Differentiability and Differentiation MCQ Questions & Answers in Calculus | Maths

$$\eqalign{ & x = {e^{y + x}} \cr & \Rightarrow \log \,x = y + x \cr & \therefore \,\frac{1}{x} = \frac{{dy}}{{dx}} + 1 \cr} $$

$$f\left( x \right) = {e^{ - x}}$$   is one such function
$$\eqalign{ & {\text{Here}}\,\,f\left( 0 \right) = 1,\,\,f'\left( 0 \right) = - 1,\,f\left( x \right) > 0,\,\forall \,x \cr & \therefore \,f''\left( x \right) > 0\,\forall \,x \cr} $$

$$g\left( x \right) = f\left( {f\left( x \right)} \right)$$
In the neighborhood of $$x = 0,$$
$$\eqalign{ & f\left( x \right) = \left| {\log \,2 - \sin \,x} \right| = \left( {\log \,2 - \sin \,x} \right) \cr & \backslash \,\,g\left( x \right) = \left| {\log \,2 - \sin \,\left| {\log \,2 - \sin \,x} \right|} \right| \cr & = \left( {\log \,2 - \sin \left( {\log \,2 - \sin \,x} \right)} \right) \cr & \backslash \,\,g\left( x \right) = {\text{is differentiable}} \cr & {\text{and }}g'\left( x \right) = - \cos \left( {\log \,2 - \sin \,x} \right)\left( { - \cos \,x} \right) \cr & \Rightarrow g'\left( 0 \right) = \cos \left( {\log \,2} \right) \cr} $$

$$\eqalign{ & \mathop {\lim }\limits_{x \to 0} \frac{{\sin \left( {\pi \,{{\cos }^2}\,x} \right)}}{{{x^2}}} \cr & = \mathop {\lim }\limits_{x \to 0} \frac{{\sin \left( {\pi - \pi \,{{\sin }^2}\,x} \right)}}{{{x^2}}}\,\,\,\,\,\,\,\,\,\left[ {\sin \left( {\pi - \theta } \right) = \sin \,\theta } \right] \cr & = \mathop {\lim }\limits_{x \to 0} \frac{{\sin \left( {\pi \,{{\sin }^2}\,x} \right)}}{{\pi \,{{\sin }^2}\,x}} \times \frac{{\left( {\pi \,{{\sin }^2}\,x} \right)}}{{{x^2}}} = \pi \cr} $$

We have $$f:R \to R,$$   a differentiable function and $$f\left( 1 \right) = 4$$
NOTE THIS STEP
$$\eqalign{ & \mathop {\lim }\limits_{x \to 1} \int_4^{f\left( x \right)} {\frac{{2t}}{{x - 1}}} dt \cr & = \mathop {\lim }\limits_{x \to 1} \left[ {\frac{{{t^2}}}{{x - 1}}} \right]_4^{f\left( x \right)} \cr & = \mathop {\lim }\limits_{x \to 1} \frac{{{{\left( {f\left( x \right)} \right)}^2} - 16}}{{x - 1}} \cr & = \mathop {\lim }\limits_{x \to 1} \frac{{f\left( x \right) - 4}}{{x - 1}}.\mathop {\lim }\limits_{x \to 1} \left( {f\left( x \right) + 4} \right) \cr & = f'\left( 1 \right).\left( {f\left( 1 \right) + 4} \right) \cr & = 8f'\left( 1 \right)\,\,\,\,\,\,\,\,\,\left[ {{\text{Using }}f\left( 1 \right) = 4} \right] \cr} $$

$$\eqalign{ & h\left( x \right) = f\left\{ {g\left( x \right)} \right\} = \left[ {g\left( x \right)} \right] = \left[ {\left| x \right|} \right] \cr & h'\left( { - 1 + 0} \right) = \mathop {\lim }\limits_{h \to 0} \frac{{\left[ {\left| { - 1 + h} \right|} \right] - \left[ {\left| { - 1} \right|} \right]}}{h} \cr & = \mathop {\lim }\limits_{h \to 0} \frac{{\left[ {1 - h} \right] - \left[ 1 \right]}}{h} \cr & = \mathop {\lim }\limits_{h \to 0} \frac{{0 - 1}}{h} = - \infty \cr & h'\left( { - 1 - 0} \right) = \mathop {\lim }\limits_{h \to 0} \frac{{\left[ {\left| { - 1 - h} \right|} \right] - \left[ {\left| { - 1} \right|} \right]}}{{ - h}} \cr & = \mathop {\lim }\limits_{h \to 0} \frac{{\left[ {1 + h} \right] - \left[ 1 \right]}}{{ - h}} \cr & = \mathop {\lim }\limits_{h \to 0} \frac{{1 - 1}}{{ - h}} = \mathop {\lim }\limits_{h \to 0} = 0 \cr} $$

