), & x < 3 \\ 0, & x = 3 \\ x - 3, & x > 3 \end{cases} \)
For continuity:
\( \text{LHL} = \lim_{x \to 3^-} f(x) = \lim_{h \to 0} f(3 - h) \)
\( = \lim_{h \to 0} [-(3 - h - 3)] = \lim_{h \to 0} h = 0 \)
\( \text{RHL} = \lim_{x \to 3^+} f(x) = \lim_{h \to 0} f(3 + h) \)
\( = \lim_{h \to 0} (3 + h - 3) = \lim_{h \to 0} h = 0 \)
Also, \( f(3) = 0 \)
Since, \( \text{LHL} = \text{RHL} = f(3) \)
\( \therefore f(x) \) is continuous at \( x = 3 \).
For differentiability:
\( \text{LHD} = \lim_{h \to 0} \frac{f(3 - h) - f(3)}{-h} \)
\( = \lim_{h \to 0} \frac{-(3 - h - 3) - 0}{-h} \)
\( = \lim_{h \to 0} \frac{h}{-h} = -1 \)
\( \text{RHD} = \lim_{h \to 0} \frac{f(3 + h) - f(3)}{h} \)
\( = \lim_{h \to 0} \frac{(3 + h - 3) - 0}{h} \)
\( = \lim_{h \to 0} \frac{h}{h} = 1 \)
Since, \( \text{LHD} \neq \text{RHD} \)
\( \therefore f(x) \) is not differentiable at \( x = 3 \).

Question. Test the continuity and differentiability at \( x = 1 \) of the function \( f \) defined by \( f(x) = \begin{cases} 3^x, & -1 \le x < 1 \\ 4 - x, & 1 \le x < 4 \end{cases} \)
Answer: For continuity at \( x = 1 \):
\( \lim_{x \to 1^-} f(x) = \lim_{x \to 1^-} 3^x = 3^1 = 3 \)
\( \lim_{x \to 1^+} f(x) = \lim_{x \to 1^+} (4 - x) = 4 - 1 = 3 \)
and \( f(1) = 4 - 1 = 3 \)
\( \therefore \lim_{x \to 1^-} f(x) = \lim_{x \to 1^+} f(x) = f(1) \)
So, \( f(x) \) is continuous at \( x = 1 \).
For differentiability at \( x = 1 \):
\( Lf'(1) = \lim_{h \to 0^-} \frac{f(1 - h) - f(1)}{-h} = \lim_{h \to 0} \frac{3^{1 - h} - 3}{-h} = \lim_{h \to 0} \frac{3(3^{-h} - 1)}{-h} = 3 \log 3 \)
\( Rf'(1) = \lim_{h \to 0^+} \frac{f(1 + h) - f(1)}{h} = \lim_{h \to 0} \frac{4 - (1 + h) - 3}{h} = \lim_{h \to 0} \frac{-h}{h} = -1 \)
Since, \( Lf'(1) \neq Rf'(1) \), \( f(x) \) is continuous at \( x = 1 \) but not differentiable at \( x = 1 \).

Question. For what value of \( \lambda \) the function defined by \( f(x) = \begin{cases} \lambda(x^2 + 2), & \text{if } x \le 0 \\ 4x + 6, & \text{if } x > 0 \end{cases} \) is continuous at \( x = 0 \)? Hence, check the differentiability of \( f(x) \) at \( x = 0 \).
Answer: Given function is continuous at \( x = 0 \).
\( \text{LHL} = \lim_{x \to 0^-} f(x) = \lim_{h \to 0} \lambda((0-h)^2 + 2) = 2\lambda \)
\( \text{RHL} = \lim_{x \to 0^+} f(x) = \lim_{h \to 0} [4(0+h) + 6] = 6 \)
and \( f(0) = 2\lambda \)
Since the function is continuous at \( x = 0 \), we have:
\( 2\lambda = 6 \Rightarrow \lambda = 3 \)
For differentiability at \( x = 0 \) with \( \lambda = 3 \):
\( \text{LHD} = \lim_{h \to 0} \frac{f(0 - h) - f(0)}{-h} = \lim_{h \to 0} \frac{3((-h)^2 + 2) - 6}{-h} = \lim_{h \to 0} \frac{3h^2}{-h} = 0 \)
\( \text{RHD} = \lim_{h \to 0} \frac{f(0 + h) - f(0)}{h} = \lim_{h \to 0} \frac{4(0 + h) + 6 - 6}{h} = \lim_{h \to 0} \frac{4h}{h} = 4 \)
Since, \( \text{LHD} \neq \text{RHD} \), \( f(x) \) is not differentiable at \( x = 0 \).

