CBSE Class 12 Mathematics Continuity And Differentiability Worksheet Set 06

CBSE Class 12 Mathematics Continuity And Differentiability Worksheet 6. 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.

Inverse Trigonometric Diff.

Question. \( y = \cos^{-1}\left(\frac{2x - 3\sqrt{1-x^2}}{\sqrt{13}}\right) \). Find \( \frac{dy}{dx} \).
Answer: We have, \( y = \cos^{-1}\left(\frac{2x - 3\sqrt{1-x^2}}{\sqrt{13}}\right) \)
put \( x = \sin\theta \)
\( y = \cos^{-1}\left(\frac{2\sin\theta - 3\cos\theta}{\sqrt{13}}\right) \)
\( \Rightarrow y = \cos^{-1}\left(\frac{2}{\sqrt{13}}\sin\theta - \frac{3}{\sqrt{13}}\cdot\cos\theta\right) /> let \( \sin\alpha = \frac{2}{\sqrt{13}} \) & \( \cos\alpha = \frac{3}{\sqrt{13}} \dots\dots \left\{\cos\alpha = \sqrt{1 - \sin^2\alpha} = \sqrt{1 - \frac{4}{13}} = \frac{3}{\sqrt{13}}\right\} \)
\( \Rightarrow y = \cos^{-1}(\sin\alpha\cdot\sin\theta - \cos\alpha\cdot\cos\theta) \)
\( \Rightarrow y = \cos^{-1}(-(\cos\theta\cdot\cos\alpha - \sin\theta\cdot\sin\alpha)) \)
\( \Rightarrow y = \cos^{-1}(-\cos(\theta + \alpha)) \dots\dots \{\cos A \cdot \cos B - \sin A \cdot \sin B = \cos(A + B)\} \)
\( \Rightarrow y = \pi - \cos^{-1}(\cos(\theta + \alpha)) \dots\dots \{\cos^{-1}(-x) = \pi - \cos^{-1}x\} \)
\( \Rightarrow y = \pi - (\theta + \alpha) \)
\( \Rightarrow y = \pi - \sin^{-1}x - \sin^{-1}\left(\frac{2}{\sqrt{13}}\right) \quad \{\text{Constant}\} \)
Diff w.r.t x
\( \frac{dy}{dx} = -\frac{1}{\sqrt{1-x^2}} \) Ans.

Question. \( y = \sin^{-1}\left(\frac{2^{x+1}}{1+4^x}\right) \). Find \( \frac{dy}{dx} \).
Answer: \( y = \sin^{-1}\left(\frac{2^{x+1}}{1+4^x}\right) \)
\( \Rightarrow y = \sin^{-1}\left(\frac{2\cdot 2^x}{1+(2^x)^2}\right) \)
Put \( 2^x = \tan\theta \)
\( \Rightarrow y = \sin^{-1}\left(\frac{2\tan\theta}{1+\tan^2\theta}\right) \)
\( \Rightarrow y = \sin^{-1}(\sin(2\theta)) \)
\( \Rightarrow y = 2\theta \)
\( \Rightarrow y = 2\tan^{-1}(2^x) \)
Diff w.r.t. x
\( \Rightarrow \frac{dy}{dx} = 2\cdot\frac{1}{1+(2^x)^2}\cdot 2^x \cdot \log 2 \dots\dots \left\{\frac{d}{dx}(a^x) = a^x\log a\right\} \)
\( \Rightarrow \frac{dy}{dx} = \frac{2^{x+1}\cdot\log 2}{1+4^x} \) Ans

