Differentiability and Differentiation MCQ Questions & Answers in Calculus | Maths

Let $$f\left( x \right) = {a_n}{x^n} + {a_{n - 1}}{x^{n - 1}} + ..... + {a_2}{x^2} + {a_1}x + {a_0}$$          which is a polynomial function in $$x$$ of degree $$n.$$
Hence, $$f\left( x \right)$$  is continuous and differentiable for all $$x.$$
Let $$\alpha < \beta $$
We are given, $$f\left( \alpha \right) = 0 = f\left( \beta \right)$$
By Rolle's theorem, $$f'\left( c \right) = 0$$   for some value $$c,\,\alpha < c < \beta .$$
Hence, the equation $$f'\left( x \right) = n{a_n}{x^{n - 1}} + \left( {n - 1} \right){a_{n - 1}}{x^{n - 2}} + ..... + {a_1} = 0$$          has at least one root between $$\alpha $$ and $$\beta .$$

In the neighborhood for $$x = \frac{{2\pi }}{3},$$  we have
$$\eqalign{ & \cos \,x < 0{\text{ and }}\sin \,x > 0 \cr & \therefore \,y = - \cos \,x + \sin \,x \cr & \Rightarrow \frac{{dy}}{{dx}} = \sin \,x + \cos \,x \cr & \Rightarrow {\left[ {\frac{{dy}}{{dx}}} \right]_{x = \frac{{2\pi }}{3}}} = \sin \frac{{2\pi }}{3} + \cos \frac{{2\pi }}{3} \cr & = \frac{{\sqrt 3 }}{2} - \frac{1}{2} \cr & = \frac{{\sqrt 3 - 1}}{2} \cr} $$

$$\eqalign{ & f\left( x \right) = \left( {{x^2} - 1} \right)\left| {\left( {x - 1} \right)\left( {x - 2} \right)} \right| \cr & f'\left( {{1^ + }} \right) = \mathop {\lim }\limits_{h \to 0} \frac{{\left( {{{\left( {1 + h} \right)}^2} - 1} \right)\left| {h \cdot \left( {1 + h - 2} \right)} \right| - 0}}{h} = 0, \cr & f'\left( {{1^ - }} \right) = \mathop {\lim }\limits_{h \to 0} \frac{{\left( {{{\left( {1 - h} \right)}^2} - 1} \right)\left| { - h \cdot \left( {1 - h - 2} \right)} \right| - 0}}{{ - h}} = 0 \cr & {\text{Hence, it is differentiable at }}x = 0 \cr & {\text{For, }}f\left( x \right) = \sin \left( {\left| {x - 1} \right|} \right) - \left| {x - 1} \right| \cr & f'\left( {{0^ + }} \right) = \mathop {\lim }\limits_{h \to 0} \frac{{\sin \,h - h - 0}}{h} = 0, \cr & f'\left( {{0^ - }} \right) = \mathop {\lim }\limits_{h \to 0} \frac{{\sin \,\left| { - h} \right| - \left| { - h} \right|}}{{ - h}} = 0 = \mathop {\lim }\limits_{h \to 0} \frac{{\sin \,h - h}}{{ - h}} = 0 \cr & {\text{Hence, }}f\left( x \right){\text{ is differentiable at }}x = 0 \cr & {\text{For }}f\left( x \right) = \tan \left( {\left| {x - 1} \right|} \right) + \left| {x - 1} \right| \cr & f'\left( {{0^ + }} \right) = \mathop {\lim }\limits_{h \to 0} \frac{{\tan \,h + h - 0}}{h} = 2, \cr & f'\left( {{0^ - }} \right) = \mathop {\lim }\limits_{h \to 0} \frac{{\tan \,\left| { - h} \right| + \left| { - h} \right|}}{{ - h}} = \mathop {\lim }\limits_{h \to 0} \frac{{\tan \,h + h}}{{ - h}} = - 2 \cr} $$