Question. Find the values of \( a \) and \( b \) so that the following function is differentiable for all values of \( x \): \( f(x) = \begin{cases} ax + b, & x > -1 \\ bx^2 - 3, & x \le -1 \end{cases} \)
Answer: Given, \( f(x) = \begin{cases} ax + b, & x > -1 \\ bx^2 - 3, & x \le -1 \end{cases} \) is differentiable at \( x = -1 \).
\( \therefore Lf'(-1) = Rf'(-1) = f'(-1) \) ...(i)
Here, \( Lf'(-1) = \lim_{h \to 0} \frac{f(-1 - h) - f(-1)}{-h} \)
\( = \lim_{h \to 0} \frac{[b(-1 - h)^2 - 3] - [b(-1)^2 - 3]}{-h} \)
\( = \lim_{h \to 0} \frac{[b(1 + h^2 + 2h) - 3] - [b - 3]}{-h} \)
\( = \lim_{h \to 0} \frac{b + bh^2 + 2bh - 3 - b + 3}{-h} \)
\( = \lim_{h \to 0} \frac{h(bh + 2b)}{-h} = - \lim_{h \to 0} (2b + bh) = -2b \)
and \( Rf'(-1) = \lim_{h \to 0} \frac{f(-1 + h) - f(-1)}{h} \)
\( = \lim_{h \to 0} \frac{[a(-1 + h) + b] - (b - 3)}{h} \)
\( = \lim_{h \to 0} \frac{-a + ah + b - b + 3}{h} = \lim_{h \to 0} \frac{ah - a + 3}{h} \)
Clearly, for \( Rf'(-1) \) to exist, \( (-a + 3) \) should be equal to \( 0 \), i.e.,
\( -a + 3 = 0 \Rightarrow a = 3 \) ...(ii)
Now, \( Rf'(-1) = \lim_{h \to 0} \frac{ah}{h} = a \)
From Eq. (i), we get:
\( Lf'(-1) = Rf'(-1) \Rightarrow -2b = a \Rightarrow b = -\frac{a}{2} \Rightarrow b = -\frac{3}{2} \)
Hence, \( a = 3 \) and \( b = -\frac{3}{2} \).

Question. Find \( a \) and \( b \), if the function given by \( f(x) = \begin{cases} ax^2 + b, & \text{if } x < 1 \\ 2x + 1, & \text{if } x \ge 1 \end{cases} \) is differentiable at \( x = 1 \).
Answer: We have, \( f(x) = \begin{cases} ax^2 + b, & \text{if } x < 1 \\ 2x + 1, & \text{if } x \ge 1 \end{cases} \) is differentiable at \( x = 1 \).
\( \therefore f(x) \) is also continuous at \( x = 1 \).
Hence, \( \lim_{x \to 1^+} f(x) = \lim_{x \to 1^+} (2x + 1) = 2(1) + 1 = 3 \)
and \( \lim_{x \to 1^-} f(x) = \lim_{x \to 1^-} (ax^2 + b) = a(1)^2 + b = a + b \)
Thus, \( f(1) = 3 \).
As \( f(x) \) is continuous at \( x = 1 \), we have:
\( a + b = 3 \) ...(i)
Now, \( Lf'(1) = \lim_{h \to 0^-} \frac{f(1 - h) - f(1)}{-h} \)
\( = \lim_{h \to 0} \frac{a(1 - h)^2 + b - 3}{-h} \)
\( = \lim_{h \to 0} \frac{a(1 + h^2 - 2h) + b - 3}{-h} \)
\( = \lim_{h \to 0} \frac{a + ah^2 - 2ah + b - 3}{-h} \)
\( = \lim_{h \to 0} \frac{ah^2 - 2ah}{-h} \) [since \( a + b = 3 \)]
\( = \lim_{h \to 0} (-ah + 2a) = 2a \)
And \( Rf'(1) = \lim_{h \to 0^+} \frac{f(1 + h) - f(1)}{h} \)
\( = \lim_{h \to 0} \frac{2(1 + h) + 1 - 3}{h} = \lim_{h \to 0} \frac{2h}{h} = 2 \)
As \( f(x) \) is differentiable at \( x = 1 \), so \( Lf'(1) = Rf'(1) \).
So, we get:
\( 2a = 2 \Rightarrow a = 1 \)
\( \therefore a = 1 \) and \( b = 2 \) [from Eq. (i)].

