CBSE Class 12 Mathematics Continuity And Differentiability Worksheet Set 04

CBSE Class 12 Mathematics Continuity And Differentiability Worksheet 4. 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 \( \cos^{-1}\left(\frac{x^2-y^2}{x^2+y^2}\right) = \tan^{-1} a \), show that \( \frac{dy}{dx} = \frac{y}{x} \).
Answer: We have, \( \cos^{-1} \left(\frac{x^2-y^2}{x^2+y^2}\right) = \tan^{-1} a \)
\( \Rightarrow \frac{x^2-y^2}{x^2+y^2} = \cos(\tan^{-1}a) \)
\( \Rightarrow \frac{x^2-y^2}{x^2+y^2} = k \quad \dots \{\text{where } k = \cos(\tan^{-1}a) = \text{constant}\} \dots(1) \)
\( \Rightarrow x^2 - y^2 = k(x^2 + y^2) \)
Diff w.r.t. x
\( \Rightarrow 2x - 2y\frac{dy}{dx} = k\left(2x + 2y\frac{dy}{dx}\right) \)
\( \Rightarrow x - y\frac{dy}{dx} = k\left(x + y\frac{dy}{dx}\right) \)
\( \Rightarrow x - y\frac{dy}{dx} = kx + ky\frac{dy}{dx} \)
\( \Rightarrow x - kx = ky\frac{dy}{dx} + y\frac{dy}{dx} \)
\( \Rightarrow x(1 - k) = y\frac{dy}{dx}(k + 1) \)
\( \Rightarrow \frac{dy}{dx} = \frac{x(1-k)}{y(k+1)} \)
replace \( k \) by \dots \( \{\text{from eq. (1)}\} \)
\( \Rightarrow \frac{dy}{dx} = \frac{x\left[1-\frac{x^2-y^2}{x^2+y^2}\right]}{y\left[\frac{x^2-y^2}{x^2+y^2}+1\right]} \)
\( \Rightarrow \frac{dy}{dx} = \frac{x\left[\frac{x^2+y^2-x^2+y^2}{x^2+y^2}\right]}{y\left[\frac{x^2-y^2+x^2+y^2}{x^2+y^2}\right]} \)
\( \Rightarrow \frac{x(2y^2)}{y(2x^2)} \)
\( \Rightarrow \frac{dy}{dx} = \frac{y}{x} \) Proved.

Question. If \( \sqrt{1-x^6} + \sqrt{1-y^6} = a(x^3 - y^3) \) show that \( \frac{dy}{dx} = \frac{x^2}{y^2}\sqrt{\frac{1-y^6}{1-x^6}} \).
Answer: We have, \( \sqrt{1-x^6} + \sqrt{1-y^6} = a(x^3 - y^3) \)
put \( x^3 = \sin A \) and \( y^3 = \sin B \)
\( \Rightarrow \sqrt{1-\sin^2 A} + \sqrt{1-\sin^2 B} = a(\sin A - \sin B) \)
\( \Rightarrow \cos A + \cos B = a(\sin A - \sin B) \)
\( \Rightarrow 2\cos\left(\frac{A+B}{2}\right)\cos\left(\frac{A-B}{2}\right) = a \cdot 2\cos\left(\frac{A+B}{2}\right)\sin\left(\frac{A-B}{2}\right) \)
\( \Rightarrow \cos\left(\frac{A-B}{2}\right) = a\sin\left(\frac{A-B}{2}\right) \)
\( \Rightarrow \cot\left(\frac{A-B}{2}\right) = a /> \( \Rightarrow \frac{A-B}{2} = \cot^{-1}a \)
\( \Rightarrow A - B = 2\cot^{-1}a \)
replace \( A \) by \( \sin^{-1}x^3 \) & \( B \) by \( \sin^{-1}y^3 \)
\( \therefore \sin^{-1}x^3 - \sin^{-1}y^3 = 2\cot^{-1}a \)
Diff w.r.t. x {RHS is constant}
\( \Rightarrow \frac{1}{\sqrt{1-x^6}}\cdot (3x^2) - \frac{1}{\sqrt{1-y^6}}\cdot \left(3y^2\frac{dy}{dx}\right) = 0 \)
\( \Rightarrow \frac{x^2}{\sqrt{1-x^6}} = \frac{y^2}{\sqrt{1-y^6}}\frac{dy}{dx} \)
\( \Rightarrow \frac{dy}{dx} = \frac{x^2}{y^2}\sqrt{\frac{1-y^6}{1-x^6}} \) (Proved)

