CBSE Class 12 Mathematics Continuity And Differentiability Worksheet Set 02

CBSE Class 12 Mathematics Continuity And Differentiability Worksheet 2. The Continuity And Differentiability questions in the worksheets have been specifically designed by best mathematics teachers so that the students can practise them to clear their Continuity And Differentiability concepts and get better marks in class 12 mathematics tests and examinations. Students can free download these Continuity And Differentiability worksheets in pdf and practice them. This will help them to get better marks in examinations. Also refer to other worksheets for the Continuity And Differentiability chapter and other subjects too. Use them for better understanding of the subjects.

Question. If \( x = 2\cos\theta - \cos(2\theta) \) and \( y = 2\sin\theta - \sin(2\theta) \). Show that \( \frac{dy}{dx} = \tan \left(\frac{3\theta}{2}\right) \).
Answer: We have, \( x = 2\cos\theta - \cos(2\theta) \)
Diff w.r.t. \( \theta \)
\( \frac{dx}{d\theta} = -2\sin\theta + 2\sin(2\theta) \)
\( \Rightarrow \frac{dx}{d\theta} = 2(\sin(2\theta) - \sin\theta) \)
and \( y = 2\sin\theta - \sin(2\theta) \)
Diff w.r.t \( \theta \)
\( \frac{dy}{d\theta} = 2\cos\theta - 2\cos(2\theta) \)
\( \Rightarrow \frac{dy}{d\theta} = 2(\cos\theta - \cos(2\theta)) \)
Now, \( \frac{dy}{dx} = \frac{dy/d\theta}{dx/d\theta} = \frac{2(\cos\theta - \cos(2\theta))}{2(\sin(2\theta) - \sin\theta)} \)
\( \Rightarrow \frac{dy}{dx} = \frac{-2\sin\left(\frac{3\theta}{2}\right)\sin\left(\frac{-\theta}{2}\right)}{2\cos\left(\frac{3\theta}{2}\right)\sin\left(\frac{\theta}{2}\right)} \ldots \{\cos A - \cos B \text{ & } \sin A - \sin B \text{ formula}\} \)
\( \Rightarrow \frac{dy}{dx} = \frac{\sin(3\theta/2)}{\cos(3\theta/2)} \ldots \{\sin(-\theta) = -\sin\theta\} \)
\( \Rightarrow \frac{dy}{dx} = \tan \left(\frac{3\theta}{2}\right) \) Ans.

Question. \( x = \left(t + \frac{1}{t}\right)^a \) and \( y = a^{t + \frac{1}{t}} \). Find \( \frac{dy}{dx} \).
Answer: We have, \( x = \left(t + \frac{1}{t}\right)^a \)
Diff w.r.t. \( t \)
\( \frac{dx}{dt} = a \left(t + \frac{1}{t}\right)^{a-1} \cdot \left(1 - \frac{1}{t^2}\right) \)
and \( y = a^{t + \frac{1}{t}} \)
Diff w.r.t. \( t \)
\( \frac{dy}{dt} = a^{t + \frac{1}{t}} \cdot \log a \cdot \left(1 - \frac{1}{t^2}\right) \ldots \left\{\frac{d}{dx}(a^x) = a^x \log a\right\} \)
Now,
\( \frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{a^{t + \frac{1}{t}} \cdot \log a \cdot \left(1 - \frac{1}{t^2}\right)}{a \left(t + \frac{1}{t}\right)^{a-1} \cdot \left(1 - \frac{1}{t^2}\right)} \)
\( \frac{dy}{dx} = \frac{a^{t + \frac{1}{t}} \cdot \log a}{a \left(t + \frac{1}{t}\right)^{a-1}} \) Ans.