Question. \( y = \sin^{-1}(x\sqrt{1-x} - \sqrt{x}\sqrt{1-x^2}) \). Find \( \frac{dy}{dx} \).
Answer: We have, \( y = \sin^{-1}(x\sqrt{1-x} - \sqrt{x}\sqrt{1-x^2}) \)
put \( x = \sin A \) and \( \sqrt{x} = \sin B \)
\( \Rightarrow y = \sin^{-1}(\sin A\sqrt{1 - \sin^2 B} - \sin B\sqrt{1 - \sin^2 A}) \)
\( \Rightarrow y = \sin^{-1}(\sin A\cdot\cos B - \sin B\cdot\cos A)  /> \( \Rightarrow y = \sin^{-1}(\sin(A - B)) \)
\( \Rightarrow y = A - B \)
\( \Rightarrow y = \sin^{-1}x - \sin^{-1}\sqrt{x} \)
Diff w.r.t. x
\( \Rightarrow \frac{dy}{dx} = \frac{1}{\sqrt{1-x^2}} - \frac{1}{\sqrt{1-x}}\cdot\frac{1}{2\sqrt{x}} \)
\( \Rightarrow \frac{dy}{dx} = \frac{1}{\sqrt{1-x^2}} - \frac{1}{2\sqrt{x-x^2}} \) Ans.

Diff. Of A Function w.r.t. Another Function

Question. Diff. \( \tan^{-1}\left(\frac{\sqrt{1-x^2}}{x}\right) \) w.r.t. \( \cos^{-1}(2x\sqrt{1-x^2}) \).
Answer: Let \( u = \tan^{-1}\left(\frac{\sqrt{1-x^2}}{x}\right) \)
put \( x = \sin\theta \)
\( \Rightarrow u = \tan^{-1}\left(\frac{\sqrt{1 - \sin^2\theta}}{\sin\theta}\right) \)
\( \Rightarrow u = \tan^{-1}\left(\frac{\cos\theta}{\sin\theta}\right) \)
\( \Rightarrow u = \tan^{-1}(\cot\theta) \)
\( \Rightarrow u = \tan^{-1}\left(\tan\left(\frac{\pi}{2} - \theta\right)\right) \)
\( \Rightarrow u = \frac{\pi}{2} - \theta \)
\( \Rightarrow u = \frac{\pi}{2} - \sin^{-1}x \)
Diff w.r.t. x
\( \frac{du}{dx} = -\frac{1}{\sqrt{1-x^2}} \)
let \( v = \cos^{-1}(2x\sqrt{1-x^2}) \)
put \( x = \sin\theta \)
\( \Rightarrow v = \cos^{-1}(2\sin\theta\sqrt{1-\sin^2\theta}) \)
\( \Rightarrow v = \cos^{-1}(2\sin\theta\cdot\cos\theta)  /> \( \Rightarrow v = \cos^{-1}(\sin(2\theta))  /> \( \Rightarrow v = \cos^{-1}\left(\cos\left(\frac{\pi}{2} - 2\theta\right)\right) \)
\( \Rightarrow v = \frac{\pi}{2} - 2\theta \)
\( \Rightarrow v = \frac{\pi}{2} - 2\sin^{-1}x \)
Diff w.r.t. x
\( \frac{dv}{dx} = 0 - \frac{2}{\sqrt{1-x^2}} = -\frac{2}{\sqrt{1-x^2}} \)
Now \( \frac{du}{dv} = \frac{du/dx}{dv/dx} \)
\( \Rightarrow \frac{du}{dv} = \frac{-\frac{1}{\sqrt{1-x^2}}}{-\frac{2}{\sqrt{1-x^2}}} = \frac{1}{2} \)