Question. Find the values of \( a \) and \( b \), if the function \( f \) defined by \( f(x) = \begin{cases} x^2 + 3x + a, & x \le 1 \\ bx + 2, & x > 1 \end{cases} \) is differentiable at \( x = 1 \).
Answer: Given, \( f(x) = \begin{cases} x^2 + 3x + a, & x \le 1 \\ bx + 2, & x > 1 \end{cases} \) is differentiable at \( x = 1 \).
\( \therefore Lf'(1) = Rf'(1) \) ...(i)
Here, \( Lf'(1) = \lim_{h \to 0} \frac{f(1 - h) - f(1)}{-h} \)
\( = \lim_{h \to 0} \frac{(1 - h)^2 + 3(1 - h) + a - (4 + a)}{-h} \)
\( = \lim_{h \to 0} \frac{1 + h^2 - 2h + 3 - 3h + a - 4 - a}{-h} \)
\( = \lim_{h \to 0} \frac{h^2 - 5h}{-h} = \lim_{h \to 0} (5 - h) = 5 \)
and \( Rf'(1) = \lim_{h \to 0} \frac{f(1 + h) - f(1)}{h} \)
\( = \lim_{h \to 0} \frac{b(1 + h) + 2 - (4 + a)}{h} \)
\( = \lim_{h \to 0} \frac{b + bh + 2 - 4 - a}{h} = \lim_{h \to 0} \frac{bh + b - a - 2}{h} \)
Clearly, for \( Rf'(1) \) to exist, \( b - a - 2 \) should be equal to \( 0 \), i.e.,
\( b - a - 2 = 0 \) ...(ii)
Now, \( Rf'(1) = \lim_{h \to 0} \frac{bh}{h} = \lim_{h \to 0} b = b \)
From Eq. (i), we get:
\( Lf'(1) = Rf'(1) \Rightarrow 5 = b \Rightarrow b = 5 \)
Now, on substituting \( b = 5 \) in Eq. (ii), we get:
\( 5 - a - 2 = 0 \Rightarrow a = 3 \)
Hence, \( a = 3 \) and \( b = 5 \).

Question. If \( y = \frac{\sin(ax + b)}{\cos(cx + d)} \), then find \( \frac{dy}{dx} \).
Answer: Given, \( y = \frac{\sin(ax + b)}{\cos(cx + d)} \)
On differentiating both sides, w.r.t. \( x \), we get:
\( \frac{dy}{dx} = \frac{\cos(cx + d) \frac{d}{dx}[\sin(ax + b)] - \sin(ax + b) \frac{d}{dx}[\cos(cx + d)]}{\{\cos(cx + d)\}^2} \) [by quotient rule]
\( = \frac{\cos(cx + d) \cos(ax + b)(a) - \sin(ax + b) [-\sin(cx + d)(c)]}{\cos^2(cx + d)} \)
\( = \frac{a \cos(cx + d) \cos(ax + b) + c \sin(ax + b) \sin(cx + d)}{\cos^2(cx + d)} \)

