Differentiability and Differentiation MCQ Questions & Answers in Calculus | Maths

$${\text{As }}f\left( 1 \right) = - 2\,\,\&\,\, f'\left( x \right) \geqslant 2\,\forall \,x \in \left[ {1,\,6} \right]$$
Applying Lagrange’s mean value theorem
$$\eqalign{ & \frac{{f\left( 6 \right) - f\left( 1 \right)}}{5} = f'\left( c \right) \geqslant 2 \cr & \Rightarrow f\left( 6 \right) \geqslant 10 + f\left( 1 \right) \cr & \Rightarrow f\left( 6 \right) \geqslant 10 - 2 \cr & \Rightarrow f\left( 6 \right) \geqslant 8 \cr} $$

Let $$x = \sec \,\theta $$
$$\eqalign{ & {\text{Then }}y = {\tan ^{ - 1}}\sqrt {\frac{{\sec \,\theta + 1}}{{\sec \,\theta - 1}}} \cr & = {\tan ^{ - 1}}\sqrt {\frac{{1 + \cos \,\theta }}{{1 - \cos \,\theta }}} \cr & = {\tan ^{ - 1}}\left( {\cot \frac{\theta }{2}} \right) \cr & \therefore \,\,y = {\tan ^{ - 1}}\left\{ {\tan \left( {\frac{\pi }{2} - \frac{\theta }{2}} \right)} \right\} = \frac{\pi }{2} - \frac{1}{2}{\sec ^{ - 1}}x \cr & \therefore \,\,\frac{{dy}}{{dx}} = - \frac{1}{2}.\frac{1}{{\left| x \right|\sqrt {{x^2} - 1} }} \cr} $$

The given function is
\[\begin{array}{l} f\left( x \right) = \left\{ \begin{array}{l} {\tan ^{ - 1}}x\,\,\,\,\,\,\,\,{\rm{if}}\,\,\left| x \right| \le 1\\ \frac{1}{2}\left( {\left| x \right| - 1} \right)\,\,\,{\rm{if}}\,\,\left| x \right| > 1 \end{array} \right.\\ f\left( x \right) = \left\{ \begin{array}{l} \frac{1}{2}\left( { - x - 1} \right)\,\,\,{\rm{if}}\,x < - 1\\ {\tan ^{ - 1}}x\,\,\,\,\,\,\,\,\,{\rm{if}} - 1 \le x \le 1\\ \frac{1}{2}\left( {x - 1} \right)\,\,\,{\rm{if}}\,x > 1 \end{array} \right. \end{array}\]
$$\eqalign{ & {\text{Clearly,}}\,\, \cr & {\text{L}}{\text{.H}}{\text{.L}}{\text{. at}}\left( {x = - 1} \right) = \mathop {\lim }\limits_{h \to 0} f\left( { - 1 - h} \right) = 0 \cr & {\text{R}}{\text{.H}}{\text{.L}}{\text{.at }}\left( {x = - 1} \right) = \mathop {\lim }\limits_{h \to 0} f\left( { - 1 + h} \right) \cr & = \mathop {\lim }\limits_{h \to 0} \,{\tan ^{ - 1}}\left( { - 1 + h} \right) = \frac{{3\pi }}{4} \cr & \therefore {\text{L}}{\text{.H}}{\text{.L}}{\text{.}} \ne {\text{R}}{\text{.H}}{\text{.L}}{\text{. at }}x = - 1 \cr} $$
$$\therefore f\left( x \right)$$   is discontinuous at $$x =-1$$
Also we can prove in the same way, that $$f\left( x \right)$$  is discontinuous at $$x = 1$$
$$\therefore f'\left( x \right)$$   can not be found for $$x = \pm 1$$   or domain of $$f'\left( x \right) = R - \left\{ { - 1,\,1} \right\}$$

Differentiating w.r.t. $$x,\,{e^{xy}} + x.{e^{xy}}\left( {y + x\frac{{dy}}{{dx}}} \right) - \frac{{dy}}{{dx}} = 2\sin \,x\cos \,x$$
$$\eqalign{ & {\text{or }}\frac{{dy}}{{dx}}\left\{ {{x^2}.{e^{xy}} - 1} \right\} = \sin \,2x - {e^{xy}} - xy.{e^{xy}} \cr & {\text{When }}x = 0,\,0 - y = 0,\,{\text{i}}{\text{.e}}{\text{., }}y = 0 \cr & \therefore \,\frac{{dy}}{{dx}} = \frac{{0 - 1 - 0}}{{0 - 1}} = 1 \cr} $$

