CBSE Class 12 Mathematics Continuity And Differentiability Worksheet Set 13

Selected NCERT Questions

Question. Let \( A = \begin{bmatrix} 3 & 7 \\ 2 & 5 \end{bmatrix} \) and \( B = \begin{bmatrix} 6 & 8 \\ 7 & 9 \end{bmatrix} \), verify that \( (AB)^{-1} = B^{-1}A^{-1} \).
Answer: Sol. \( AB = \begin{bmatrix} 3 & 7 \\ 2 & 5 \end{bmatrix} \begin{bmatrix} 6 & 8 \\ 7 & 9 \end{bmatrix} = \begin{bmatrix} 67 & 87 \\ 47 & 61 \end{bmatrix} \)
\( |AB| = 67 \times 61 - 47 \times 87 = 4087 - 4089 = -2 \)
\( \therefore AB \) is invertible.
Cofactors of elements of determinant \( AB \) are,
\( AB_{11} = 61, AB_{12} = -47, AB_{21} = -87, AB_{22} = 67 \)
\( Adj(AB) = \begin{bmatrix} 61 & -47 \\ -87 & 67 \end{bmatrix}^T = \begin{bmatrix} 61 & -87 \\ -47 & 67 \end{bmatrix} \)
\( (AB)^{-1} = \frac{1}{|AB|} Adj(AB) = \frac{1}{-2} \begin{bmatrix} 61 & -87 \\ -47 & 67 \end{bmatrix} = \frac{1}{2} \begin{bmatrix} -61 & 87 \\ 47 & -67 \end{bmatrix} \)
\( |A| = \begin{vmatrix} 3 & 7 \\ 2 & 5 \end{vmatrix} = 15 - 14 = 1 \)
Cofactors of elements of \( A \) are
\( A_{11} = 5, A_{12} = -2, A_{21} = -7, A_{22} = 3 \)
\( Adj A = \begin{bmatrix} 5 & -2 \\ -7 & 3 \end{bmatrix}^T = \begin{bmatrix} 5 & -7 \\ -2 & 3 \end{bmatrix} \)
\( A^{-1} = \frac{1}{|A|} Adj A = \frac{1}{1} \begin{bmatrix} 5 & -7 \\ -2 & 3 \end{bmatrix} = \begin{bmatrix} 5 & -7 \\ -2 & 3 \end{bmatrix} \)
\( |B| = \begin{vmatrix} 6 & 8 \\ 7 & 9 \end{vmatrix} = 54 - 56 = -2 \)
Cofactors of elements of \( B \) are
\( B_{11} = 9, B_{12} = -7, B_{21} = -8, B_{22} = 6 \)
\( Adj B = \begin{bmatrix} 9 & -7 \\ -8 & 6 \end{bmatrix}^T = \begin{bmatrix} 9 & -8 \\ -7 & 6 \end{bmatrix} \)
\( B^{-1} = \frac{1}{|B|} Adj A = \frac{1}{-2} \begin{bmatrix} 9 & -8 \\ -7 & 6 \end{bmatrix} = \frac{1}{2} \begin{bmatrix} -9 & 8 \\ 7 & -6 \end{bmatrix} \)
\( B^{-1}A^{-1} = \frac{1}{2} \begin{bmatrix} -9 & 8 \\ 7 & -6 \end{bmatrix} \begin{bmatrix} 5 & -7 \\ -2 & 3 \end{bmatrix} = \frac{1}{2} \begin{bmatrix} -45 - 16 & 63 + 24 \\ 35 + 12 & -49 - 18 \end{bmatrix} = \frac{1}{2} \begin{bmatrix} -61 & 87 \\ 47 & -67 \end{bmatrix} \)
Thus, \( (AB)^{-1} = B^{-1}A^{-1} \).

Question. If \( A = \begin{bmatrix} 2 & -1 & 1 \\ -1 & 2 & -1 \\ 1 & -1 & 2 \end{bmatrix} \), then verify that \( A^3 - 6A^2 + 9A - 4I = 0 \) and hence find \( A^{-1} \). 
Answer: Sol. \( A^2 = \begin{bmatrix} 2 & -1 & 1 \\ -1 & 2 & -1 \\ 1 & -1 & 2 \end{bmatrix} \begin{bmatrix} 2 & -1 & 1 \\ -1 & 2 & -1 \\ 1 & -1 & 2 \end{bmatrix} \) [\( \because A^2 = A.A \)]
\( = \begin{bmatrix} 4 + 1 + 1 & -2 - 2 - 1 & 2 + 1 + 2 \\ -2 - 2 - 1 & 1 + 4 + 1 & -1 - 2 - 2 \\ 2 + 1 + 2 & -1 - 2 - 2 & 1 + 1 + 4 \end{bmatrix} = \begin{bmatrix} 6 & -5 & 5 \\ -5 & 6 & -5 \\ 5 & -5 & 6 \end{bmatrix} \)
\( A^3 = \begin{bmatrix} 2 & -1 & 1 \\ -1 & 2 & -1 \\ 1 & -1 & 2 \end{bmatrix} \begin{bmatrix} 6 & -5 & 5 \\ -5 & 6 & -5 \\ 5 & -5 & 6 \end{bmatrix} \) [\( \because A^3 = A.A^2 \)]
\( = \begin{bmatrix} 12 + 5 + 5 & -10 - 6 - 5 & 10 + 5 + 6 \\ -6 - 10 - 5 & 5 + 12 + 5 & -5 - 10 - 6 \\ 6 + 5 + 10 & -5 - 6 - 10 & 5 + 5 + 12 \end{bmatrix} = \begin{bmatrix} 22 & -21 & 21 \\ -21 & 22 & -21 \\ 21 & -21 & 22 \end{bmatrix} \)
Now, \( A^3 - 6A^2 + 9A - 4I = \begin{bmatrix} 22 & -21 & 21 \\ -21 & 22 & -21 \\ 21 & -21 & 22 \end{bmatrix} - 6 \begin{bmatrix} 6 & -5 & 5 \\ -5 & 6 & -5 \\ 5 & -5 & 6 \end{bmatrix} + 9 \begin{bmatrix} 2 & -1 & 1 \\ -1 & 2 & -1 \\ 1 & -1 & 2 \end{bmatrix} - 4 \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} \)
\( = \begin{bmatrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix} = 0 \) (zero matrix)
\( \therefore A^3 - 6A^2 + 9A - 4I = 0 \)
Pre-multiplying by \( A^{-1} \), we get
\( A^{-1}A.A^2 - 6A^{-1}.A.A + 9A^{-1}.A - 4A^{-1}I = 0 \)
\( \implies \) \( A^2 - 6A + 9I - 4A^{-1} = 0 \)
\( \implies \) \( 4A^{-1} = A^2 - 6A + 9I \)
\( 4A^{-1} = \begin{bmatrix} 6 & -5 & 5 \\ -5 & 6 & -5 \\ 5 & -5 & 6 \end{bmatrix} - 6 \begin{bmatrix} 2 & -1 & 1 \\ -1 & 2 & -1 \\ 1 & -1 & 2 \end{bmatrix} + 9 \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} \)
\( 4A^{-1} = \begin{bmatrix} 3 & 1 & -1 \\ 1 & 3 & 1 \\ -1 & 1 & 3 \end{bmatrix} \)
\( A^{-1} = \frac{1}{4} \begin{bmatrix} 3 & 1 & -1 \\ 1 & 3 & 1 \\ -1 & 1 & 3 \end{bmatrix} = \begin{bmatrix} 3/4 & 1/4 & -1/4 \\ 1/4 & 3/4 & 1/4 \\ -1/4 & 1/4 & 3/4 \end{bmatrix} \)

Question. If \( A = \begin{bmatrix} 2 & -3 & 5 \\ 3 & 2 & -4 \\ 1 & 1 & -2 \end{bmatrix} \), find \( A^{-1} \). Using \( A^{-1} \) solve the system of equations. \( 2x - 3y + 5z = 11 \); \( 3x + 2y - 4z = -5 \); \( x + y - 2z = -3 \).
Answer: Sol. \( |A| = \begin{vmatrix} 2 & -3 & 5 \\ 3 & 2 & -4 \\ 1 & 1 & -2 \end{vmatrix} = 2(-4+4) + 3(-6+4) + 5(3-2) = 0 - 6 + 5 = -1 \neq 0 \)
Therefore \( A \) is non-singular matrix and \( A^{-1} \) exists.
\( C_{11} = 0; C_{12} = 2; C_{13} = 1; C_{21} = -1; C_{22} = -9; C_{23} = -5; C_{31} = 2; C_{32} = 23; C_{33} = 13 \)
\( adj A = \begin{bmatrix} 0 & 2 & 1 \\ -1 & -9 & -5 \\ 2 & 23 & 13 \end{bmatrix}^T = \begin{bmatrix} 0 & -1 & 2 \\ 2 & -9 & 23 \\ 1 & -5 & 13 \end{bmatrix} \)
\( \therefore A^{-1} = \frac{1}{|A|} adj A = \frac{1}{-1} \begin{bmatrix} 0 & -1 & 2 \\ 2 & -9 & 23 \\ 1 & -5 & 13 \end{bmatrix} = \begin{bmatrix} 0 & 1 & -2 \\ -2 & 9 & -23 \\ -1 & 5 & -13 \end{bmatrix} \)
The given system can be expressed as \( AX = B \), where
\( A = \begin{bmatrix} 2 & -3 & 5 \\ 3 & 2 & -4 \\ 1 & 1 & -2 \end{bmatrix}, X = \begin{bmatrix} x \\ y \\ z \end{bmatrix}, B = \begin{bmatrix} 11 \\ -5 \\ -3 \end{bmatrix} \)
Now, \( AX = B \)
\( \implies \) \( X = A^{-1}B \)
\( \begin{bmatrix} x \\ y \\ z \end{bmatrix} = \begin{bmatrix} 0 & 1 & -2 \\ -2 & 9 & -23 \\ -1 & 5 & -13 \end{bmatrix} \begin{bmatrix} 11 \\ -5 \\ -3 \end{bmatrix} \)
\( \implies \) \( \begin{bmatrix} x \\ y \\ z \end{bmatrix} = \begin{bmatrix} 0 - 5 + 6 \\ -22 - 45 + 69 \\ -11 - 25 + 39 \end{bmatrix} = \begin{bmatrix} 1 \\ 2 \\ 3 \end{bmatrix} \)
On equating, we get \( x = 1, y = 2 \) and \( z = 3 \)