Question. If \( x\sqrt{1 + y} + y\sqrt{1 + x} = 0 \), (\( x \neq y \)), then prove that \( \frac{dy}{dx} = -\frac{1}{(1 + x)^2} \).
Answer: Given equation is \( x\sqrt{1 + y} + y\sqrt{1 + x} = 0 \), where \( x \neq y \).
We first convert the given equation into \( y = f(x) \) form.
Clearly, \( x\sqrt{1 + y} = -y\sqrt{1 + x} \)
On squaring both sides, we get:
\( x^2(1 + y) = y^2(1 + x) \)
\( \Rightarrow x^2 + x^2y = y^2 + y^2x \)
\( \Rightarrow x^2 - y^2 = y^2x - x^2y \)
\( \Rightarrow (x - y)(x + y) = -xy(x - y) \) [since \( a^2 - b^2 = (a-b)(a+b) \)]
\( \Rightarrow (x - y)(x + y) + xy(x - y) = 0 \)
\( \Rightarrow (x - y)(x + y + xy) = 0 \)
Either \( x - y = 0 \) or \( x + y + xy = 0 \)
But it is given that \( x \neq y \).
So, \( y + xy + x = 0 \)
\( \Rightarrow y(1 + x) = -x \Rightarrow y = -\frac{x}{1 + x} \) ...(i)
On differentiating both sides w.r.t. \( x \), we get:
\( \frac{dy}{dx} = \frac{(1 + x) \frac{d}{dx}(-x) - (-x) \frac{d}{dx}(1 + x)}{(1 + x)^2} \) [using quotient rule]
\( \frac{dy}{dx} = \frac{(1 + x)(-1) + x(1)}{(1 + x)^2} = \frac{-1 - x + x}{(1 + x)^2} = \frac{-1}{(1 + x)^2} \)
Hence proved.

Question. If \( \sin y = x \cos(a + y) \), then show that \( \frac{dy}{dx} = \frac{\cos^2(a + y)}{\cos a} \). Also, show that \( \frac{dy}{dx} = \cos a \), when \( x = 0 \).
Answer: Given, \( \sin y = x \cos(a + y) \) ...(i)
\( \Rightarrow x = \frac{\sin y}{\cos(a + y)} \)
On differentiating both sides w.r.t. \( y \), we get:
\( \frac{dx}{dy} = \frac{\cos(a + y) \frac{d}{dy}(\sin y) - \sin y \frac{d}{dy}[\cos(a + y)]}{\cos^2(a + y)} \) [using quotient rule]
\( \frac{dx}{dy} = \frac{\cos(a + y)\cos y + \sin y \sin(a + y)}{\cos^2(a + y)} \)
\( = \frac{\cos(a + y - y)}{\cos^2(a + y)} \) [since \( \cos A \cos B + \sin A \sin B = \cos(A - B) \)]
\( \Rightarrow \frac{dx}{dy} = \frac{\cos a}{\cos^2(a + y)} \)
\( \Rightarrow \frac{dy}{dx} = \frac{\cos^2(a + y)}{\cos a} \)
Put \( x = 0 \) in Eq. (i), we get \( \sin y = 0 \Rightarrow y = 0 \).
Now, \( \frac{dy}{dx} = \frac{\cos^2(a + 0)}{\cos a} = \frac{\cos^2 a}{\cos a} = \cos a \).
Hence proved.

Question. If \( y = \tan^{-1}\left(\frac{a}{x}\right) + \log \sqrt{\frac{x - a}{x + a}} \), then prove that \( \frac{dy}{dx} = \frac{2a^3}{x^4 - a^4} \).
Answer: Given, \( y = \tan^{-1}\left(\frac{a}{x}\right) + \log \sqrt{\frac{x - a}{x + a}} \)
\( = \tan^{-1}\left(\frac{a}{x}\right) + \frac{1}{2}[\log(x - a) - \log(x + a)] \)
\( \therefore \frac{dy}{dx} = \frac{1}{1 + \frac{a^2}{x^2}} \cdot \left(-\frac{a}{x^2}\right) + \frac{1}{2}\left[\frac{1}{x - a} - \frac{1}{x + a}\right] \)
\( = \frac{-a}{x^2 + a^2} + \frac{1}{2}\left[\frac{x + a - x + a}{(x - a)(x + a)}\right] \)
\( = \frac{-a}{x^2 + a^2} + \frac{a}{x^2 - a^2} = \frac{-a(x^2 - a^2) + a(x^2 + a^2)}{(x^2 + a^2)(x^2 - a^2)} = \frac{2a^3}{x^4 - a^4} \)
Hence proved.