Question. Diff \( \sin^{-1}(2ax\sqrt{1-a^2x^2}) \) w.r.t. \( \sqrt{1-a^2x^2} \).
Answer: Let \( u = \sin^{-1}(2ax\sqrt{1-a^2x^2}) \)
put \( ax = \sin\theta \)
\( u = \sin^{-1}(2\sin\theta\sqrt{1-\sin^2\theta}) \)
\( \Rightarrow u = \sin^{-1}(2\sin\theta\cdot\cos\theta) \)
\( \Rightarrow u = \sin^{-1}(\sin(2\theta)) \)
\( \Rightarrow u = 2\theta \)
\( \Rightarrow u = 2\sin^{-1}(ax) \)
Diff w.r.t. x
\( \frac{du}{dx} = \frac{2}{\sqrt{1-a^2x^2}}\cdot(a) = \frac{2a}{\sqrt{1-a^2x^2}} \)
let \( v = \sqrt{1-a^2x^2} \)
Diff w.r.t. x
\( \frac{dv}{dx} = \frac{1}{2\sqrt{1-a^2x^2}}(-2a^2x) \)
\( \frac{dv}{dx} = \frac{-a^2x}{\sqrt{1-a^2x^2}} \)
Now \( \frac{du}{dv} = \frac{du/dx}{dv/dx} = \frac{\frac{2a}{\sqrt{1-a^2x^2}}}{\frac{-a^2x}{\sqrt{1-a^2x^2}}} \)
\( \Rightarrow \frac{2a}{-a^2x} \)
\( \therefore \frac{du}{dv} = \frac{-2}{ax} \) Ans.

Question. Diff. \( \tan^{-1}\left(\frac{\cos x}{1+\sin x}\right) \) w.r.t. \( \sec^{-1}x \).
Answer: Let \( u = \tan^{-1}\left(\frac{\cos x}{1+\sin x}\right) \)
\( \Rightarrow u = \tan^{-1}\left(\frac{\sin\left(\frac{\pi}{2}-x\right)}{1+\cos\left(\frac{\pi}{2}-x\right)}\right) \)
\( \Rightarrow u = \tan^{-1}\left(\frac{2\sin\left(\frac{\pi}{4}-\frac{x}{2}\right)\cdot\cos\left(\frac{\pi}{4}-\frac{x}{2}\right)}{2\cos^2\left(\frac{\pi}{4}-\frac{x}{2}\right)}\right) \)
\( \Rightarrow u = \tan^{-1}\left(\tan\left(\frac{\pi}{4}-\frac{x}{2}\right)\right) \)
\( \Rightarrow u = \frac{\pi}{4} - \frac{x}{2} \)
Diff w.r.t. x
\( \frac{du}{dx} = -\frac{1}{2} \)
let \( v = \sec^{-1}x \)
Diff w.r.t. x
\( \frac{dv}{dx} = \frac{1}{x\sqrt{x^2-1}} \)
Now \( \frac{du}{dv} = \frac{du/dx}{dv/dx} = \frac{-1/2}{\frac{1}{x\sqrt{x^2-1}}}  /> \( \frac{du}{dv} = \frac{-x\sqrt{x^2-1}}{2} \) Ans.

Continuity & Differentiability

Question. If \( f(x) \) is continuous at \( x = 1 \). Find the values of a and b. \[ f(x) = \begin{cases} 3ax + b & ; \ x > 1 \\ 11 & ; \ x = 1 \\ 5ax - 2b & ; \ x < 1 \end{cases} \]
Answer: LHL = \( (5ax - 2b) \)
Put \( x = 1 - h \) and \( h \to 0 \)
LHL = \( (5a(1 - h) - 2b) \)
\( \Rightarrow \) LHL = \( 5a - 2b \)
RHL = \( (3ax + b) \)
put \( x = 1 + h \) and \( h \to 0 \)
RHL = \( [3a(1 + h) + b] \)
\( \Rightarrow \) RHL = \( [3a + b] \)
Now \( f(1) = 11 \)
Since \( f(x) \) is continuous at \( x = 1 \)
\( \therefore \text{LHL} = \text{RHL} = f(1) \)
\( \Rightarrow 5a - 2b = 3a + b = 11 \)
consider \( 5a - 2b = 11 \)
and \( 3a + b = 11 \)
solving these equations we get
\( a = 3 \) and \( b = 2 \)
\( \therefore f(x) \) is continuous at \( x = 1 \) for \( a = 3 \) & \( b = 2 \)