Very Short Answer Questions

Question. For what value of 'k' is the function \( f(x) = \begin{cases} \frac{\sin 5x}{3x} + \cos x, & \text{if } x \neq 0 \\ k, & \text{if } x = 0 \end{cases} \) continuous at \( x = 0 \)?
Answer: Sol. \( \lim_{x \to 0^+} f(x) = \lim_{h \to 0} f(0 + h) \)
\( = \lim_{h \to 0} f(h) = \lim_{h \to 0} \left( \frac{\sin 5h}{3h} + \cos h \right) \)
\( = \lim_{h \to 0} \frac{\sin 5h}{5h} \times \frac{5}{3} + \lim_{h \to 0} \cos h = 1 \times \frac{5}{3} + 1 \) [\( \because h \to 0 \implies 5h \to 0 \)]
\( \implies \) \( \lim_{x \to 0^+} f(x) = \frac{8}{3} \)
Also, \( f(0) = k \)
Since, \( f(x) \) is continuous at \( x = 0 \).
\( \lim_{x \to 0^+} f(x) = f(0) \)
\( \implies \) \( \frac{8}{3} = k \)

Question. Determine value of the constant 'k' so that the function \( f(x) = \begin{cases} \frac{kx}{|x|}, & \text{if } x < 0 \\ 3, & \text{if } x \geq 0 \end{cases} \) is continuous at \( x = 0 \). 
Answer: Sol. \( \because f(x) \) is continuous at \( x = 0 \)
\( \implies \) \( \lim_{x \to 0^+} f(x) = \lim_{x \to 0^-} f(x) = f(0) \)
Now, \( \lim_{x \to 0^-} f(x) = \lim_{h \to 0^-} f(0 - h) \)
\( = \lim_{h \to 0^-} f(-h) = \lim_{h \to 0} \frac{k(-h)}{|-h|} = \lim_{h \to 0} \frac{-kh}{h} = -k \)
Also, \( f(0) = 3 \)
\( \because \lim_{x \to 0^-} f(x) = f(0) \)
\( \implies \) \( -k = 3 \)
\( \implies \) \( k = -3 \)

Question. If \( y = 2\sqrt{\sec(e^{2x})} \); then find \( \frac{dy}{dx} \).
Answer: Sol. Given, \( y = 2\sqrt{\sec(e^{2x})} \)
\( \therefore \frac{dy}{dx} = \frac{d}{dx} \left( 2\sqrt{\sec(e^{2x})} \right) = 2 \times \frac{1}{2\sqrt{\sec(e^{2x})}} \times \sec(e^{2x}) \tan(e^{2x}) \times 2e^{2x} \)
\( = 2\sqrt{\sec(e^{2x})} \tan(e^{2x}) . e^{2x} = 2e^{2x} \sqrt{\sec(e^{2x})} \tan(e^{2x}) \)

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

Question. If \( y = \text{cosec}(\cot \sqrt{x}) \), then find \( \frac{dy}{dx} \).
Answer: Sol. \( y = \text{cosec}(\cot \sqrt{x}) \)
\( \therefore \frac{dy}{dx} = \frac{d}{dx} (\text{cosec}(\cot \sqrt{x})) \)
\( = -\text{cosec}(\cot \sqrt{x}) \cot(\cot \sqrt{x}) \times \left( -\text{cosec}^2(\sqrt{x}) \times \frac{1}{2\sqrt{x}} \right) \)
\( = \frac{\text{cosec}(\cot \sqrt{x}) \cot(\cot \sqrt{x}) \times \text{cosec}^2(\sqrt{x})}{2\sqrt{x}} \)

Question. If \( y = 5e^{7x} + 6e^{-7x} \), show that \( \frac{d^2y}{dx^2} = 49y \).
Answer: Sol. \( \because y = 5e^{7x} + 6e^{-7x} \)
\( \implies \) \( \frac{dy}{dx} = 35e^{7x} - 42e^{-7x} \)
\( \implies \) \( \frac{d^2y}{dx^2} = 245e^{7x} + 294e^{-7x} = 49(5e^{7x} + 6e^{-7x}) = 49y \)

Question. If \( y = \log(\cos e^x) \), then find \( \frac{dy}{dx} \).
Answer: Sol. \( \because y = \log(\cos e^x) \)
\( \implies \) \( \frac{dy}{dx} = \frac{d}{dx}(\log(\cos e^x)) = \frac{1}{\cos(e^x)} \times -\sin(e^x) \times e^x = \frac{-e^x \sin(e^x)}{\cos(e^x)} = -e^x \tan(e^x) \)
\( \therefore \frac{dy}{dx} = -e^x \tan(e^x) \)

Question. Find the derivative of \( \log_{10} x \) with respect to \( x \).
Answer: Sol. Let \( y = \log_{10} x = \log_{10} e . \log_e x \)
\( \therefore \frac{dy}{dx} = \log_{10} e \frac{1}{x} = \frac{\log_{10} e}{x} \) [\( \because \frac{d}{dx} \log_e x = \frac{1}{x} \)]

Short Answer Questions

Question. Find \( \frac{dy}{dx} \) at \( t = \frac{2\pi}{3} \) when \( x = 10(t - \sin t) \) and \( y = 12(1 - \cos t) \).
Answer: Sol. Given, \( x = 10(t - \sin t) \) and \( y = 12(1 - \cos t) \)
\( \because x = 10(t - \sin t) \implies \frac{dx}{dt} = 10(1 - \cos t) \) (Differentiating w.r.t. t)
Again \( y = 12(1 - \cos t) \implies \frac{dy}{dt} = 12(0 + \sin t) = 12 \sin t \) (Differentiating w.r.t. t)
Now, \( \frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}} = \frac{12 \sin t}{10(1 - \cos t)} \)
\( \therefore \left. \frac{dy}{dx} \right|_{t = \frac{2\pi}{3}} = \frac{12 \sin \frac{2\pi}{3}}{10(1 - \cos \frac{2\pi}{3})} = \frac{6}{5} \times \frac{\sin(\pi - \frac{\pi}{3})}{(1 - \cos(\pi - \frac{\pi}{3}))} \)
\( = \frac{6}{5} \times \frac{\sin \frac{\pi}{3}}{1 + \cos \frac{\pi}{3}} = \frac{6}{5} \times \frac{\frac{\sqrt{3}}{2}}{1 + \frac{1}{2}} = \frac{6\sqrt{3}}{5 \times 3} = \frac{2\sqrt{3}}{5} \)

Question. Find \( \frac{dy}{dx} \) at \( x = 1, y = \frac{\pi}{4} \) if \( \sin^2 y + \cos xy = K \). 
Answer: Sol. \( \sin^2 y + \cos xy = K \)
Differentiating w.r.t. \( x \), we get
\( 2 \sin y \cos y \frac{dy}{dx} + (-\sin xy)(x \frac{dy}{dx} + y) = 0 \)
\( \implies \) \( \sin 2y \frac{dy}{dx} - x \sin xy \frac{dy}{dx} - y \sin xy = 0 \)
\( \implies \) \( \frac{dy}{dx} = \frac{y \sin xy}{(\sin 2y - x \sin xy)} \)
\( \therefore \left. \frac{dy}{dx} \right|_{x=1, y=\frac{\pi}{4}} = \frac{\frac{\pi}{4} \sin \frac{\pi}{4}}{\sin \frac{\pi}{2} - \sin \frac{\pi}{4}} = \frac{\frac{\pi}{4} . \frac{1}{\sqrt{2}}}{1 - \frac{1}{\sqrt{2}}} = \frac{\pi}{4(\sqrt{2} - 1)} \)

Question. If \( (1 + x)^n = C_0 + C_1x + C_2x^2 + \dots + C_nx^n \), then prove that
(i) \( C_1 + 2C_2 + \dots + nC_n = n \cdot 2^{n-1} \)
(ii) \( C_1 - 2C_2 + 3C_3 - \dots + (-1)^{n-1} nC_n = 0 \) 
Answer: Sol. We have, \( (1 + x)^n = C_0 + C_1x + C_2x^2 + \dots + C_nx^n \)
Differentiating both sides with respect to \( x \), we have
\( n(1 + x)^{n-1} = C_1 + 2C_2x + 3C_3x^2 + \dots + nC_nx^{n-1} \)
Putting \( x = 1 \) and \( x = -1 \) successively, we have
\( C_1 + 2C_2 + 3C_3 + \dots + nC_n = n \cdot 2^{n-1} \) and \( C_1 - 2C_2 + 3C_3 - \dots + (-1)^{n-1} nC_n = 0 \)