Question. If \( \cos y = x\cos(a + y) \) show that \( \frac{dy}{dx} = \frac{\cos^2(a+y)}{\sin a} \).
Answer: We have, \( \cos y = x\cos(a + y) \dots(1) \)
Diff w.r.t. x {product rule on RHS}
\( -\sin y \cdot \frac{dy}{dx} = -x\sin(a+y)\frac{dy}{dx} + \cos(a+y)\cdot 1 \)
\( \Rightarrow x\sin(a+y)\frac{dy}{dx} - \sin y \frac{dy}{dx} = \cos(a+y) \)
\( \Rightarrow \frac{dy}{dx}(x\sin(a+y) - \sin y) = \cos(a+y) \)
\( \Rightarrow \frac{dy}{dx} = \frac{\cos(a+y)}{x\sin(a+y) - \sin y} \)
\( \Rightarrow \frac{dy}{dx} = \frac{\cos(a+y)}{\frac{\cos y}{\cos(a+y)}\sin(a+y) - \sin y} \dots \{\text{from eq. } x = \frac{\cos y}{\cos(a+y)}\} \)
\( \Rightarrow \frac{dy}{dx} = \frac{\cos^2(a+y)}{\sin(a+y)\cos y - \cos(a+y)\sin y} \)
\( \Rightarrow \frac{dy}{dx} = \frac{\cos^2(a+y)}{\sin(a+y-y)} \dots \{\sin A\cos B - \cos A\sin B = \sin(A-B)\} \)
\( \Rightarrow \frac{dy}{dx} = \frac{\cos^2(a+y)}{\sin a} \) (Proved)

Question. If \( x\sin(a + y) + \sin a \cdot \cos(a + y) = 0 \), show that \( \frac{dy}{dx} = \frac{\sin^2(a+y)}{\sin a} \).
Answer: We have, \( x\sin(a + y) + \sin a \cdot \cos(a + y) = 0 \dots(1) \)
Diff w.r.t. x {\( \sin a \) constant}
\( x\cos(a+y)\frac{dy}{dx} + \sin(a+y)\cdot 1 + \sin a[-\sin(a+y)]\frac{dy}{dx} = 0 \)
\( \Rightarrow x\cos(a+y)\frac{dy}{dx} + \sin(a+y) - \sin a\sin(a+y)\frac{dy}{dx} = 0 \)
\( \Rightarrow \frac{dy}{dx}(x\cos(a+y) - \sin a\sin(a+y)) = -\sin(a+y) \)
\( \Rightarrow \frac{dy}{dx} = \frac{-\sin(a+y)}{x\cos(a+y) - \sin a\sin(a+y)} \)
\( \Rightarrow \frac{dy}{dx} = \frac{-\sin(a+y)}{\frac{-\sin a\cos(a+y)}{\sin(a+y)}\cos(a+y) - \sin a\sin(a+y)} \dots \{\text{from eq. (1) value of } x\} \)
\( \Rightarrow \frac{dy}{dx} = \frac{-\sin^2(a+y)}{-\sin a\cos^2(a+y) - \sin a\sin^2(a+y)} \)
\( \Rightarrow \frac{dy}{dx} = \frac{-\sin^2(a+y)}{-\sin a(\cos^2(a+y) + \sin^2(a+y))} \)
\( \Rightarrow \frac{dy}{dx} = \frac{\sin^2(a+y)}{\sin a} \dots \{\cos^2\theta + \sin^2\theta = 1\} \) (Proved)