Question. If \( \log(\sqrt{1 + x^2} - x) = y\sqrt{1 + x^2} \), then show that \( (1 + x^2)\frac{dy}{dx} + xy + 1 = 0 \).
Answer: Given, \( \log(\sqrt{1 + x^2} - x) = y\sqrt{1 + x^2} \)
On differentiating both sides w.r.t. \( x \), we get:
\( \frac{1}{\sqrt{1 + x^2} - x} \frac{d}{dx}(\sqrt{1 + x^2} - x) = y \frac{d}{dx}(\sqrt{1 + x^2}) + \sqrt{1 + x^2} \frac{dy}{dx} \)
\( \Rightarrow \frac{1}{\sqrt{1 + x^2} - x} \left[\frac{2x}{2\sqrt{1 + x^2}} - 1\right] = y \left[\frac{x}{\sqrt{1 + x^2}}\right] + \sqrt{1 + x^2} \frac{dy}{dx} \)
\( \Rightarrow \frac{1}{\sqrt{1 + x^2} - x} \left[\frac{x - \sqrt{1 + x^2}}{\sqrt{1 + x^2}}\right] = \frac{xy}{\sqrt{1 + x^2}} + \sqrt{1 + x^2} \frac{dy}{dx} \)
\( \Rightarrow \frac{-1}{\sqrt{1 + x^2}} = \frac{xy + (1 + x^2)\frac{dy}{dx}}{\sqrt{1 + x^2}} \)
\( \Rightarrow -1 = xy + (1 + x^2)\frac{dy}{dx} \)
\( \Rightarrow (1 + x^2)\frac{dy}{dx} + xy + 1 = 0 \)
Hence proved.

Question. If \( y = \cos^{-1}\left(\frac{3x + 4\sqrt{1 - x^2}}{5}\right) \), then find \( \frac{dy}{dx} \).
Answer: We have, \( y = \cos^{-1}\left(\frac{3x + 4\sqrt{1 - x^2}}{5}\right) \)
Let \( x = \cos \theta \Rightarrow \theta = \cos^{-1} x \)
\( y = \cos^{-1}\left(\frac{3\cos \theta + 4\sqrt{1 - \cos^2 \theta}}{5}\right) \)
\( = \cos^{-1}\left(\frac{3}{5}\cos \theta + \frac{4}{5}\sin \theta\right) \) [since \( 1 - \cos^2 \theta = \sin^2 \theta \)]
Let us assume a right angled triangle with base 3 and perpendicular 4.
Then, \( \frac{3}{5} = \cos \alpha \Rightarrow \cos \alpha = \text{base}/\text{hypotenuse} \)
and \( \frac{4}{5} = \sin \alpha \Rightarrow \sin \alpha = \text{perpendicular}/\text{hypotenuse} \)
\( \therefore y = \cos^{-1}[\cos \alpha \cos \theta + \sin \alpha \sin \theta] \)
\( \Rightarrow y = \cos^{-1}[\cos(\theta - \alpha)] = \theta - \alpha \)
\( y = \cos^{-1} x - \cos^{-1}\left(\frac{3}{5}\right) \)
\( \therefore \frac{dy}{dx} = \frac{-1}{\sqrt{1 - x^2}} \)