Question. If \( y = (\cos x)^{(\cos x)^{(\cos x) \dots \infty}} \), then show that \( \frac{dy}{dx} = \frac{y^2 \tan x}{y \log \cos x - 1} \). 
Answer: Sol. We have, \( y = (\cos x)^{(\cos x)^{(\cos x) \dots \infty}} \)
\( y = (\cos x)^y \)
\( \therefore \log y = \log (\cos x)^y \)
\( \implies \) \( \log y = y \log \cos x \)
On differentiating w.r.t. \( x \), we get
\( \frac{1}{y} \frac{dy}{dx} = y \cdot \frac{d}{dx} \log \cos x + \log \cos x \cdot \frac{dy}{dx} \)
\( \implies \) \( \frac{1}{y} \frac{dy}{dx} = y \frac{1}{\cos x} (-\sin x) + \log \cos x \frac{dy}{dx} \)
\( \implies \) \( \frac{dy}{dx} \left[ \frac{1}{y} - \log \cos x \right] = \frac{-y \sin x}{\cos x} = -y \tan x \)
\( \therefore \frac{dy}{dx} = \frac{-y^2 \tan x}{(1 - y \log \cos x)} = \frac{y^2 \tan x}{y \log \cos x - 1} \)

Long Answer Questions-I

Question. Find the values of \( p \) and \( q \), for which
\( f(x) = \begin{cases} \frac{1 - \sin^3 x}{3 \cos^2 x}, & \text{if } x < \frac{\pi}{2} \\ p, & \text{if } x = \frac{\pi}{2} \\ \frac{q(1 - \sin x)}{(\pi - 2x)^2}, & \text{if } x > \frac{\pi}{2} \end{cases} \) is continuous at \( x = \frac{\pi}{2} \). 
Answer: Sol. We have, \( f(x) = \begin{cases} \frac{1 - \sin^3 x}{3 \cos^2 x}, & \text{if } x < \frac{\pi}{2} \\ p, & \text{if } x = \frac{\pi}{2} \\ \frac{q(1 - \sin x)}{(\pi - 2x)^2}, & \text{if } x > \frac{\pi}{2} \end{cases} \) is continuous at \( x = \frac{\pi}{2} \).
Now, \( \lim_{h \to 0} f\left( \frac{\pi}{2} + h \right) \) [Let \( x = \frac{\pi}{2} + h, x \to \frac{\pi}{2}^+ \implies h \to 0 \)]
\( = \lim_{h \to 0} \frac{q \{ 1 - \sin(\frac{\pi}{2} + h) \}}{\{ \pi - 2(\frac{\pi}{2} + h) \}^2} = \lim_{h \to 0} \frac{q(1 - \cos h)}{(\pi - \pi - 2h)^2} = \lim_{h \to 0} \frac{q(1 - \cos h)}{4h^2} \)
\( = \lim_{h \to 0} \frac{q.2 \sin^2 \frac{h}{2}}{4h^2} = \lim_{h \to 0} \frac{q. \sin^2 \frac{h}{2}}{2h^2} = q \cdot \lim_{h \to 0} \left( \frac{\sin \frac{h}{2}}{\frac{h}{2}} \right)^2 \times \frac{1}{8} = \frac{q}{8} \)
Again \( \lim_{h \to 0} f\left( \frac{\pi}{2} - h \right) \) [Let \( x = \frac{\pi}{2} - h, x \to \frac{\pi}{2}^- \implies h \to 0 \)]
\( = \lim_{h \to 0} \frac{1 - \sin^3 (\frac{\pi}{2} - h)}{3 \cos^2 (\frac{\pi}{2} - h)} = \lim_{h \to 0} \frac{1 - \cos^3 h}{3 \sin^2 h} = \lim_{h \to 0} \frac{(1 - \cos h)(1 + \cos h + \cos^2 h)}{3 \sin^2 h} \)
\( = \lim_{h \to 0} \frac{2 \sin^2 \frac{h}{2} \cdot (1 + 1 + 1)}{3 \sin^2 h} = \lim_{h \to 0} \frac{2 \sin^2 \frac{h}{2} \cdot 3}{3 \sin^2 h} = \lim_{h \to 0} \frac{2 \sin^2 \frac{h}{2}}{\sin^2 h} \)
Dividing \( N' \) and \( D' \) by \( h^2 \), we get
\( = \lim_{h \to 0} \frac{2 \cdot \frac{\sin^2 \frac{h}{2}}{h^2}}{\frac{\sin^2 h}{h^2}} = \lim_{h \to 0} \frac{2 \cdot \frac{\sin^2 \frac{h}{2}}{4 \frac{h^2}{4}}}{\frac{\sin^2 h}{h^2}} = \frac{1}{2} \frac{\lim_{h \to 0} (\frac{\sin \frac{h}{2}}{\frac{h}{2}})^2}{\lim_{h \to 0} (\frac{\sin h}{h})^2} = \frac{1}{2} \cdot \frac{1}{1} = \frac{1}{2} \)
Also \( f(\frac{\pi}{2}) = p \)
\( \because f(x) \) is continuous at \( x = \frac{\pi}{2} \)
\( \implies \) \( \lim_{h \to 0} f(\frac{\pi}{2} + h) = \lim_{h \to 0} f(\frac{\pi}{2} - h) = f(\frac{\pi}{2}) \)
\( \implies \) \( \frac{q}{8} = \frac{1}{2} = p \)
\( \implies \) \( p = \frac{1}{2} \) and \( q = 4 \)

Question. Show that the function \( f(x) = 2x - |x| \) is continuous but not differentiable at \( x = 0 \). 
Answer: Sol. Here \( f(x) = 2x - |x| \)
For continuity at \( x = 0 \)
\( \lim_{x \to 0^+} f(x) = \lim_{h \to 0} f(0 + h) = \lim_{h \to 0} f(h) \)
\( = \lim_{h \to 0} \{ 2h - |h| \} = \lim_{h \to 0} (2h - h) \)
\( = \lim_{h \to 0} h = 0 \) ...(i)
\( \lim_{x \to 0^-} f(x) = \lim_{h \to 0} f(0 - h) = \lim_{h \to 0} f(-h) \)
\( = \lim_{h \to 0} \{ 2(-h) - |-h| \} = \lim_{h \to 0} \{-2h - h\} \)
\( = \lim_{h \to 0} (-3h) = 0 \) ...(ii)
Also, \( f(0) = 2 \times 0 - |0| = 0 \) ...(iii)
(i), (ii) and (iii) \( \implies \lim_{x \to 0^+} f(x) = \lim_{x \to 0^-} f(x) = f(0) \)
Hence, \( f(x) \) is continuous at \( x = 0 \).
For differentiability at \( x = 0 \)
LHD \( = \lim_{h \to 0} \frac{f(0 - h) - f(0)}{-h} = \lim_{h \to 0} \frac{f(-h) - f(0)}{-h} \)
\( = \lim_{h \to 0} \frac{\{ 2(-h) - |-h| \} - \{ 2 \times 0 - |0| \}}{-h} = \lim_{h \to 0} \frac{-2h - h}{-h} \)
\( = \lim_{h \to 0} \frac{-3h}{-h} = \lim_{h \to 0} 3 = 3 \) ...(iv)
Again RHD \( = \lim_{h \to 0} \frac{f(0 + h) - f(0)}{h} \)
\( = \lim_{h \to 0} \frac{f(h) - f(0)}{h} = \lim_{h \to 0} \frac{2h - |h| - \{ 2 \times 0 - |0| \}}{h} = \lim_{h \to 0} \frac{2h - h}{h} \)
\( = \lim_{h \to 0} \frac{1}{h} = 1 \) ...(v)
From (iv) and (v), we get
LHD \( \neq \) RHD i.e., function \( f(x) = 2x - |x| \) is not differentiable at \( x = 0 \).
Hence, \( f(x) \) is continuous but not differentiable at \( x = 0 \).