Question. \( x = \frac{\sin^3 t}{\sqrt{\cos(2t)}} \) and \( y = \frac{\cos^3 t}{\sqrt{\cos(2t)}} \). Find \( \frac{dy}{dx} \).
Answer: We have, \( x = \frac{\sin^3 t}{\sqrt{\cos(2t)}} \)
taking log on both sides
\( \log x = 3\log(\sin t) - \frac{1}{2}\log(\cos(2t)) \ldots \{\text{using log properties}\} \)
Diff w.r.t. 't'
\( \frac{1}{x} \cdot \frac{dx}{dt} = 3 \cdot \frac{1}{\sin t} \cdot \cos t - \frac{1}{2} \cdot \frac{1}{\cos(2t)} (-\sin(2t)) \cdot 2 \)
\( \Rightarrow \frac{dx}{dt} = x \left[ \frac{3}{\tan t} + \tan(2t) \right] \)
and \( y = \frac{\cos^3 t}{\sqrt{\cos(2t)}} \)
taking log on both sides
\( \log y = 3\log(\cos t) - \frac{1}{2}\log(\cos(2t)) \ldots \{\text{using log properties}\} \)
Diff w.r.t. 't'
\( \frac{dy}{dx} = \frac{\tan^2 t}{\tan^3 t} \left[ \frac{-1+3\tan^2 t}{3-\tan^2 t} \right] \)
\( = \frac{1}{\tan t} \left( \frac{-1+3\tan^2 t}{3-\tan^2 t} \right) \)
\( = \frac{-1+3\tan^2 t}{3\tan t - \tan^3 t} \)
\( = \frac{-(1-3\tan^2 t)}{3\tan t - \tan^3 t} \)
\( = \frac{-1}{\tan(3t)} \ldots \left\{\tan(3\theta) = \frac{3\tan\theta-\tan^3\theta}{1-3\tan^2\theta}\right\} \)
\( \frac{dy}{dx} = -\cot(3t) \) Ans.

Higher Order Derivative

Question. If \( y = \sin^{-1} x \), then show that \( (1 - x^2) \frac{d^2 y}{dx^2} - x \frac{dy}{dx} = 0 \).
Answer: We have, \( y = \sin^{-1} x \)
\( \frac{dy}{dx} = \frac{1}{\sqrt{1-x^2}} \)
\( \Rightarrow \sqrt{1-x^2} \cdot \frac{dy}{dx} = 1 \)
Diff again w.r.t. x (product rule)
\( \sqrt{1-x^2} \cdot \frac{d^2 y}{dx^2} + \frac{dy}{dx} \cdot \frac{1}{2\sqrt{1-x^2}} (-2x) = 0 \)
\( \Rightarrow \sqrt{1-x^2} \frac{d^2 y}{dx^2} - \frac{x}{\sqrt{1-x^2}} \cdot \frac{dy}{dx} = 0 \)
LCM
\( \Rightarrow \frac{(1-x^2) \frac{d^2 y}{dx^2} - x \frac{dy}{dx}}{\sqrt{1-x^2}} = 0 \)
\( \Rightarrow (1-x^2) \frac{d^2 y}{dx^2} - x \frac{dy}{dx} = 0 \) Hence Proved

Question. \( y = (\cot^{-1} x)^2 \) Show that \( (x^2 + 1)^2 \cdot y_2 + 2x(x^2 + 1)y_1 = 2 \).
Answer: We have, \( y = (\cot^{-1} x)^2 \)
Diff w.r.t. x
\( \frac{dy}{dx} = 2(\cot^{-1} x) \cdot \left( \frac{-1}{1+x^2} \right) \ldots \left\{\frac{d}{dx}(\cot^{-1} x) = \frac{-1}{1+x^2}\right\} \)
\( \Rightarrow (1+x^2) \cdot \frac{dy}{dx} = -2\cot^{-1} x \)
Diff again w.r.t. (product rule on LHS)
\( (1+x^2) \frac{d^2 y}{dx^2} + \frac{dy}{dx} (2x) = \frac{2}{1+x^2} \)
\( \Rightarrow (1+x^2)^2 \frac{d^2 y}{dx^2} + 2x(1+x^2) \frac{dy}{dx} = 2 \) (Proved)