Question. If \( y = \tan^{-1}\left(\frac{\sqrt{1 + x^2} - \sqrt{1 - x^2}}{\sqrt{1 + x^2} + \sqrt{1 - x^2}}\right) \), then find \( \frac{dy}{dx} \).
Answer: We have, \( y = \tan^{-1}\left(\frac{\sqrt{1 + x^2} - \sqrt{1 - x^2}}{\sqrt{1 + x^2} + \sqrt{1 - x^2}}\right) \)
Let us assume, \( x^2 = \cos \theta \Rightarrow \theta = \cos^{-1} x^2 \)
\( y = \tan^{-1}\left(\frac{\sqrt{1 + \cos \theta} - \sqrt{1 - \cos \theta}}{\sqrt{1 + \cos \theta} + \sqrt{1 - \cos \theta}}\right) \)
\( = \tan^{-1}\left(\frac{\sqrt{2\cos^2\frac{\theta}{2}} - \sqrt{2\sin^2\frac{\theta}{2}}}{\sqrt{2\cos^2\frac{\theta}{2}} + \sqrt{2\sin^2\frac{\theta}{2}}}\right) \) [since \( 1 - \cos \theta = 2\sin^2\frac{\theta}{2} \) and \( 1 + \cos \theta = 2\cos^2\frac{\theta}{2} \)]
\( = \tan^{-1}\left(\frac{\sqrt{2}\cos\frac{\theta}{2} - \sqrt{2}\sin\frac{\theta}{2}}{\sqrt{2}\cos\frac{\theta}{2} + \sqrt{2}\sin\frac{\theta}{2}}\right) \)
On dividing numerator and denominator by \( \sqrt{2}\cos\frac{\theta}{2} \), we get:
\( y = \tan^{-1}\left[\frac{1 - \tan\frac{\theta}{2}}{1 + \tan\frac{\theta}{2}}\right] = \tan^{-1}\left[\tan\left(\frac{\pi}{4} - \frac{\theta}{2}\right)\right] = \frac{\pi}{4} - \frac{\theta}{2} \)
\( y = \frac{\pi}{4} - \frac{1}{2}\cos^{-1} x^2 \)
On differentiating w.r.t. \( x \), we get:
\( \frac{dy}{dx} = 0 - \frac{1}{2}\frac{-1}{\sqrt{1 - (x^2)^2}} \cdot \frac{d}{dx}(x^2) \) [by chain rule]
\( = \frac{1}{2\sqrt{1 - x^4}}(2x) = \frac{x}{\sqrt{1 - x^4}} \)

Question. Differentiate \( \sin^{-1}\left(\frac{2^{x+1} \cdot 3^x}{1 + (36)^x}\right) \) w.r.t. \( x \).
Answer: Let \( y = \sin^{-1}\left[\frac{2^{x+1} \cdot 3^x}{1 + (36)^x}\right] = \sin^{-1}\left[\frac{2 \cdot 2^x \cdot 3^x}{1 + (6^x)^2}\right] \)
\( = \sin^{-1}\left[\frac{2 \cdot 6^x}{1 + (6^x)^2}\right] \)
Put \( 6^x = \tan \theta \)
\( y = \sin^{-1}\left(\frac{2\tan \theta}{1 + \tan^2 \theta}\right) = \sin^{-1}(\sin 2\theta) = 2\theta = 2\tan^{-1}(6^x) \)
On differentiating both sides w.r.t. \( x \), we get:
\( \frac{dy}{dx} = 2\frac{d}{dx}[\tan^{-1}(6^x)] = \frac{2}{1 + (6^x)^2} \cdot 6^x \log 6 \)
\( = \frac{2 \cdot 6^x \log 6}{1 + (36)^x} = \frac{2^{x+1} \cdot 3^x \log 6}{1 + (36)^x} \)

Question. Differentiate \( x^{\sin x} \) w.r.t. \( x \).
Answer: Let \( y = x^{\sin x} \)
On taking log both sides, we get:
\( \log y = \sin x(\log x) \)
On differentiating both sides w.r.t. \( x \), we get:
\( \frac{1}{y} \frac{dy}{dx} = \sin x \frac{d}{dx}(\log x) + \log x \frac{d}{dx}(\sin x) \)
\( \Rightarrow \frac{dy}{dx} = x^{\sin x} \left[\frac{\sin x}{x} + \log x \cos x\right] \)