Question. Find the value of 'a' for which the function \( f \) defined as
\( f(x) = \begin{cases} \frac{a \sin \frac{\pi}{2}(x+1),}{x^3} & \text{if } x \leq 0 \\ \frac{\tan x - \sin x}{x^3}, & \text{if } x > 0 \end{cases} \) is continuous at \( x = 0 \). 
Answer: Sol. \( \because f(x) \) is continuous at \( x = 0 \).
\( \implies \) (LHL of \( f(x) \) at \( x = 0 \)) = (RHL of \( f(x) \) at \( x = 0 \)) = \( f(0) \)
\( \implies \lim_{x \to 0^-} f(x) = \lim_{x \to 0^+} f(x) = f(0) \) ...(i)
Now, \( \lim_{x \to 0^-} f(x) = \lim_{x \to 0} a \sin \frac{\pi}{2}(x + 1) \) [\( \because f(x) = a \sin \frac{\pi}{2}(x + 1) \), if \( x \leq 0 \)]
\( = \lim_{x \to 0} a \sin \left( \frac{\pi}{2} + \frac{\pi}{2}x \right) = \lim_{x \to 0} a \cos \frac{\pi}{2}x = a . \cos 0 = a \) ...(ii)
Again, \( \lim_{x \to 0^+} f(x) = \lim_{x \to 0} \frac{\tan x - \sin x}{x^3} \) [\( \because f(x) = \frac{\tan x - \sin x}{x^3} \), if \( x > 0 \)]
\( = \lim_{x \to 0} \frac{\frac{\sin x}{\cos x} - \sin x}{x^3} = \lim_{x \to 0} \frac{\sin x - \sin x \cos x}{x^3 \cos x} = \lim_{x \to 0} \frac{\sin x (1 - \cos x)}{x^3 \cos x} \)
\( = \lim_{x \to 0} \frac{1}{\cos x} \cdot \frac{\sin x}{x} \cdot \frac{2 \sin^2 \frac{x}{2}}{x^2} \) [\( \because 1 - \cos x = 2 \sin^2 \frac{x}{2} \)]
\( = \frac{1}{1} \cdot 1 \cdot \frac{1}{2} \lim_{x \to 0} \left( \frac{\sin \frac{x}{2}}{\frac{x}{2}} \right)^2 = \frac{1}{2} \cdot \lim_{x \to 0} \frac{\sin \frac{x}{2}}{\frac{x}{2}} = \frac{1}{2} \times 1 = \frac{1}{2} \) ...(iii)
Also, \( f(0) = a \sin \frac{\pi}{2}(0 + 1) = a \sin \frac{\pi}{2} = a \) ...(iv)
\( \because f \) is continuous at \( x = 0 \).
\( \therefore \) (i), (ii), (iii) and (iv) \( \implies a = \frac{1}{2} \)

Question. If \( f(x) = \begin{cases} \frac{\sin(a + 1)x + 2 \sin x}{x}, & \text{if } x < 0 \\ 2, & \text{if } x = 0 \\ \frac{\sqrt{1 + bx} - 1}{x}, & \text{if } x > 0 \end{cases} \) is continuous at \( x = 0 \), then find the values of \( a \) and \( b \).
Answer: Sol. We have, \( f(x) = \begin{cases} \frac{\sin(a + 1)x + 2 \sin x}{x}, & x < 0 \\ 2, & x = 0 \\ \frac{\sqrt{1 + bx} - 1}{x}, & x > 0 \end{cases} \)
Since, \( f(x) \) is continuous at \( x = 0 \)
\( \implies \lim_{x \to 0^+} f(x) = \lim_{x \to 0^-} f(x) = f(0) \) ...(i)
Now, \( \lim_{x \to 0^+} f(x) = \lim_{h \to 0} f(0 + h) \) [Let \( x = 0 + h, h \) is +ve small quantity \( x \to 0^+ \implies h \to 0 \)]
\( = \lim_{h \to 0} f(h) = \lim_{h \to 0} \frac{\sqrt{1 + bh} - 1}{h} = \lim_{h \to 0} \frac{\sqrt{1 + bh} - 1}{h} \times \frac{\sqrt{1 + bh} + 1}{\sqrt{1 + bh} + 1} \)
\( = \lim_{h \to 0} \frac{1 + bh - 1}{h(\sqrt{1 + bh} + 1)} = \lim_{h \to 0} \frac{bh}{h(\sqrt{1 + bh} + 1)} = \lim_{h \to 0} \frac{b}{\sqrt{1 + bh} + 1} = \frac{b}{2} \)
Again \( \lim_{x \to 0^-} f(x) = \lim_{h \to 0} f(0 - h) \) [Let \( x = 0 - h, h \) is +ve small quantity \( x \to 0^- \implies h \to 0 \)]
\( = \lim_{h \to 0} f(-h) = \lim_{h \to 0} \frac{\sin(a + 1)(-h) + 2 \sin(-h)}{-h} \)
\( = \lim_{h \to 0} \frac{-\sin(a + 1)h - 2 \sin h}{-h} = \lim_{h \to 0} \left[ \frac{\sin(a + 1)h}{h} + \frac{2 \sin h}{h} \right] \)
\( = \lim_{h \to 0} \frac{\sin(a + 1)h}{h} + 2 \lim_{h \to 0} \frac{\sin h}{h} = \lim_{h \to 0} \frac{\sin(a + 1)h}{h \cdot (a + 1)} \times (a + 1) + 2 \lim_{h \to 0} \frac{\sin h}{h} \)
\( = 1 \times (a + 1) + 2 = a + 3 \)
Also \( f(0) = 2 \)
Now from (i) \( \frac{b}{2} = a + 3 = 2 \implies b = 4, a = -1 \)

Question. Show that the function \( 'f' \) defined by \( f(x) = \begin{cases} 3x - 2, & 0 < x \leq 1 \\ 2x^2 - x, & 1 < x \leq 2 \\ 5x - 4, & x > 2 \end{cases} \) is continuous at \( x = 2 \), but not differentiable. 
Answer: Sol. For continuity:
\( \lim_{x \to 2^-} f(x) = \lim_{h \to 0} f(2 - h) \) [Let \( x = 2 - h, \implies x \to 2^- \implies h \to 0 \)]
\( = \lim_{h \to 0} \{ 2(2 - h)^2 - (2 - h) \} = \lim_{h \to 0} \{ 2(4 + h^2 - 4h) - (2 - h) \} \)
\( = \lim_{h \to 0} (8 + 2h^2 - 8h - 2 + h) = 6 \)
\( \lim_{x \to 2^+} f(x) = \lim_{h \to 0} f(2 + h) \) [Let \( x = 2 + h, \implies x \to 2^+ \implies h \to 0 \)]
\( = \lim_{h \to 0} \{ 5(2 + h) - 4 \} = 6 \)
\( f(2) = 2(2)^2 - 2 = 6 \)
\( \implies \lim_{x \to 2^-} f(x) = \lim_{x \to 2^+} f(x) = f(2) \)
\( \implies f(x) \) is continuous at \( x = 2 \).
For Differentiability:
LHD (at \( x = 2 \)) = \( \lim_{h \to 0} \frac{f(2 - h) - f(2)}{-h} \)
\( = \lim_{h \to 0} \frac{2(2 - h)^2 - (2 - h) - \{ 2 \cdot 2^2 - 2 \}}{-h} = \lim_{h \to 0} \frac{8 + 2h^2 - 8h - 2 + h - 6}{-h} \)
\( = \lim_{h \to 0} \frac{2h^2 - 7h}{-h} = \lim_{h \to 0} \frac{2h - 7}{-1} = 7 \)
RHD (at \( x = 2 \)) = \( \lim_{h \to 0} \frac{f(2 + h) - f(2)}{h} \)
\( = \lim_{h \to 0} \frac{5(2 + h) - 4 - \{ 2 \cdot 2^2 - 2 \}}{h} = \lim_{h \to 0} \frac{10 + 5h - 4 - 6}{h} = \lim_{h \to 0} \frac{5h}{h} = 5 \)
\( \because \) LHD \( \neq \) RHD (at \( x = 2 \))
Hence, \( f(x) \) is not differentiable at \( x = 2 \).

Question. Show that the function \( f(x) = |x - 3|, x \in \mathbb{R} \), is continuous but not differentiable at \( x = 3 \). 
Answer: Sol. Here, \( f(x) = |x - 3| \implies f(x) = \begin{cases} -(x - 3), & x < 3 \\ 0, & x = 3 \\ (x - 3), & x > 3 \end{cases} \)
For Continuity:
Now, \( \lim_{x \to 3^+} f(x) = \lim_{h \to 0} f(3 + h) \) [Let \( x = 3 + h \) and \( x \to 3^+ \implies h \to 0 \)]
\( = \lim_{h \to 0} \{ (3 + h) - 3 \} = \lim_{h \to 0} h = 0 \) ...(i)
\( \lim_{x \to 3^-} f(x) = \lim_{h \to 0} f(3 - h) \) [Let \( x = 3 - h \) and \( x \to 3^- \implies h \to 0 \)]
\( = \lim_{h \to 0} -\{ (3 - h) - 3 \} = \lim_{h \to 0} h = 0 \) ...(ii)
Also, \( f(3) = 0 \) ...(iii)
From equation (i), (ii) and (iii), we get
\( \lim_{x \to 3^+} f(x) = \lim_{x \to 3^-} f(x) = f(3) \)
Hence, \( f(x) \) is continuous at \( x = 3 \).
For Differentiability:
RHD \( = \lim_{h \to 0} \frac{f(3 + h) - f(3)}{h} = \lim_{h \to 0} \frac{(3 + h - 3) - 0}{h} = \lim_{h \to 0} \frac{h}{h} = \lim_{h \to 0} 1 = 1 \) ...(iv)
LHD \( = \lim_{h \to 0} \frac{f(3 - h) - f(3)}{-h} = \lim_{h \to 0} \frac{-(3 - h - 3) - 0}{-h} = \lim_{h \to 0} \frac{h}{-h} = \lim_{h \to 0} (-1) = -1 \) ...(v)
Equation (iv) and (v) \( \implies \) RHD \( \neq \) LHD at \( x = 3 \).
Hence, \( f(x) \) is not differentiable at \( x = 3 \).
Therefore, \( f(x) = |x - 3|, x \in \mathbb{R} \) is continuous but not differentiable at \( x = 3 \).