Question. If \( x^2 + y^2 = t - \frac{1}{t} \) and \( x^4 + y^4 = t^2 + \frac{1}{t^2} \) then show that \( \frac{dy}{dx} = \frac{1}{x^3y} \).
Answer: We have, \( x^2 + y^2 = t - \frac{1}{t} \dots(1) \)
& \( x^4 + y^4 = t^2 + \frac{1}{t^2} \dots(2) \)
squaring both sides in eq. (1)
\( x^4 + y^4 + 2x^2y^2 = t^2 + \frac{1}{t^2} - 2 \)
\( \Rightarrow t^2 + \frac{1}{t^2} + 2x^2y^2 = t^2 + \frac{1}{t^2} - 2 \dots \{\text{from eq. (2)}\} \)
\( \Rightarrow 2x^2y^2 = -2 /> \( \Rightarrow x^2y^2 = -1 /> \( \Rightarrow y^2 = \frac{-1}{x^2} \)
Diff w.r.t. 'x'
\( \Rightarrow 2y\frac{dy}{dx} = \frac{2}{x^3} \)
\( \Rightarrow \frac{dy}{dx} = \frac{1}{x^3y} \) Ans.

Question. If \( y = \log\left(\tan\left(\frac{\pi}{4} + \frac{x}{2}\right)\right) \). Find \( \frac{dy}{dx} \).
Answer: We have, \( y = \log\left(\tan\left(\frac{\pi}{4} + \frac{x}{2}\right)\right) \)
Diff w.r.t. x
\( \Rightarrow \frac{dy}{dx} = \frac{1}{\tan\left(\frac{\pi}{4} + \frac{x}{2}\right)} \cdot \sec^2\left(\frac{\pi}{4} + \frac{x}{2}\right) \cdot \left(\frac{1}{2}\right) \)
\( \Rightarrow \frac{dy}{dx} = \frac{\cos\left(\frac{\pi}{4} + \frac{x}{2}\right)}{\sin\left(\frac{\pi}{4} + \frac{x}{2}\right)} \times \frac{1}{\cos^2\left(\frac{\pi}{4} + \frac{x}{2}\right)} \times \frac{1}{2} \)
\( \Rightarrow \frac{dy}{dx} = \frac{1}{2\sin\left(\frac{\pi}{4} + \frac{x}{2}\right)\cos\left(\frac{\pi}{4} + \frac{x}{2}\right)} /> \( \Rightarrow \frac{dy}{dx} = \frac{1}{\sin\left(\frac{\pi}{2} + x\right)} \dots \{2\sin\theta\cos\theta = \sin(2\theta)\} /> \( \Rightarrow \frac{dy}{dx} = \frac{1}{\cos x} = \sec x \) Ans.

Question. If \( y = \log_7(\log_7 x) \). Find \( \frac{dy}{dx} \).
Answer: We have, \( y = \log_7(\log_7 x) \)
\( \Rightarrow y = \frac{\log(\log_7 x)}{\log 7} \dots \left\{\text{change of base } \log_a b = \frac{\log b}{\log a}\right\} \)
Diff w.r.t x
\( \Rightarrow \frac{dy}{dx} = \frac{1}{\log 7} \cdot \frac{1}{\log_7 x} \cdot \frac{d}{dx}(\log_7 x) \)
\( \Rightarrow \frac{dy}{dx} = \frac{1}{\log 7} \cdot \frac{1}{\log_7 x} \cdot \frac{d}{dx}\left(\frac{\log x}{\log 7}\right) \)
\( \Rightarrow \frac{dy}{dx} = \frac{1}{\log 7} \cdot \frac{1}{\log_7 x} \cdot \frac{1}{\log 7} \cdot \frac{1}{x} \)
\( \Rightarrow \frac{dy}{dx} = \frac{1}{x(\log 7)^2 \log_7 x} \) Ans.

Question. If \( y = \sqrt{\frac{1-x}{1+x}} \) show that \( (1 - x^2)\frac{dy}{dx} + y = 0 \).
Answer: We have, \( y = \sqrt{\frac{1-x}{1+x}} \)
taking log on both sides
\( \Rightarrow \log y = \frac{1}{2}[\log(1-x) - \log(1+x)] \dots \{\text{using log prop.}\} /> Diff w.r.t. x
\( \Rightarrow \frac{1}{y} \cdot \frac{dy}{dx} = \frac{1}{2}\left[\frac{-1}{1-x} - \frac{1}{1+x}\right] \)
\( \Rightarrow \frac{1}{y}\frac{dy}{dx} = \frac{1}{2}\left[\frac{-1-x-1+x}{1-x^2}\right] \)
\( \Rightarrow \frac{1}{y}\frac{dy}{dx} = \frac{1}{2}\left(\frac{-2}{1-x^2}\right) \)
\( \Rightarrow (1-x^2)\frac{dy}{dx} = -y \)
\( \Rightarrow (1-x^2)\frac{dy}{dx} + y = 0 \) (Proved)