$$\eqalign{ & {\text{Let }}t = \cos \,2\theta \cr & {\text{Then }}{e^x} = \frac{{\sqrt {1 + \cos \,2\theta } - \sqrt {1 - \cos \,2\theta } }}{{\sqrt {1 + \cos \,2\theta } + \sqrt {1 - \cos \,2\theta } }} \cr & = \frac{{\cos \,\theta - \sin \,\theta }}{{\cos \,\theta + \sin \,\theta }} \cr & = \frac{{1 - \tan \,\theta }}{{1 + \tan \,\theta }} \cr & = \tan \left( {\frac{\pi }{4} - \theta } \right) \cr & \tan \frac{y}{2} = \sqrt {\frac{{1 - \cos \,2\theta }}{{1 + \cos \,2\theta }}} = \tan \,\theta \cr & {\text{At }}t = \frac{1}{2},\,\cos \,2\theta = \frac{1}{2}\,\,\,\,\,\,\, \Rightarrow \theta = \frac{\pi }{6} \cr & {\text{Then }}x = \log \,\tan \frac{\pi }{{12}},\,\,y = \frac{\pi }{3} \cr & {\text{Differentiating w}}{\text{.r}}{\text{.t}}{\text{.}}\,\theta ,\,{e^x}{\text{.}}\frac{{dx}}{{d\theta }} = - {\sec ^2}\left( {\frac{\pi }{4} - \theta } \right) \cr & {\text{and }}\frac{1}{2}{\sec ^2}\frac{y}{2}.\frac{{dy}}{{d\theta }} = {\sec ^2}\theta \cr & \therefore \,\,\frac{{dy}}{{dx}} = \frac{{\frac{{dy}}{{d\theta }}}}{{\frac{{dx}}{{d\theta }}}} = \frac{{2{{\sec }^2}\theta .{{\cos }^2}\frac{y}{2}}}{{ - {e^{ - x}}.{{\sec }^2}\left( {\frac{\pi }{4} - \theta } \right)}} \cr & {\text{At }}t = \frac{1}{2},\,{\text{i}}{\text{.e}}{\text{., }}\theta = \frac{\pi }{6},\,\frac{{dy}}{{dx}} = \frac{{2{{\sec }^2}\frac{\pi }{6}.{{\cos }^2}\frac{\pi }{6}}}{{ - {e^{ - \log \,\tan \,\frac{\pi }{{12}}}}.{{\sec }^2}\frac{\pi }{{12}}}} \cr & \therefore \,\frac{{dy}}{{dx}} = \frac{2}{{ - \cot \,\frac{\pi }{{12}}.{{\sec }^2}\frac{\pi }{{12}}}} \cr & = - 2\sin \,\frac{\pi }{{12}}.\cos \frac{\pi }{{12}} \cr & = - \sin \frac{\pi }{6} \cr & = - \frac{1}{2} \cr} $$

$$f\left( x \right) = \left| {x - 1} \right|$$
Redefined the function $$f\left( x \right)$$
\[\begin{array}{l} f\left( x \right) = \left\{ \begin{array}{l} 1 - x,\,\,\,\,\,x < 1\\ x - 1,\,\,\,\,\,x > 1 \end{array} \right.\\ f'\left( x \right) = \left\{ \begin{array}{l} - 1\,;\,\,\,\,\,x < 1\\ \,\,\,1\,;\,\,\,\,\,x > 1 \end{array} \right.\\ \therefore f'\left( 2 \right) = 1 \end{array}\]

$$\eqalign{ & {\text{RH derivative}} = \mathop {\lim }\limits_{h \to 0} \frac{{{e^{ - \,\frac{1}{{{{\left( {0 + h} \right)}^2}}}}} - 0}}{h} \cr & = \mathop {\lim }\limits_{h \to 0} \frac{1}{h}.{e^{ - \,\frac{1}{{{h^2}}}}} \cr & = \mathop {\lim }\limits_{h \to 0} \frac{{\frac{1}{h}}}{{{e^{\frac{1}{{{h^2}}}}}}} \cr & = \mathop {\lim }\limits_{h \to 0} \frac{{ - \frac{1}{{{h^2}}}}}{{{e^{\frac{1}{{{h^2}}}}}.\left( { - \frac{2}{{{h^3}}}} \right)}} \cr & = \mathop {\lim }\limits_{h \to 0} \frac{h}{{2{e^{\frac{1}{{{h^2}}}}}}} \cr & = \frac{0}{\infty } \cr & = 0 \cr & {\text{LH derivative}} = \mathop {\lim }\limits_{h \to 0} \frac{{{e^{ - \,\frac{1}{{{{\left( {0 - h} \right)}^2}}}}} - 0}}{{ - h}} \cr & = \mathop {\lim }\limits_{h \to 0} - \frac{1}{h}.{e^{ - \,\frac{1}{{{h^2}}}}} \cr & = \mathop {\lim }\limits_{h \to 0} \frac{{ - \frac{1}{h}}}{{{e^{\frac{1}{{{h^2}}}}}}} \cr & = \mathop {\lim }\limits_{h \to 0} \frac{{\frac{1}{{{h^2}}}}}{{{e^{\frac{1}{{{h^2}}}}}.\left( { - \frac{2}{{{h^3}}}} \right)}} \cr & = \mathop {\lim }\limits_{h \to 0} \frac{h}{{ - 2{e^{\frac{1}{{{h^2}}}}}}} \cr & = \frac{0}{{ - \infty }} \cr & = 0 \cr} $$

