Differentiability and Differentiation MCQ Questions & Answers in Calculus | Maths

$$f\left( x \right)$$  is differentiable at $$x=1$$  also.
$$\eqalign{ & \therefore \mathop {\lim }\limits_{h \to 0} \frac{{f\left( {1 + h} \right) - f\left( 1 \right)}}{h} = \mathop {\lim }\limits_{h \to 0} \frac{{f\left( {1 - h} \right) - f\left( 1 \right)}}{{ - h}} \cr & {\text{Now,}}\,\,\mathop {\lim }\limits_{h \to 0} \frac{{f\left( {1 + h} \right) - f\left( 1 \right)}}{h} \cr & = \mathop {\lim }\limits_{h \to 0} \frac{{{{\left( {1 + h} \right)}^2} + b\left( {1 + h} \right) + c - 1}}{h} \cr & = \mathop {\lim }\limits_{h \to 0} \frac{{{h^2} + \left( {2 + b} \right)h + b + c}}{h}. \cr & \mathop {\lim }\limits_{h \to 0} \frac{{f\left( {1 + h} \right) - f\left( 1 \right)}}{{ - h}} = \mathop {\lim }\limits_{h \to 0} \frac{{1 - h - 1}}{{ - h}} = 1. \cr} $$
The two limits can be equal if $$2+b=1,\,b+c=0$$

$$\eqalign{ & {\text{First we check continuity at }}x = 2 \cr & {\text{L}}{\text{.H}}{\text{.L}}{\text{.}} = \mathop {\lim }\limits_{h \to 0} f\left( {2 - h} \right) \cr & = \mathop {\lim }\limits_{h \to 0} 3\left( {2 - h} \right) - 2 \cr & = \mathop {\lim }\limits_{h \to 0} 4 - 3h \cr & = 4 \cr & {\text{R}}{\text{.H}}{\text{.L}}{\text{.}} = \mathop {\lim }\limits_{h \to 0} f\left( {2 + h} \right) \cr & = \mathop {\lim }\limits_{h \to 0} {\left( {2 + h} \right)^2} \cr & = 4 \cr & {\text{Also, }}f\left( 2 \right) = {\left( 2 \right)^2} = 4 \cr & {\text{Since, L}}{\text{.H}}{\text{.L}}{\text{.}} = {\text{R}}{\text{.H}}{\text{.L}}{\text{.}} = f\left( 2 \right) \cr & \therefore \,f\left( x \right)\,{\text{is continuous at }}2 \cr & {\text{Now, we check for differentiability}} \cr & {\text{L}}{\text{.H}}{\text{.D}}{\text{. at }}x = 2 \cr & f'\left( x \right) = 3x - 2 \cr & f'\left( x \right) = 3 \cr & {\left. {f'\left( x \right)} \right|_{x = 2}} = 3 \cr & {\text{R}}{\text{.H}}{\text{.D}}{\text{. at }}x = 2 \cr & f'\left( x \right) = {x^2} \cr & f'\left( x \right) = 2x \cr & {\left. {f'\left( x \right)} \right|_{x = 2}} = 4 \cr & {\text{Since L}}{\text{.H}}{\text{.D}}{\text{.}} \ne {\text{R}}{\text{.H}}{\text{.D}}{\text{.}} \cr & \therefore \,f\left( x \right){\text{ is not dervable at }}x = 2 \cr} $$

$$\eqalign{ & h\left( x \right) = g\left\{ {f\left( x \right)} \right\} = \left[ {f\left( x \right) + 1} \right] = \left[ {\sin \,x + 1} \right] \cr & h'\left( {\frac{\pi }{2} + 0} \right) = \mathop {\lim }\limits_{h \to 0} \frac{{\left[ {\sin \left( {\frac{\pi }{2} + h} \right) + 1} \right] - \left[ {\sin \frac{\pi }{2} + 1} \right]}}{h} \cr & = \mathop {\lim }\limits_{h \to 0} \frac{{\left[ {\cos \,h + 1} \right] - \left[ 2 \right]}}{h} \cr & = \mathop {\lim }\limits_{h \to 0} \frac{{1 - 2}}{h} \cr & = \mathop {\lim }\limits_{h \to 0} \frac{{ - 1}}{h} = - \infty \cr & h'\left( {\frac{\pi }{2} - 0} \right) = \mathop {\lim }\limits_{h \to 0} \frac{{\left[ {\sin \left( {\frac{\pi }{2} - h} \right) + 1} \right] - \left[ {\sin \frac{\pi }{2} + 1} \right]}}{{ - h}} \cr & = \mathop {\lim }\limits_{h \to 0} \frac{{\left[ {\cos \,h + 1} \right] - \left[ 2 \right]}}{{ - h}} \cr & = \mathop {\lim }\limits_{h \to 0} \frac{1}{h} = + \infty \cr} $$

Let $$\cos \,\alpha = \frac{5}{{13}}.$$    Then $$\sin \,\alpha = \frac{{12}}{{13}}.$$
So, $$y = {\cos ^{ - 1}}\left\{ {\cos \,\alpha .\cos \,x - \sin \,\alpha .\sin \,x} \right\}$$
$$\therefore \,y = {\cos ^{ - 1}}\left\{ {\cos \left( {x + \alpha } \right)} \right\} = x + \alpha $$         ($$\because x + \alpha $$   is in the first or the second quadrant).