Question. If \( y = e^{a\cos^{-1} x} \), show that \( (1-x^2) \frac{d^2 y}{dx^2} - x \frac{dy}{dx} - a^2 y = 0 \).
Answer: We have, \( y = e^{a\cos^{-1} x} \)
taking log on both sides
\( \log y = a\cos^{-1} x \cdot \log e \)
\( \Rightarrow \log y = a\cos^{-1} x \ldots \{\because \log e = 1\} \)
Diff w.r.t. x
\( \frac{1}{y} \cdot \frac{dy}{dx} = \frac{-a}{\sqrt{1-x^2}} \)
\( \Rightarrow \sqrt{1-x^2} \frac{dy}{dx} = -ay \ldots (1) \)
Diff again w.r.t. x {product rule on LHS}
\( \Rightarrow \sqrt{1-x^2} \cdot \frac{d^2 y}{dx^2} + \frac{dy}{dx} \cdot \frac{1}{2\sqrt{1-x^2}} (-2x) = -a \frac{dy}{dx} \)
\( \Rightarrow \sqrt{1-x^2} \cdot \frac{d^2 y}{dx^2} - \frac{x}{\sqrt{1-x^2}} \cdot \frac{dy}{dx} = -a \frac{dy}{dx} \)
LCM \( (\sqrt{1-x^2}) \)
\( \Rightarrow \frac{(1-x^2)\frac{d^2 y}{dx^2} - x\frac{dy}{dx}}{\sqrt{1-x^2}} = -a \frac{dy}{dx} \)
\( \Rightarrow (1-x^2) y_2 - x y_1 = -a\sqrt{1-x^2} \cdot \frac{dy}{dx} \)
\( \Rightarrow (1-x^2)y_2 - x y_1 = -a(-ay) \ldots \{\text{from eq. (1)}\} \)
\( \Rightarrow (1-x^2)y_2 - x y_1 = a^2 y \) (Proved)

Question. If \( y = A\cos(\log x) + B\sin(\log x) \), show that \( x^2 \frac{d^2 y}{dx^2} + x \frac{dy}{dx} + y = 0 \).
Answer: We have, \( y = A\cos(\log x) + B\sin(\log x) \ldots (1) \)
Diff w.r.t. x
\( \frac{dy}{dx} = -A\sin(\log x) \cdot \frac{1}{x} + B\cos(\log x) \cdot \frac{1}{x} \)
\( \Rightarrow \frac{dy}{dx} = \frac{-A\sin(\log x) + B\cos(\log x)}{x} \)
\( \Rightarrow x \frac{dy}{dx} = -A\sin(\log x) + B\cos(\log x) \)
Diff w.r.t. x
\( x \cdot \frac{d^2 y}{dx^2} + \frac{dy}{dx} = -A\cos(\log x) \cdot \frac{1}{x} - B\sin(\log x) \cdot \frac{1}{x} \)
\( \Rightarrow x^2 \frac{d^2 y}{dx^2} + x \frac{dy}{dx} = -[A\cos(\log x) + B\sin(\log x)] \)
\( \Rightarrow x^2 \frac{d^2 y}{dx^2} + x \frac{dy}{dx} = -y \ldots \{\text{from eq. (1)}\} \)
\( \Rightarrow x^2 \frac{d^2 y}{dx^2} + x \frac{dy}{dx} + y = 0 \) (Proved)

Question. If \( y = \frac{\sin^{-1} x}{\sqrt{1-x^2}} \), show that \( (1-x^2)y_2 - 3xy_1 - y = 0 \).
Answer: We have, \( y = \frac{\sin^{-1} x}{\sqrt{1-x^2}} \)
\( \Rightarrow y\sqrt{1-x^2} = \sin^{-1} x \)
Diff w.r.t. x (product rule on LHS)
\( y \cdot \frac{1}{2\sqrt{1-x^2}} (-2x) + \sqrt{1-x^2} \cdot \frac{dy}{dx} = \frac{1}{\sqrt{1-x^2}} \)
\( \Rightarrow \frac{-xy}{\sqrt{1-x^2}} + \sqrt{1-x^2} \cdot \frac{dy}{dx} = \frac{1}{\sqrt{1-x^2}} \)
\( \Rightarrow \frac{-xy + (1-x^2)dy/dx}{\sqrt{1-x^2}} = \frac{1}{\sqrt{1-x^2}} \)
\( \Rightarrow (1-x^2)\frac{dy}{dx} - xy = 1 \)
Diff again w.r.t. x
\( (1-x^2) \frac{d^2 y}{dx^2} + \frac{dy}{dx}(-2x) - \left[ x \frac{dy}{dx} + y \cdot 1 \right] = 0 \)
\( \Rightarrow (1-x^2) \frac{d^2 y}{dx^2} - 2x \frac{dy}{dx} - x \frac{dy}{dx} - y = 0 \)
\( \Rightarrow (1-x^2) \frac{d^2 y}{dx^2} - 3x \frac{dy}{dx} - y = 0 \) (Proved)