Question. The function \( f(x) \) is continuous on \( [0, 8] \). Find the value of 'a' and 'b'. \[ f(x) = \begin{cases} x^2 + ax + b & ; \ 0 \le x < 2 \\ 3x + 2 & ; \ 2 \le x \le 4 \\ 2ax + 5b & ; \ 4 < x \le 8 \end{cases} \]
Answer: Since \( f(x) \) is also continuous in \( [0, 8] \)
\( \therefore f(x) \) is also continuous at \( x = 2 \) and \( x = 4 \)
continuously at \( x = 2 \)
LHL = \( (x^2 + ax + b) \)
put \( x = 2 - h \) and \( h \to 0 \)
\( \therefore \text{LHL} = [(2 - h)^2 + a(2 - h) + b] \)
LHL = \( 4 + 2a + b \)
RHL = \( (3x + 2) \)
put \( x = 2 + h \) & \( h \to 0 \)
\( \Rightarrow \text{RHL} = (3(2 + h) + 2) \)
\( \Rightarrow \text{RHL} = 8 \)
\( f(2) = 3(2) + 2 = 8 \)
we have , \( \text{LHL} = \text{RHL} = f(2) \)
\( \Rightarrow 4 + 2a + b = 8 = 8 \)
\( \Rightarrow 2a + b = 4 \dots\dots(1) \)
continuity at \( x = 4 \)
LHL = \( (3x + 2) \)
put \( x = 4 - h \) and \( h \to 0 \)
LHL = \( (3(4 - h) + 2) \)
\( \therefore \text{LHL} = 14 \)
RHL = \( (2ax + 5b) \)
put \( x = 4 + h \) and \( h \to 0 \)
\( \Rightarrow \text{RHL} = (2a(4 + h) + 5b) \)
\( \Rightarrow \text{RHL} = 8a + 5b \)
Now \( f(4) = 3(4) + 2 = 14 \)
we have, \( \text{LHL} = \text{RHL} = f(4) \)
\( \Rightarrow 14 = 8a + 5b = 14  /> \( \Rightarrow 8a + 5b = 14 \dots\dots(2) \)
solving (1) & (2)
we get \( a = 3 \) & \( b = -2 \)
\( \dots f(x) \) is continuous in \( [0, 8] \) for \( a = 3 \) & \( b = -2 \) (Ans.)