$$\eqalign{ & f'\left( {2{x^2} + \frac{\pi }{2}} \right) = \frac{{df\left( {2{x^2} + \frac{\pi }{2}} \right)}}{{dx}} = \frac{{df\left( {2{x^2} + \frac{\pi }{2}} \right)}}{{d\left( {2{x^2} + \frac{\pi }{2}} \right)}}.\frac{{d\left( {2{x^2} + \frac{\pi }{2}} \right)}}{{dx}} \cr & \because \,f'\left( x \right) = \frac{{df\left( x \right)}}{{dx}} = \sin \,x + \sin \,4x.\cos \,x,{\text{ we get}} \cr & \frac{{df\left( {2{x^2} + \frac{\pi }{2}} \right)}}{{d\left( {2{x^2} + \frac{\pi }{2}} \right)}} = \sin \left( {2{x^2} + \frac{\pi }{2}} \right) + \sin \,4\left( {2{x^2} + \frac{\pi }{2}} \right).\cos \left( {2{x^2} + \frac{\pi }{2}} \right) \cr & = \cos \,2{x^2} + \sin \,8{x^2}.\left( { - \sin \,2{x^2}} \right) \cr & \therefore \,f'\left( {2{x^2} + \frac{\pi }{2}} \right) = \left( {\cos \,2{x^2} - \sin \,8{x^2}.\sin \,2{x^2}} \right).4x \cr & \therefore \,{\text{at }}x = \sqrt {\frac{\pi }{2}} ,\,f'\left( {2{x^2} + \frac{\pi }{2}} \right) \cr & = \left( {\cos \,\pi - \sin \,4\pi .\sin \,\pi } \right).4.\sqrt {\frac{\pi }{2}} \cr & = - 2\sqrt {2\pi } \cr} $$

Given $$f\left( x \right) = \frac{x}{{1 + \left| x \right|}}$$
\[ = \left\{ \begin{array}{l} \frac{x}{{1 - x}},\,\,\,\,x < 0\\ \frac{x}{{1 + x}},\,\,\,\,x \ge 0 \end{array} \right.\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left( {\,\because\left| x \right| = \left\{ \begin{array}{l} \,\,\,x\,\,\,\,{\rm{if}}\,x \ge 0\\ - x\,\,\,\,{\rm{if}}\,x < 0 \end{array} \right.} \right)\]
$$\eqalign{ & \therefore {\text{ L}}{\text{.H}}{\text{.D}}{\text{.}} = f'\left( {{0^ - }} \right) = \mathop {\lim }\limits_{h \to 0} \frac{{f\left( {0 - h} \right) - f\left( 0 \right)}}{{ - h}} \cr & = \mathop {\lim }\limits_{h \to 0} \frac{{f\left( { - h} \right) - f\left( 0 \right)}}{{ - h}} \cr & = \mathop {\lim }\limits_{h \to 0} \frac{{\frac{{ - h}}{{1 + \left| { - h} \right|}} - 0}}{{ - h}} \cr & = \mathop {\lim }\limits_{h \to 0} \frac{{\frac{{ - h}}{{1 + h}} - 0}}{{ - h}} \cr & = \mathop {\lim }\limits_{h \to 0} \frac{1}{{1 + h}} \cr & = 1 \cr & {\text{and R}}{\text{.H}}{\text{.D}}{\text{.}} = f'\left( {{0^ + }} \right) = \mathop {\lim }\limits_{h \to 0} \frac{{f\left( {0 + h} \right) - f\left( 0 \right)}}{h} \cr & = \mathop {\lim }\limits_{h \to 0} \frac{{\frac{h}{{1 + h}} - 0}}{h} \cr & = \mathop {\lim }\limits_{h \to 0} \frac{1}{{1 + h}} \cr & = 1 \cr & {\text{Since, L}}{\text{.H}}{\text{.D}}{\text{.}} = {\text{R}}{\text{.H}}{\text{.D}}{\text{.}} \cr & \therefore \,f\left( x \right){\text{ is differentiable at }}x = 0 \cr & {\text{Hence, }}f\left( x \right){\text{ is differentiable in }}\left( { - \infty ,\,\infty } \right){\text{. }} \cr} $$

$$\mathop {\lim }\limits_{x \to 1} \frac{{\sqrt {f\left( x \right)} - 1}}{{\sqrt x - 1}}\,\,\,\,\,\,\left( {\frac{0}{0}} \right)$$     form using L'Hospital rule
$$\eqalign{ & = \mathop {\lim }\limits_{x \to 1} \frac{{\frac{1}{{2\sqrt {f\left( x \right)} }}f'\left( x \right)}}{{\frac{1}{{2\sqrt x }}}} \cr & = \frac{{f'\left( 1 \right)}}{{\sqrt {f\left( 1 \right)} }} \cr & = \frac{2}{1} \cr & = 2 \cr} $$