Question. Discuss the continuity and differentiability of the function
\( f(x) = |x| + |x - 1| \) in the interval (-1, 2). 
Answer: Sol. Given function is \( f(x) = |x| + |x - 1| \)
Function is also written as
\( f(x) = \begin{cases} -x - (x - 1), & \text{if } -1 < x < 0 \\ 1, & \text{if } 0 \leq x < 1 \\ x + (x - 1), & \text{if } x \geq 1 \end{cases} \implies f(x) = \begin{cases} -2x + 1, & \text{if } x < 0 \\ 1, & \text{if } 0 \leq x < 1 \\ 2x - 1, & \text{if } x \geq 1 \end{cases} \)
Obviously, in given function we need to discuss the continuity and differentiability of the function \( f(x) \) at \( x = 0 \) or 1 only.
For continuity at \( x = 0 \)
\( \lim_{x \to 0^+} f(x) = \lim_{h \to 0} f(0 + h) \) [Let \( x = 0 + h \) and \( x \to 0^+ \implies h \to 0 \)]
\( = \lim_{h \to 0} f(h) = \lim_{h \to 0} 1 \) [\( \because h \) is very small positive quantity]
\( = 1 \) ...(i)
\( \lim_{x \to 0^-} f(x) = \lim_{h \to 0} f(0 - h) \) [Let \( x = 0 - h \) and \( x \to 0^- \implies h \to 0 \)]
\( = \lim_{h \to 0} f(-h) = \lim_{h \to 0} \{ -2(-h) + 1 \} = \lim_{h \to 0} (2h + 1) \)
\( = 1 \) ...(ii)
Also, \( f(0) = 1 \) ...(iii)
(i), (ii) and (iii) \( \implies \lim_{x \to 0^+} f(x) = \lim_{x \to 0^-} f(x) = f(0) \)
Hence, \( f(x) \) is continuous at \( x = 0 \).
For differentiability at \( x = 0 \)
RHD \( = \lim_{h \to 0} \frac{f(0 + h) - f(0)}{h} \)
\( = \lim_{h \to 0} \frac{f(h) - f(0)}{h} \) [\( \because h \) is very small positive quantity]
\( = \lim_{h \to 0} \frac{1 - 1}{h} = \lim_{h \to 0} \frac{0}{h} = \lim_{h \to 0} 0 = 0 \) [\( \because |h| = h, |0| = 0 \)]
RHD = 0 ...(iv)
LHD \( = \lim_{h \to 0} \frac{f(0 - h) - f(0)}{-h} = \lim_{h \to 0} \frac{f(-h) - f(0)}{-h} \)
\( = \lim_{h \to 0} \frac{-2(-h) + 1 - 1}{-h} = \lim_{h \to 0} \frac{2h}{-h} = \lim_{h \to 0} (-2) \)
LHD = -2 ...(v)
(iv) and (v) \( \implies \) RHD \( \neq \) LHD at \( x = 0 \).
Hence, \( f(x) \) is not differentiable at \( x = 0 \) but continuous at \( x = 0 \).
Similarly, we can prove \( f(x) \) is not differentiable at \( x = 1 \) but continuous at \( x = 1 \). (Do yourself)

Question. Show that the function \( f(x) = |x - 1| + |x + 1| \), for all \( x \in R \), is not differentiable at the points \( x = -1 \) and \( x = 1 \).
Answer: Sol. Here, given function is \( f(x) = |x - 1| + |x + 1| \)
\( \implies f(x) = \begin{cases} -(x - 1) - (x + 1), & x < -1 \\ 2, & x = -1 \\ -(x - 1) + (x + 1), & -1 < x < 1 \\ 2 & x = 1 \\ (x - 1) + (x + 1) & x > 1 \end{cases} \implies \begin{cases} -2x, & \text{if } x < -1 \\ 2, & \text{if } x = -1 \\ 2, & \text{if } -1 < x < 1 \\ 2 & \text{if } x = 1 \\ 2x, & \text{if } x > 1 \end{cases} \implies \begin{cases} -2x & \text{if } x < -1 \\ 2 & \text{if } -1 \leq x \leq 1 \\ 2x & \text{if } x > 1 \end{cases} \)
For \( x = -1 \)
RHD \( = \lim_{h \to 0} \frac{f(-1 + h) - f(-1)}{h} = \lim_{h \to 0} \frac{2 - 2}{h} = \lim_{h \to 0} 0 = 0 \)
LHD \( = \lim_{h \to 0} \frac{f(-1 - h) - f(-1)}{-h} = \lim_{h \to 0} \frac{-2(-1 - h) - 2}{-h} = \lim_{h \to 0} \frac{2 + 2h - 2}{-h} = \lim_{h \to 0} \frac{-2h}{h} = -2 \)
i.e., RHD \( \neq \) LHD.
Hence, \( f(x) \) is not differentiable at \( x = -1 \).
For \( x = 1 \)
RHD \( = \lim_{h \to 0} \frac{f(1 + h) - f(1)}{h} = \lim_{h \to 0} \frac{2(1 + h) - 2}{h} = \lim_{h \to 0} \frac{2 + 2h - 2}{h} = \lim_{h \to 0} \frac{2h}{h} = 2 \)
LHD \( = \lim_{h \to 0} \frac{f(1 - h) - f(1)}{-h} = \lim_{h \to 0} \frac{2 - 2}{-h} = \lim_{h \to 0} 0 = 0 \)
RHD \( \neq \) LHD.
Hence, \( f(x) \) not differentiable at \( x = 1 \).

Question. Find 'a' and 'b', if the function given by \( f(x) = \begin{cases} ax^2 + b, & \text{if } x < 1 \\ 2x + 1, & \text{if } x \geq 1 \end{cases} \) is differentiable at \( x = 1 \). 
Answer: Sol. Since, \( f \) is differentiable at 1 \( \implies \) \( f \) is also continuous at 1.
Now \( \lim_{x \to 1^+} f(x) = \lim_{h \to 0} f(1 + h) \) [Here \( h \) is +ve and very small quantity]
\( = \lim_{h \to 0} \{ 2(1 + h) + 1 \} = 2 + 1 = 3 \)
\( \lim_{x \to 1^-} f(x) = \lim_{h \to 0} f(1 - h) = \lim_{h \to 0} \{ a(1 - h)^2 + b \} = a + b \)
Since \( f(x) \) is continuous at \( x = 1 \)
\( \implies \) \( a + b = 3 \) ...(i)
Again, since \( f \) is differentiable
\( \implies \) LHD (at \( x = 1 \)) = RHD (at \( x = 1 \)) \( \implies \lim_{h \to 0} \frac{f(1 - h) - f(1)}{-h} = \lim_{h \to 0} \frac{f(1 + h) - f(1)}{h} \)
\( \implies \lim_{h \to 0} \frac{a(1 - h)^2 + b - 3}{-h} = \lim_{h \to 0} \frac{2(1 + h) + 1 - 3}{h} \) \( \implies \lim_{h \to 0} \frac{a - 2ah + ah^2 + b - 3}{-h} = \lim_{h \to 0} \frac{2 + 2h + 1 - 3}{h} \)
\( \implies \lim_{h \to 0} \frac{-2ah + ah^2 + (a + b) - 3}{-h} = \lim_{h \to 0} \frac{2h}{h} \) \( \implies \lim_{h \to 0} \frac{-2ah + ah^2 + 3 - 3}{-h} = 2 \)
\( \implies \lim_{h \to 0} \frac{ah(2 - h)}{h} = 2 \implies 2a = 2 \implies a = 1 \implies b = 2 \) [From equation (i)]

DERIVATIVES

Question. If \( y^x = e^{y - x} \), then prove that \( \frac{dy}{dx} = \frac{(1 + \log y)^2}{\log y} \) 
Answer: Sol. Given, \( y^x = e^{y - x} \)
Taking logarithm both sides, we get \( \log y^x = \log e^{y - x} \)
\( \implies \) \( x \cdot \log y = (y - x) \cdot \log e \implies x \cdot \log y = (y - x) \)
\( \implies \) \( x(1 + \log y) = y \implies x = \frac{y}{1 + \log y} \)
Differentiating both sides with respect to \( y \), we get
\( \frac{dx}{dy} = \frac{(1 + \log y) \cdot 1 - y \cdot (0 + \frac{1}{y})}{(1 + \log y)^2} = \frac{1 + \log y - 1}{(1 + \log y)^2} = \frac{\log y}{(1 + \log y)^2} \)
\( \implies \frac{dy}{dx} = \frac{(1 + \log y)^2}{\log y} \)

Question. Differentiate \( \tan^{-1} \frac{3x - x^3}{1 - 3x^2}, |x| < \frac{1}{\sqrt{3}} \) w.r.t \( \tan^{-1} \frac{x}{\sqrt{1 - x^2}} \). 
Answer: Sol. Let \( u = \tan^{-1} \frac{3x - x^3}{1 - 3x^2} \) and \( v = \tan^{-1} \frac{x}{\sqrt{1 - x^2}} \) then find \( \frac{du}{dv} \).
Put \( x = \tan \theta \implies \theta = \tan^{-1} x \), we get
\( u = \tan^{-1} \left( \frac{3 \tan \theta - \tan^3 \theta}{1 - 3 \tan^2 \theta} \right) = \tan^{-1}(\tan 3\theta) = 3\theta = 3 \tan^{-1} x \)
\( v = \tan^{-1} \left( \frac{\tan \theta}{\sqrt{1 - \tan^2 \theta}} \right) = \tan^{-1} \left( \frac{\sin \theta}{\sqrt{1 - \sin^2 \theta}} \right) = \tan^{-1} \left( \frac{\sin \theta}{\cos \theta} \right) = \tan^{-1}(\tan \theta) = \theta = \tan^{-1} x \)
Now, \( \frac{du}{dx} = \frac{3}{1 + x^2} \) and \( \frac{dv}{dx} = \frac{1}{1 + x^2} \)
\( \therefore \frac{du}{dv} = \frac{du}{dx} \times \frac{dx}{dv} = \frac{3}{1 + x^2} \times \frac{1 + x^2}{1} = 3 \)