Question. If \( y = \frac{x}{2}\sqrt{a^2-x^2} + \frac{a^2}{2}\sin^{-1}\left(\frac{x}{a}\right) \). Find \( \frac{dy}{dx} \).
Answer: We have, \( y = \frac{x}{2}\sqrt{a^2-x^2} + \frac{a^2}{2}\sin^{-1}\left(\frac{x}{a}\right) \)
Diff w.r.t. x
\( \frac{dy}{dx} = \frac{x}{2}\cdot\frac{1}{2\sqrt{a^2-x^2}}\cdot(-2x) + \sqrt{a^2-x^2}\cdot\frac{1}{2} + \frac{a^2}{2}\cdot\frac{1}{\sqrt{1-\frac{x^2}{a^2}}}\cdot\left(\frac{1}{a}\right) \)
\( \Rightarrow \frac{dy}{dx} = \frac{-x^2}{2\sqrt{a^2-x^2}} + \frac{\sqrt{a^2-x^2}}{2} + \frac{a^2}{2}\cdot\frac{a}{\sqrt{a^2-x^2}}\cdot\frac{1}{a} \)
\( \Rightarrow \frac{dy}{dx} = \frac{-x^2}{2\sqrt{a^2-x^2}} + \frac{\sqrt{a^2-x^2}}{2} + \frac{a^2}{2\sqrt{a^2-x^2}} \)
\( \Rightarrow \frac{dy}{dx} = \frac{-x^2+a^2-x^2+a^2}{2\sqrt{a^2-x^2}} \)
\( \Rightarrow \frac{dy}{dx} = \frac{2a^2-2x^2}{2\sqrt{a^2-x^2}} = \frac{2(a^2-x^2)}{2\sqrt{a^2-x^2}} /> \( \Rightarrow \frac{dy}{dx} = \sqrt{a^2-x^2} \) (Ans)

Question. If \( y = \frac{x\sin^{-1}x}{\sqrt{1-x^2}} \), show that \( (1-x^2)\frac{dy}{dx} = x + \frac{y}{x} \).
Answer: We have, \( y = \frac{x\sin^{-1}x}{\sqrt{1-x^2}} \dots(1) \)
\( \Rightarrow y\sqrt{1-x^2} = x\sin^{-1}x \)
Diff w.r.t. (Product rule on both sides)
\( y \cdot \frac{1}{2\sqrt{1-x^2}}(-2x) + \sqrt{1-x^2}\cdot\frac{dy}{dx} = x\cdot\frac{1}{\sqrt{1-x^2}} + \sin^{-1}x\cdot 1 \)
\( \Rightarrow \frac{-xy}{\sqrt{1-x^2}} + \sqrt{1-x^2}\cdot\frac{dy}{dx} = \frac{x}{\sqrt{1-x^2}} + \sin^{-1}x\cdot 1 \)
\( \Rightarrow \frac{-xy + (1-x^2)dy/dx}{\sqrt{1-x^2}} = \frac{x+\sqrt{1-x^2}\sin^{-1}x}{\sqrt{1-x^2}} \)
\( \Rightarrow -xy + (1-x^2)\frac{dy}{dx} = x + \sqrt{1-x^2}\cdot\left(\frac{y\sqrt{1-x^2}}{x}\right) \dots\dots\text{from (1), } \sin^{-1}x = \frac{y\sqrt{1-x^2}}{x} \)
\( \Rightarrow -xy + (1-x^2)\frac{dy}{dx} = x + \frac{y(1-x^2)}{x} \)
\( \Rightarrow (1-x^2)\frac{dy}{dx} = \frac{x^2+y-x^2y+x^2y}{x} \)
\( \Rightarrow \frac{dy}{dx} = \frac{x^2+y-x^2y+x^2y}{x} \)
\( \Rightarrow \frac{x^2+y}{x} \)
\( \Rightarrow (1-x^2)\frac{dy}{dx} = x + \frac{y}{x} \) (proved)