Question. If \( f(x) \) is continuous at \( x = \frac{\pi}{2} \). Find the value of a and b. \[ f(x) = \begin{cases} \frac{1-\sin^3 x}{3\cos^2 x} & ; \ x < \frac{\pi}{2} \\ a & ; \ x = \frac{\pi}{2} \\ \frac{b(1-\sin x)}{(\pi-2x)^2} & ; \ x > \frac{\pi}{2} \end{cases} \]
Answer: LHL = \( \left[\frac{1-\sin^3 x}{3\cos^2 x}\right] \)
put \( x = \frac{\pi}{2} - h \) and \( h \to 0 \)
\( \therefore \text{LHL} = \left[\frac{1-\sin^3\left(\frac{\pi}{2}-h\right)}{3\cos^2\left(\frac{\pi}{2}-h\right)}\right] \)
\( = \left[\frac{1-\cos^3 h}{3\sin^2 h}\right] \)
\( = \left[\frac{(1-\cos h)(1+\cos^2 h+\cos h)}{3(1-\cos^2 h)}\right] \dots\dots \{a^3 - b^3 = (a - b)(a^2 + b^2 + ab)\} \)
\( = \left[\frac{(1-\cos h)(1+\cos^2 h+\cos h)}{3(1+\cos h)(1-\cos h)}\right] \)
\( = \left[\frac{1+\cos^2 h + \cos h}{3(1+\cos h)}\right] \)
LHL = \( \frac{1+1+1}{3(1+1)} = \frac{1}{2} \)
RHL = \( \left[\frac{b(1-\sin x)}{(\pi-2x)^2}\right] \)
put \( x = \frac{\pi}{2} + h \) & \( h \to 0 \)
RHL = \( \left[\frac{b\left(1-\sin\left(\frac{\pi}{2}+h\right)\right)}{\left(\pi-2\left(\frac{\pi}{2}+h\right)\right)^2}\right] \)
\( = \left[\frac{b(1-\cos h)}{(\pi-\pi-2h)^2}\right] \)
\( = \left(\frac{b\cdot 2\sin^2(h/2)}{4h^2}\right) \)
\( \Rightarrow \text{RHL} = \left[\frac{2b\cdot\sin^2(h/2)}{4\cdot\frac{h^2}{4}\cdot 4}\right] \)
\( = \frac{2b}{16}\left(\frac{\sin^2(h/2)}{h^2/4}\right)  /> RHL = \( \frac{b}{8} \dots\dots \left\{\left(\frac{\sin^2 x}{x^2}\right) = 1\right\} \)
\( f\left(\frac{\pi}{2}\right) = a \)
since \( f(x) \) is continuous at \( x = \frac{\pi}{2} \)
\( \therefore \text{LHL} = \text{RHL} = f\left(\frac{\pi}{2}\right) \)
\( \Rightarrow \frac{1}{2} = \frac{b}{8} = a \)
\( \Rightarrow b = 4 \) and \( a = \frac{1}{2} \)
\( \therefore f(x) \) is continuous at \( x = \frac{\pi}{2} \) if \( a = \frac{1}{2} \) and \( b = 4 \) Ans.

Question. Find the value of 'a' so that \( f(x) \) is continuous at \( x = 0 \). \[ f(x) = \begin{cases} a\sin\left(\frac{\pi}{2}(x+1)\right) & ; \ x \le 0 \\ \frac{\tan x - \sin x}{x^3} & ; \ x > 0 \end{cases} \]
Answer: LHL = \( \left[a\sin\frac{\pi}{2}(x + 1)\right] \)
put \( x = 0 - h = -h \) and \( h \to 0 \)
\( \therefore \text{LHL} = a\sin\left(\frac{\pi}{2}\right) \)
LHL = \( a \)
RHL = \( \left(\frac{\tan x - \sin x}{x^3}\right) \)
put \( x = 0 + h = h \) and \( h \to 0 \)
\( \therefore \text{RHL} = \left[\frac{\tan h - \sin h}{h^3}\right] \)
\( = \left[\frac{\frac{\sin h}{\cos h} - \sin h}{h^3}\right] \)
\( = \left[\frac{\sin h - \sin h\cdot\cos h}{h^3\cdot\cos h}\right] \)
\( = \left[\frac{\sin h(1-\cos h)}{h^3\cdot\cos h}\right] \)
\( = \left[\frac{\sin h\cdot 2\sin^2(h/2)}{h^3\cdot\cos h}\right] \)
\( = \left[\frac{\sin h}{h} \cdot \frac{2\sin^2(h/2)}{\frac{h^2}{4}\cdot 4} \cdot \frac{1}{\cos h}\right] \)
\( = \frac{2}{4}\left(\frac{\sin h}{h}\right) \cdot \left(\frac{\sin^2(h/2)}{h^2/4}\right) \cdot \left(\frac{1}{\cos h}\right)  /> \( = \frac{1}{2}(1)(1)(1) \dots\dots \left\{\left(\frac{\sin x}{x}\right) = 1 \text{ & } (\cos x) = 1\right\} \)
RHL = \( \frac{1}{2} \)
\( f(0) = a\sin\frac{\pi}{2}(0 + 1) = a\sin\frac{\pi}{2} = a \)
since \( f(x) \) is continuous at \( x = 0 \)
LHL = RHL = \( f(0) \)
\( \Rightarrow a = \frac{1}{2} = a \)
\( \therefore a = \frac{1}{2} \) Ans.

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