Question. If \( (\cos x)^y = (\cos y)^x \), then find \( \frac{dy}{dx} \). 
Answer: Sol. Given, \( (\cos x)^y = (\cos y)^x \)
Taking logarithm both sides, we get \( \log(\cos x)^y = \log(\cos y)^x \)
\( \implies \) \( y \cdot \log(\cos x) = x \cdot \log(\cos y) \) [\( \because \log m^n = n \log m \)]
Differentiating both sides, we get
\( y \cdot \frac{1}{\cos x} (-\sin x) + \log(\cos x) \cdot \frac{dy}{dx} = x \cdot \frac{1}{\cos y} \cdot (-\sin y) \cdot \frac{dy}{dx} + \log(\cos y) \)
\( \implies \) \( -\frac{y \sin x}{\cos x} + \log(\cos x) \cdot \frac{dy}{dx} = -\frac{x \sin y}{\cos y} \frac{dy}{dx} + \log(\cos y) \)
\( \implies \) \( \left[ \log(\cos x) + \frac{x \sin y}{\cos y} \right] \frac{dy}{dx} = \log(\cos y) + \frac{y \sin x}{\cos x} \)
\( \implies \) \( \left[ \log(\cos x) + x \tan y \right] \frac{dy}{dx} = \log(\cos y) + y \tan x \)
\( \implies \frac{dy}{dx} = \frac{\log(\cos y) + y \tan x}{\log(\cos x) + x \tan y} \)

Question. Find the value of \( \frac{dy}{dx} \) at \( \theta = \frac{\pi}{4} \), if \( x = ae^\theta (\sin \theta - \cos \theta) \) and \( y = ae^\theta (\sin \theta + \cos \theta) \). 
Answer: Sol. Given, \( x = ae^\theta (\sin \theta - \cos \theta) \) and \( y = ae^\theta (\sin \theta + \cos \theta) \)
Taking \( x = ae^\theta (\sin \theta - \cos \theta) \)
Differentiating with respect to \( \theta \), we get
\( \frac{dx}{d\theta} = ae^\theta (\cos \theta + \sin \theta) + a (\sin \theta - \cos \theta) . e^\theta = ae^\theta (\cos \theta + \sin \theta + \sin \theta - \cos \theta) \)
\( = 2 ae^\theta \sin \theta \) ...(i)
Again, \( y = ae^\theta (\sin \theta + \cos \theta) \)
Differentiating with respect to \( \theta \), we get
\( \frac{dy}{d\theta} = ae^\theta (\cos \theta - \sin \theta) + a (\sin \theta + \cos \theta) . e^\theta = ae^\theta (\cos \theta - \sin \theta + \sin \theta + \cos \theta) \)
\( = 2 ae^\theta . \cos \theta \) ...(ii)
\( \therefore \frac{dy}{dx} = \frac{\frac{dy}{d\theta}}{\frac{dx}{d\theta}} = \frac{2 ae^\theta . \cos \theta}{2 ae^\theta . \sin \theta} \) [From (i) and (ii)]
\( \implies \frac{dy}{dx} = \cot \theta \implies \left. \frac{dy}{dx} \right|_{\theta = \frac{\pi}{4}} = \cot \frac{\pi}{4} = 1 \)

Question. Differentiate \( \tan^{-1} \left[ \frac{\sqrt{1 + x^2} - 1}{x} \right] \) with respect to \( x \). 
Answer: Sol. Let \( y = \tan^{-1} \left[ \frac{\sqrt{1 + x^2} - 1}{x} \right] \)
Put \( x = \tan \theta \implies \theta = \tan^{-1} x \)
Now, \( y = \tan^{-1} \left[ \frac{\sqrt{1 + \tan^2 \theta} - 1}{\tan \theta} \right] = \tan^{-1} \left( \frac{\sec \theta - 1}{\tan \theta} \right) \)
\( = \tan^{-1} \left( \frac{\frac{1}{\cos \theta} - 1}{\frac{\sin \theta}{\cos \theta}} \right) = \tan^{-1} \left( \frac{1 - \cos \theta}{\sin \theta} \right) \)
\( = \tan^{-1} \left( \frac{2 \sin^2 \frac{\theta}{2}}{2 \sin \frac{\theta}{2} \cdot \cos \frac{\theta}{2}} \right) = \tan^{-1} (\tan \frac{\theta}{2}) = \frac{\theta}{2} \)
\( \implies y = \frac{1}{2} \tan^{-1} x \implies \frac{dy}{dx} = \frac{1}{2(1 + x^2)} \)

Question. Differentiate the following with respect to \( x \): \( (\sin x)^x + (\cos x)^{\sin x} \). 
Answer: Sol. Let \( u = (\sin x)^x \) and \( v = (\cos x)^{\sin x} \)
\( \therefore \) Given differential equation becomes \( y = u + v \)
\( \implies \frac{dy}{dx} = \frac{du}{dx} + \frac{dv}{dx} \) ...(i)
Now, \( u = (\sin x)^x \)
Taking log on both sides, we get
\( \log u = x \log \sin x \)
Differentiating with respect to \( x \), we get
\( \frac{1}{u} \frac{du}{dx} = x \cdot \frac{1}{\sin x} . \cos x + \log \sin x \implies \frac{du}{dx} = u(x \cot x + \log \sin x) \)
\( \implies \frac{du}{dx} = (\sin x)^x \{ x \cot x + \log \sin x \} \) ...(ii)
Again \( v = (\cos x)^{\sin x} \)
Taking log on both sides, we get
\( \log v = \sin x \cdot \log \cos x \)
Differentiating both sides with respect to \( x \), we get
\( \frac{1}{v} \frac{dv}{dx} = \sin x . \frac{1}{\cos x} (-\sin x) + \log(\cos y) . \cos x \)
\( \implies \frac{dv}{dx} = v \left\{ -\frac{\sin^2 x}{\cos x} + \cos x . \log \cos x \right\} = (\cos x)^{\sin x} \left\{ \cos x \log \cos x - \frac{\sin^2 x}{\cos x} \right\} \)
\( = (\cos x)^{1 + \sin x} \{ \log(\cos x) - \tan^2 x \} \) ...(iii)
From (i), (ii) and (iii), we get
\( \frac{dy}{dx} = (\sin x)^x \{ x \cot x + \log \sin x \} + (\cos x)^{1 + \sin x} \{ \log(\cos x) - \tan^2 x \} \)

Question. If \( \cos y = x \cos(a + y) \), with \( \cos a \neq \pm 1 \), then prove that \( \frac{dy}{dx} = \frac{\cos^2(a + y)}{\sin a} \). Hence show that \( \sin a \frac{d^2y}{dx^2} + \sin 2(a + y) \frac{dy}{dx} = 0 \). 
Answer: Sol. Given, \( \cos y = x \cos(a + y) \)
\( \therefore x = \frac{\cos y}{\cos(a + y)} \)
Differentiating with respect to \( y \) on both sides, we get
\( \frac{dx}{dy} = \frac{\cos(a + y) \times (-\sin y) - \cos y \times [-\sin(a + y)]}{\cos^2(a + y)} \)
\( \implies \frac{dx}{dy} = \frac{\cos y \sin(a + y) - \sin y \cos(a + y)}{\cos^2(a + y)} \implies \frac{dx}{dy} = \frac{\sin(a + y - y)}{\cos^2(a + y)} \)
\( \implies \frac{dx}{dy} = \frac{\sin a}{\cos^2(a + y)} \therefore \frac{dy}{dx} = \frac{\cos^2(a + y)}{\sin a} \)
Differentiating both sides w.r.t. \( x \), we get
\( \frac{d^2y}{dx^2} = \frac{1}{\sin a} \left\{ -2 \cos(a + y) \cdot \sin(a + y) \cdot \frac{dy}{dx} \right\} \)
\( \implies \sin a \frac{d^2y}{dx^2} = -\sin 2(a + y) \cdot \frac{dy}{dx} \implies \sin a \frac{d^2y}{dx^2} + \sin 2(a + y) \frac{dy}{dx} = 0 \)