Question. If \( y = [\log(x + \sqrt{x^2+1})]^2 \), show that \( (1 + x^2) \frac{d^2 y}{dx^2} + x \frac{dy}{dx} = 2 \).
Answer: We have, \( y = [\log(x + \sqrt{x^2+1})]^2 \)
Diff w.r.t. x
\( \frac{dy}{dx} = 2[\log(x + \sqrt{x^2+1})] \cdot \frac{1}{x+\sqrt{x^2+1}} \cdot \left( 1 + \frac{1}{2\sqrt{x^2+1}}(2x) \right) \)
\( \Rightarrow \frac{dy}{dx} = 2\log(x + \sqrt{x^2+1}) \cdot \frac{1}{x+\sqrt{x^2+1}} \cdot \left( \frac{\sqrt{x^2+1}+x}{\sqrt{x^2+1}} \right) \)
\( \Rightarrow \frac{dy}{dx} = \frac{2\log(x+\sqrt{x^2+1})}{\sqrt{x^2+1}} \)
\( \Rightarrow \sqrt{1+x^2} \frac{dy}{dx} = 2\log(x+\sqrt{x^2+1}) \)
Diff again w.r.t. (product rule on LHS)
\( \sqrt{1+x^2} \frac{d^2 y}{dx^2} + \frac{dy}{dx} \cdot \frac{1}{2\sqrt{1+x^2}} (2x) = \frac{2}{x+\sqrt{x^2+1}} \cdot \left( 1 + \frac{2x}{2\sqrt{x^2+1}} \right) \)
\( \Rightarrow \sqrt{1+x^2} \frac{d^2 y}{dx^2} + \frac{x}{\sqrt{1+x^2}} \frac{dy}{dx} = \frac{2}{x+\sqrt{x^2+1}} \cdot \left( \frac{\sqrt{x^2+1}+x}{\sqrt{x^2+1}} \right) \)
\( \Rightarrow \sqrt{1+x^2} \frac{d^2 y}{dx^2} + \frac{x}{\sqrt{1+x^2}} \frac{dy}{dx} = \frac{2}{\sqrt{x^2+1}} \)
LCM in LHS
\( \Rightarrow (1+x^2) \frac{d^2 y}{dx^2} + x \frac{dy}{dx} = 2 \) (Proved)

Question. \( y = \csc^{-1} x \), show that \( (x^2 - 1)y_2 + (2x^2 - 1)y_1 = 0 \).
Answer: We have, \( y = \csc^{-1} x \)
Diff w.r.t. x
\( \frac{dy}{dx} = -\frac{1}{x\sqrt{x^2-1}} \)
\( \Rightarrow \left(x\sqrt{x^2-1}\right) \frac{dy}{dx} = -1 \)
Diff again w.r.t. {product rule on LHS}
\( \Rightarrow \left(x\sqrt{x^2-1}\right) \frac{d^2 y}{dx^2} + \frac{dy}{dx} \left\{ x \cdot \frac{1(2x)}{2\sqrt{x^2-1}} + \sqrt{x^2-1} \right\} = 0 \)
\( \Rightarrow x\sqrt{x^2-1} \frac{d^2 y}{dx^2} + \frac{dy}{dx} \left( \frac{x^2}{\sqrt{x^2-1}} + \sqrt{x^2-1} \right) = 0 \)
\( \Rightarrow x\sqrt{x^2-1} \frac{d^2 y}{dx^2} + \frac{dy}{dx} \left( \frac{x^2 + x^2 - 1}{\sqrt{x^2-1}} \right) = 0 \)
LCM.
\( \Rightarrow x(x^2 - 1) \frac{d^2 y}{dx^2} + \frac{dy}{dx}(2x^2 - 1) = 0 \) (Proved)

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