$$\eqalign{ & \frac{{dy}}{{dx}} = \frac{{df\left( {\frac{{2x - 1}}{{{x^2} + 1}}} \right)}}{{dx}} = \frac{{df\left( {\frac{{2x - 1}}{{{x^2} + 1}}} \right)}}{{d\left( {\frac{{2x - 1}}{{{x^2} + 1}}} \right)}}.\frac{{d\left( {\frac{{2x - 1}}{{{x^2} + 1}}} \right)}}{{dx}} \cr & = \sin {\left( {\frac{{2x - 1}}{{{x^2} + 1}}} \right)^2}.\frac{{2\left( {{x^2} + 1} \right) - \left( {2x - 1} \right)2x}}{{{{\left( {{x^2} + 1} \right)}^2}}} \cr & = \frac{{2\left( {1 + x - {x^2}} \right)}}{{{{\left( {{x^2} + 1} \right)}^2}}}\sin {\left( {\frac{{2x - 1}}{{{x^2} + 1}}} \right)^2} \cr} $$

$$\eqalign{ & f\left( 0 \right) = 0\,;\,\,f\left( x \right) = x{e^{ - \left( {\frac{1}{{\left| x \right|}} + \frac{1}{x}} \right)}} \cr & {\text{R}}{\text{.H}}{\text{.L}}{\text{. }}\mathop {\lim }\limits_{h \to 0} \left( {0 + h} \right){e^{ - \frac{2}{h}}} = \mathop {\lim }\limits_{h \to 0} \frac{h}{{{e^{\frac{2}{h}}}}} = 0 \cr & {\text{L}}{\text{.H}}{\text{.L}}{\text{. }}\mathop {\lim }\limits_{h \to 0} \left( {0 - h} \right){e^{ - \left( {\frac{1}{h}\, - \,\frac{1}{h}} \right)}} = 0 \cr} $$
Therefore, $$f\left( x \right)$$  is continuous.
$$\eqalign{ & {\text{R}}{\text{.H}}{\text{.D}}{\text{. }} = \mathop {\lim }\limits_{h \to 0} \frac{{\left( {0 + h} \right){e^{ - \left( {\frac{1}{h}\, + \frac{1}{h}} \right)}} - 0}}{h} = 0 \cr & {\text{L}}{\text{.H}}{\text{.D}}{\text{. }} = \mathop {\lim }\limits_{h \to 0} \frac{{\left( {0 - h} \right){e^{ - \left( {\frac{1}{h}\, - \,\frac{1}{h}} \right)}} - 0}}{{ - h}} = 1 \cr} $$
Therefore, L.H.D. $$ \ne $$ R.H.D.
$$f\left( x \right)$$  is not differentiable at $$x =0.$$

\[\begin{array}{l} f\left( x \right) = \max \left\{ {x,\,{x^3}} \right\}\\ \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \left\{ \begin{array}{l} \,x\,\,;\,\,\,\,\,\,\,\,\,\,\,\,x < - 1\\ {x^3}\,;\,\,\,\, - 1 \le x \le 0\\ \,x\,\,;\,\,\,\,\,\,\,0 \le x \le 1\\ {x^3}\,;\,\,\,\,\,\,\,\,\,\,\,\,x \ge 1 \end{array} \right.\\ \therefore \,f'\left( x \right) = \left\{ \begin{array}{l} \,1\,\,;\,\,\,\,\,\,\,\,\,\,\,\,x < - 1\\ 3{x^2}\,;\,\,\,\, - 1 \le x \le 0\\ \,1\,\,;\,\,\,\,\,\,\,0 \le x \le 1\\ 3{x^2}\,;\,\,\,\,\,\,\,\,\,\,\,\,x \ge 1 \end{array} \right. \end{array}\]
Clearly $$f$$ is not differentiable at $$ - 1,\,0$$  and $$1.$$

$$\eqalign{ & f'\left( {1 - 0} \right) = \mathop {\lim }\limits_{h \to 0} \frac{{\sin \,\pi \left[ {1 - h} \right] - \sin \,\pi \left[ 1 \right]}}{{ - h}} \cr & f'\left( {1 - 0} \right) = \mathop {\lim }\limits_{h \to 0} \frac{0}{{ - h}} \cr & f'\left( {1 - 0} \right) = \mathop {\lim }\limits_{h \to 0} \,0 \cr & f'\left( {1 - 0} \right) = 0 \cr} $$