Question. If \( x = a \sin 2t (1 + \cos 2t) \) and \( y = b \cos 2t (1 - \cos 2t) \), then show that \( \left. \frac{dy}{dx} \right|_{t = \frac{\pi}{4}} = \frac{b}{a} \). Also find the value of \( \left( \frac{dy}{dx} \right) \) at \( t = \frac{\pi}{3} \). 
Answer: Sol. Given, \( x = a \sin 2t (1 + \cos 2t) \) and \( y = b \cos 2t (1 - \cos 2t) \)
\( \frac{dx}{dt} = a [\sin 2t \times (-\sin 2t) \times 2 + (1 + \cos 2t) \times 2 \cos 2t] = a[- 2 \sin^2 2t + 2 \cos 2t + 2 \cos^2 2t] \)
\( = a(2 \cos 4t + 2 \cos 2t) = 2a (\cos 4t + \cos 2t) \)
Again, \( \frac{dy}{dt} = b [\cos 2t \times 2 \sin 2t + (1 - \cos 2t)(- 2 \sin 2t)] \)
\( = b[\sin 4t - 2 \sin 2t + 2 \sin 4t] = b[2 \sin 4t - 2 \sin 2t] = 2b(\sin 4t - \sin 2t) \)
\( \therefore \frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{2b(\sin 4t - \sin 2t)}{2a(\cos 4t + \cos 2t)} = \frac{b}{a} \left[ \frac{\sin 4t - \sin 2t}{\cos 4t + \cos 2t} \right] \)
So, \( \left( \frac{dy}{dx} \right)_{at \: t = \frac{\pi}{4}} = \frac{b}{a} \left[ \frac{\sin \pi - \sin \frac{\pi}{2}}{\cos \pi + \cos \frac{\pi}{2}} \right] = \frac{b}{a} \times \left( \frac{-1}{-1} \right) = \frac{b}{a} \) and
\( \left( \frac{dy}{dx} \right)_{at \: t = \frac{\pi}{3}} = \frac{b}{a} \left[ \frac{\sin \frac{4\pi}{3} - \sin \frac{2\pi}{3}}{\cos \frac{4\pi}{3} + \cos \frac{2\pi}{3}} \right] = \frac{b}{a} \left[ \frac{-\sin \frac{\pi}{3} - \sin \frac{\pi}{3}}{-\cos \frac{\pi}{3} - \cos \frac{\pi}{3}} \right] \)
\( = \frac{b}{a} \left( \frac{- 2 \sin \frac{\pi}{3}}{- 2 \cos \frac{\pi}{3}} \right) = \frac{b}{a} \tan \frac{\pi}{3} = \frac{\sqrt{3}b}{a} \)

Question. If \( \sqrt{1 - x^2} + \sqrt{1 - y^2} = a(x - y) \), then show that \( \frac{dy}{dx} = \sqrt{\frac{1 - y^2}{1 - x^2}} \). 
Answer: Sol. Given, \( \sqrt{1 - x^2} + \sqrt{1 - y^2} = a(x - y) \)
Putting \( x = \sin \alpha \implies \alpha = \sin^{-1} x \) and \( y = \sin \beta \implies \beta = \sin^{-1} y \), we get
\( \sqrt{1 - \sin^2 \alpha} + \sqrt{1 - \sin^2 \beta} = a(\sin \alpha - \sin \beta) \) \( \implies \cos \alpha + \cos \beta = a(\sin \alpha - \sin \beta) \)
\( \implies \) \( 2 \cos \frac{(\alpha + \beta)}{2} \cos \frac{(\alpha - \beta)}{2} = a . 2 \cos \frac{(\alpha + \beta)}{2} \sin \frac{(\alpha - \beta)}{2} \)
\( \implies \) \( \cot \left( \frac{\alpha - \beta}{2} \right) = a \implies \frac{\alpha - \beta}{2} = \cot^{-1} a \implies \alpha - \beta = 2 \cot^{-1} a \)
\( \implies \) \( \sin^{-1} x - \sin^{-1} y = 2 \cot^{-1} a \)
Differentiating both sides with respect to \( x \), we get
\( \frac{1}{\sqrt{1 - x^2}} - \frac{1}{\sqrt{1 - y^2}} \frac{dy}{dx} = 0 \implies \frac{dy}{dx} = \sqrt{\frac{1 - y^2}{1 - x^2}} \)

Question. Differentiate \( \tan^{-1} \frac{x}{\sqrt{1 - x^2}} \) with respect to \( \sin^{-1}(2x\sqrt{1 - x^2}) \). 
Answer: Sol. Let \( u = \tan^{-1} \left( \frac{x}{\sqrt{1 - x^2}} \right) \) and \( v = \sin^{-1}(2x\sqrt{1 - x^2}) \) then find \( \frac{du}{dv} \).
Put \( x = \sin \theta \implies \theta = \sin^{-1} x \)
Now, \( u = \tan^{-1} \left( \frac{\sin \theta}{\sqrt{1 - \sin^2 \theta}} \right) \) \( \implies \) \( u = \tan^{-1} \left( \frac{\sin \theta}{\cos \theta} \right) \) \( \implies u = \theta \)
\( \implies u = \sin^{-1} x \implies \frac{du}{dx} = \frac{1}{\sqrt{1 - x^2}} \)
Again, \( v = \sin^{-1}(2x\sqrt{1 - x^2}) \)
\( \implies v = \sin^{-1} (2 \sin \theta \sqrt{1 - \sin^2 \theta} = \sin^{-1} (2 \sin \theta \cos \theta) \)
\( \implies v = \sin^{-1}(\sin 2\theta) \implies v = 2\theta \)
\( \implies v = 2 \sin^{-1} x \implies \frac{dv}{dx} = \frac{2}{\sqrt{1 - x^2}} \)
\( \therefore \frac{du}{dv} = \frac{du}{dx} \cdot \frac{dx}{dv} = \frac{\frac{1}{\sqrt{1 - x^2}}}{\frac{2}{\sqrt{1 - x^2}}} = \frac{1}{2} \)
[Note: Here the range of \( x \) is taken as \( -\frac{1}{\sqrt{2}} < x < \frac{1}{\sqrt{2}} \)]

Question. If \( x^{13}y^7 = (x + y)^{20} \), then prove that \( \frac{dy}{dx} = \frac{y}{x} \). 
Answer: Sol. Given \( x^{13}y^7 = (x + y)^{20} \)
Taking logarithm on both sides, we get
\( \log (x^{13}y^7) = \log (x + y)^{20} \)
\( \implies \log x^{13} + \log y^7 = 20 \log (x + y) \implies 13 \log x + 7 \log y = 20 \log (x + y) \)
Differentiating both sides with respect to \( x \), we get
\( \frac{13}{x} + \frac{7}{y} \frac{dy}{dx} = \frac{20}{x + y} \left( 1 + \frac{dy}{dx} \right) \) \( \implies \frac{13}{x} - \frac{20}{x + y} = \left( \frac{20}{x + y} - \frac{7}{y} \right) \frac{dy}{dx} \)
\( \implies \frac{13x + 13y - 20x}{x(x + y)} = \left( \frac{20y - 7x - 7y}{(x + y) \cdot y} \right) \frac{dy}{dx} \implies \frac{13y - 7x}{x(x + y)} = \left( \frac{13y - 7x}{y(x + y)} \right) \frac{dy}{dx} \)
\( \implies \frac{dy}{dx} = \frac{13y - 7x}{x(x + y)} \cdot \frac{y(x + y)}{13y - 7x} \implies \frac{dy}{dx} = \frac{y}{x} \)

Question. If \( e^x + e^y = e^{x + y} \), then prove that \( \frac{dy}{dx} + e^{y - x} = 0 \).
Answer: Sol. Given, \( e^x + e^y = e^{x + y} \)
Differentiating both sides with respect to \( x \), we get
\( e^x + e^y \cdot \frac{dy}{dx} = e^{x + y} \left\{ 1 + \frac{dy}{dx} \right\} \)
\( \implies e^x + e^y \cdot \frac{dy}{dx} = e^{x + y} + e^{x + y} \cdot \frac{dy}{dx} \implies (e^{x + y} - e^y) \frac{dy}{dx} = e^x - e^{x + y} \)
\( \implies (e^x + e^y - e^y) \frac{dy}{dx} = e^x - e^x - e^y \) [\( \because e^x + e^y = e^{x + y} \) (given)]
\( \implies e^x \cdot \frac{dy}{dx} = -e^y \implies \frac{dy}{dx} = -\frac{e^y}{e^x} \)
\( \implies \frac{dy}{dx} = -e^{y - x} \implies \frac{dy}{dx} + e^{y - x} = 0 \)

Question. If \( x = e^{\cos 2t} \) and \( y = e^{\sin 2t} \), prove that \( \frac{dy}{dx} = - \frac{y \log x}{x \log y} \). 
Answer: Sol. We have \( x = e^{\cos 2t} \)
\( \frac{dx}{dt} = e^{\cos 2t} (-\sin 2t) \times 2 = -2x \sin 2t \) [Differentiating w.r.t. t]
Again \( y = e^{\sin 2t} \)
\( \frac{dy}{dt} = e^{\sin 2t} \cdot 2 \cos 2t = 2y \cos 2t \) [Differentiating w.r.t. t]
Now, \( \frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{2y \cos 2t}{-2x \sin 2t} \implies \frac{dy}{dx} = - \frac{y \cos 2t}{x \sin 2t} \)
\( \because x = e^{\cos 2t} \implies \log x = \cos 2t; y = e^{\sin 2t} \implies \log y = \sin 2t \)
\( \implies \frac{dy}{dx} = - \frac{y \log x}{x \log y} \)
Hence proved.