The given equation can be written as
$$\eqalign{ & \left( {{y^2} + 2} \right)\left( {y - \frac{{\cos \,x}}{{1 + \sin \,x}}} \right) = 0 \cr & \Rightarrow y = \frac{{\cos \,x}}{{1 + \sin \,x}}\,\,\,\,\,\,\left[ {\because \,{y^2} + 2 \ne 0} \right] \cr & = \frac{{\frac{{1 - {{\tan }^2}\frac{x}{2}}}{{1 + {{\tan }^2}\frac{x}{2}}}}}{{1 + \frac{{2\,\tan \frac{x}{2}}}{{1 + {{\tan }^2}\frac{x}{2}}}}} \cr & = \frac{{1 - {t^2}}}{{1 + {t^2} + 2t}} \cr & = \frac{{\left( {1 - t} \right)\left( {1 + t} \right)}}{{{{\left( {1 + t} \right)}^2}}} \cr & = \frac{{1 - t}}{{1 + t}} \cr & = \frac{2}{{1 + t}} - 1,{\text{ where }}t = \tan \frac{x}{2} \cr & \Rightarrow y'\left( t \right) = \frac{{ - 2}}{{{{\left( {1 + t} \right)}^2}}} \cr & {\text{At }}x = \frac{\pi }{2},\,t = 1 \cr & \therefore \,y'\left( 1 \right) = \frac{{ - 2}}{{{{\left( {1 + 1} \right)}^2}}} = - \frac{1}{2}\, \cr} $$

Since $$\frac{x}{{\left| x \right|}}$$ is not continuous function.
$$\therefore $$  it is not differentiable also.
Also, L.H.D. and R.H.D. at $$x = 0$$  not equal.
Thus, only function given in option 'B' gives differentiability for all real values of $$x.$$

$$\eqalign{ & {\text{Let }}L = \mathop {\lim }\limits_{x \to 0} \frac{{f\left( {{x^2}} \right) - f\left( x \right)}}{{f\left( x \right) - f\left( 0 \right)}} \cr & \left[ {\because f'\left( a \right) > 0,\,f\,{\text{being strictly increasing}}} \right] \cr} $$
Using L'Hospital rule, we get
$$\eqalign{ & L = \mathop {\lim }\limits_{x \to 0} \frac{{f'\left( {{x^2}} \right).2x - f'\left( x \right)}}{{f'\left( x \right)}} \cr & = \mathop {\lim }\limits_{x \to 0} \frac{{f'\left( {{x^2}} \right).2x}}{{f'\left( x \right)}} - 1 \cr & = 0 - 1 \cr & = - 1 \cr} $$

$$\left[ {n + p\sin \,x} \right]$$   is not differentiable at those points where $${n + p\sin \,x}$$   is an integer. As $$p$$ is a prime number, $${n + p\sin \,x}$$   is an integer if $$\sin \,x = 1,\, - 1,\,\frac{r}{p},$$    that is, $$x = \frac{\pi }{2},\, - \frac{\pi }{2},\,{\sin ^{ - 1}}\frac{r}{p},\,\pi - {\sin ^{ - 1}}\frac{r}{p},$$       where $$0 \leqslant r \leqslant p - 1.$$    But $$x \ne - \frac{\pi }{2},\,0.$$
$$\therefore $$ the function is not differentiable at $$x = \frac{\pi }{2},\,{\sin ^{ - 1}}\frac{r}{p},\,\pi - {\sin ^{ - 1}}\frac{r}{p},$$      where $$0 < r \leqslant p - 1.$$    So, the required number of points $$ = 1 + 2\left( {p - 1} \right) = 2p - 1.$$

$$\eqalign{ & {\text{Around}}\,{\text{ }}x = \root 3 \of {\frac{\pi }{2}} ,\,\left[ x \right] = 1. \cr & {\text{So, }}\,f\left( x \right) = \cos \left\{ {\frac{\pi }{2} - {x^3}} \right\} = \sin \,{x^3}\,{\text{around }}x = \root 3 \of {\frac{\pi }{2}} \cr & \therefore f'\left( x \right) = 3{x^2}\cos \,{x^3} \cr & \therefore f'\left( {\root 3 \of {\frac{\pi }{2}} } \right) = 3.{\left( {\frac{\pi }{2}} \right)^{\frac{2}{3}}}.\cos \frac{\pi }{2} = 0 \cr} $$

$$\eqalign{ & y = \sqrt {\log \,x + y} \cr & \Rightarrow {y^2} = \log \,x + y \cr & \Rightarrow 2y\frac{{dy}}{{dx}} = \frac{1}{x} + \frac{{dy}}{{dx}} \cr & \Rightarrow \frac{{dy}}{{dx}} = \frac{1}{{x\left( {2y - 1} \right)}} \cr} $$