Question. If \( x \in R - [-1, 1] \) then prove that the derivative of \( \sec^{-1} x \) with respect to \( x \) is \( \frac{1}{|x|\sqrt{x^2 - 1}} \). 
Answer: Sol. Let \( y = \sec^{-1} x \)
Then, \( \sec y = \sec(\sec^{-1} x) = x \)
Differentiating both sides with respect to \( x \), we have
\( \frac{d}{dx} \sec y = \frac{d}{dx}(x) \implies \sec y \tan y \frac{dy}{dx} = 1 \) [Using chain rule]
\( \implies \frac{dy}{dx} = \frac{1}{\sec y \tan y} = \frac{1}{|\sec y| . |\tan y|} \)
\( \frac{dy}{dx} = \frac{1}{|\sec y|\sqrt{\tan^2 y}} = \frac{1}{|\sec y|\sqrt{\sec^2 y - 1}} = \frac{1}{|x|\sqrt{x^2 - 1}} \)

SECOND ORDER DERIVATIVES

Question. If \( x = a \cos \theta + b \sin \theta \) and \( y = a \sin \theta - b \cos \theta \), then show that \( y^2 \frac{d^2y}{dx^2} - x \frac{dy}{dx} + y = 0 \). 
Answer: Sol. Given, \( x = a \cos \theta + b \sin \theta \implies \frac{dx}{d\theta} = -a \sin \theta + b \cos \theta \) ...(i)
Also, \( y = a \sin \theta - b \cos \theta \implies \frac{dy}{d\theta} = a \cos \theta + b \sin \theta \) ...(ii)
\( \therefore \frac{dy}{dx} = \frac{dy/d\theta}{dx/d\theta} = \frac{a \cos \theta + b \sin \theta}{-a \sin \theta + b \cos \theta} \) [From (i) and (ii)]
\( \implies \frac{dy}{dx} = \frac{a \cos \theta + b \sin \theta}{b \cos \theta - a \sin \theta} \implies \frac{dy}{dx} = \frac{x}{y} \) ...(iii)
Differentiating again with respect to \( x \), we get
\( \frac{d^2y}{dx^2} = \frac{y \cdot 1 - x \cdot \frac{dy}{dx}}{y^2} \)
\( \implies y^2 \frac{d^2y}{dx^2} = y - x \frac{dy}{dx} \implies y^2 \frac{d^2y}{dx^2} + x \frac{dy}{dx} - y = 0 \)

Question. If \( y = Pe^{ax} + Qe^{bx} \), then show that \( \frac{d^2y}{dx^2} - (a + b) \frac{dy}{dx} + aby = 0 \) 
Answer: Sol. Given \( y = Pe^{ax} + Qe^{bx} \)
On differentiating with respect to \( x \), we have
\( \frac{dy}{dx} = Pae^{ax} + Qbe^{bx} \)
Again, differentiating with respect to \( x \), we have
\( \frac{d^2y}{dx^2} = Pa^2e^{ax} + Qb^2e^{bx} \)
Now, LHS \( = \frac{d^2y}{dx^2} - (a + b) \frac{dy}{dx} + aby \)
\( = Pa^2e^{ax} + Qb^2e^{bx} - (a + b)(Pae^{ax} + Qbe^{bx}) + ab(Pe^{ax} + Qe^{bx}) \)
\( = Pa^2e^{ax} + Qb^2e^{bx} - Pa^2e^{ax} - Pabe^{ax} - Qabe^{bx} - Qb^2e^{bx} + Pabe^{ax} + Qabe^{bx} = 0 = RHS \)

Question. If \( y = \sin(\log x) \), then prove that \( x^2 \frac{d^2y}{dx^2} + x \frac{dy}{dx} + y = 0 \).
Answer: Sol. Given, \( y = \sin(\log x) \)
\( \implies \frac{dy}{dx} = \cos(\log x) \times \frac{1}{x} = \frac{\cos(\log x)}{x} \)
Again, \( \frac{d^2y}{dx^2} = \frac{x [-\sin(\log x) \times \frac{1}{x}] - \cos(\log x) \cdot 1}{x^2} = \frac{-\sin(\log x) - \cos(\log x)}{x^2} \)
Now, LHS \( = x^2 \frac{d^2y}{dx^2} + x \frac{dy}{dx} + y \)
\( = x^2 \left[ \frac{-\sin(\log x) - \cos(\log x)}{x^2} \right] + \frac{x \cos(\log x)}{x} + \sin(\log x) \)
\( = -\cos(\log x) - \sin(\log x) + \cos(\log x) + \sin(\log x) = 0 = RHS \)

Question. If \( y = 3 \cos(\log x) + 4 \sin(\log x) \), show that \( x^2 \frac{d^2y}{dx^2} + x \frac{dy}{dx} + y = 0 \). 
Answer: Sol. Given, \( y = 3 \cos(\log x) + 4 \sin(\log x) \)
Differentiating with respect to \( x \), we get
\( \frac{dy}{dx} = - \frac{3 \sin(\log x)}{x} + \frac{4 \cos(\log x)}{x} \implies y_1 = \frac{1}{x} [-3 \sin(\log x) + 4 \cos(\log x)] \)
Again differentiating with respect to \( x \), we get
\( \frac{d^2y}{dx^2} = \frac{x [ \frac{-3 \cos(\log x)}{x} - \frac{4 \sin(\log x)}{x} ] - [-3 \sin(\log x) + 4 \cos(\log x)]}{x^2} \)
\( = \frac{-3 \cos(\log x) - 4 \sin(\log x) + 3 \sin(\log x) - 4 \cos(\log x)}{x^2} \)
\( \frac{d^2y}{dx^2} = \frac{-\sin(\log x) - 7 \cos(\log x)}{x^2} \implies y_2 = \frac{-\sin(\log x) - 7 \cos(\log x)}{x^2} \)
Now, LHS \( = x^2 y_2 + x y_1 + y \)
\( = x^2 \left[ \frac{-\sin(\log x) - 7 \cos(\log x)}{x^2} \right] + x \times \frac{1}{x} [-3 \sin(\log x) + 4 \cos(\log x)] + 3 \cos(\log x) + 4 \sin(\log x) \)
\( = -\sin(\log x) - 7 \cos(\log x) - 3 \sin(\log x) + 4 \cos(\log x) + 3 \cos(\log x) + 4 \sin(\log x) = 0 = RHS \)

Question. If \( x = a(\cos t + t \sin t) \) and \( y = a(\sin t - t \cos t), 0 < t < \frac{\pi}{2} \), find \( \frac{d^2x}{dt^2}, \frac{d^2y}{dt^2} \) and \( \frac{d^2y}{dx^2} \). 
Answer: Sol. Given, \( x = a(\cos t + t \sin t) \)
Differentiating both sides with respect to \( t \), we get
\( \frac{dx}{dt} = a(-\sin t + \sin t + t \cos t) \implies \frac{dx}{dt} = at \cos t \) ...(i)
Differentiating again with respect to \( t \), we get
\( \frac{d^2x}{dt^2} = a(\cos t - t \sin t) \)
Again, \( y = a(\sin t - t \cos t) \)
Differentiating with respect to \( t \), we get
\( \frac{dy}{dt} = a(\cos t - \cos t + t \sin t) \implies \frac{dy}{dt} = at \sin t \) ...(ii)
Differentiating again with respect to \( t \), we get
\( \frac{d^2y}{dt^2} = a(\sin t + t \cos t) \)
Now, \( \frac{dy}{dx} = \frac{dy/dt}{dx/dt} \) [From (i) and (ii)]
\( \implies \frac{dy}{dx} = \frac{at \sin t}{at \cos t} \implies \frac{dy}{dx} = \tan t \)

Rolle's and Mean Value Theorem

Question. Verify Lagrange's Mean Value Theorem for the following function: \( f(x) = x^2 + 2x + 3, \text{for } [4, 6] \). 
Answer: Sol. \( f(x) = x^2 + 2x + 3 \) for [4, 6]
(i) Given function is a polynomial, hence it is continuous.
(ii) \( f'(x) = 2x + 2 \), which is differentiable.
\( f(4) = 16 + 8 + 3 = 27 \) and \( f(6) = 36 + 12 + 3 = 51 \)
\( \implies f(4) \neq f(6) \). All conditions of Mean Value Theorem are satisfied.
\( \therefore \) There exist at least one real value \( c \in (4, 6) \) such that
\( f'(c) = \frac{f(6) - f(4)}{6 - 4} = \frac{51 - 27}{2} = \frac{24}{2} = 12 \)
\( \implies 2c + 2 = 12 \) or \( c = 5 \in (4, 6) \)
Hence, Lagrange's Mean Value Theorem is verified.

Question. Discuss the applicability of Rolle's theorem on the function given by \( f(x) = \begin{cases} x^2 + 1, & \text{if } 0 \leq x \leq 1 \\ 3 - x, & \text{if } 1 \leq x \leq 2 \end{cases} \). 
Answer: Sol. We have, \( f(x) = \begin{cases} x^2 + 1, & \text{if } 0 \leq x \leq 1 \\ 3 - x, & \text{if } 1 \leq x \leq 2 \end{cases} \)
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 \( = \lim_{x \to 1^-} \frac{f(x) - f(1)}{x - 1} \)
\( = \lim_{x \to 1^-} \frac{(x^2 + 1) - (1 + 1)}{x - 1} = \lim_{x \to 1^-} \frac{x^2 - 1}{x - 1} = \lim_{x \to 1^-} (x + 1) = 2 \)
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: Sol. 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, \( f'(c) = \frac{f(4) - f(3)}{4 - 3} \)
\( \implies 2(c - 3) = \frac{(4 - 3)^2 - (3 - 3)^2}{4 - 3} \implies 2c - 6 = \frac{1 - 0}{1} \implies c = \frac{7}{2} \)
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).