$$\eqalign{ & {x^a}{y^b} = {\left( {x - y} \right)^{a + b}}{\text{ taking }}\log {\text{ both sides}}{\text{.}} \cr & \log \left( {{x^a}{y^b}} \right) = \log {\left( {x - y} \right)^{\left( {a + b} \right)}} \cr & a\log \,x + b\,\log \,y = \left( {a + b} \right)\log \left( {x - y} \right) \cr & {\text{differentiating both sides w}}{\text{.r}}{\text{.t}}{\text{. }}'x' \cr & \frac{a}{x} + \frac{b}{y}\frac{{dy}}{{dx}} = \frac{{\left( {a + b} \right)}}{{\left( {x - y} \right)}}\left[ {1 - \frac{{dy}}{{dx}}} \right] \cr & \frac{{dy}}{{dx}}\left[ {\frac{b}{y} + \frac{{a + b}}{{x - y}}} \right] = \frac{{a + b}}{{x - y}} - \frac{a}{x} \cr & \frac{{dy}}{{dx}}\left[ {\frac{{bx - by + ay + by}}{{y\left( {x - y} \right)}}} \right] = \frac{{ax + bx - ax + ay}}{{x\left( {x - y} \right)}} \cr & \frac{{dy}}{{dx}}\left[ {\frac{{bx + ay}}{y}} \right] = \frac{{bx + ay}}{x} \cr & \frac{{dy}}{{dx}} = \frac{y}{x}\,;\,\,\frac{{dy}}{{dx}} - \frac{y}{x} = 0 \cr} $$

If $$f\left( x \right)$$  is differential everywhere then $$\left| f \right|$$ is not differentiable at some point, so $$f\left| f \right|$$  is not differentiable at some point.
[ Example : $$f\left( x \right) = x$$   is differentiable everywhere but $$\left| {f\left( x \right)} \right| = \left| x \right|$$   is not differentiable at $$x = 0$$  ]

$$f\left( {x + y} \right) = f\left( x \right) \times f\left( y \right)$$
Differentiate with respect to $$x,$$ treating $$y$$ as constant
$$f'\left( {x + y} \right) = f'\left( x \right)f\left( y \right)$$
Putting $$x = 0$$  and $$y=x,$$  we get $$f'\left( x \right) = f'\left( 0 \right)f\left( x \right)\,;$$
$$ \Rightarrow f'\left( 5 \right) = 3f\left( 5 \right){\text{ }} = 3 \times 2{\text{ }} = 6$$

$$\eqalign{ & \because f''\left( x \right) - g''\left( x \right) = 0 \cr & {\text{Integrating, }}f'\left( x \right) - g'\left( x \right) = c\,; \cr & \Rightarrow f'\left( 1 \right) - g'\left( 1 \right) = c \cr & \Rightarrow 4 - 2 = c \cr & \Rightarrow c = 2 \cr & \therefore f'\left( x \right) - g'\left( x \right) = 2\,; \cr & {\text{Integrating, }}f\left( x \right) - g\left( x \right) = 2x + {c_1} \cr & \Rightarrow f\left( 2 \right) - g\left( 2 \right) = 4 + {c_1} \cr & \Rightarrow 9 - 3 = 4 + {c_1}\,; \cr & \Rightarrow {c_1} = 2 \cr & \therefore f\left( x \right) - g\left( x \right) = 2x + 2 \cr & {\text{At }}x = \frac{3}{2},\,f\left( x \right) - g\left( x \right) = 3 + 2 = 5 \cr} $$

$$\eqalign{ & {\text{RH derivative}} = \mathop {\lim }\limits_{h \to 0} \frac{{\left| {\tan \left( {\frac{\pi }{4} - \overline {\frac{\pi }{4} + h} } \right)} \right| - \left| {\tan \,0} \right|}}{h} \cr & = \mathop {\lim }\limits_{h \to 0} \frac{{\tan \,h}}{h} \cr & = 1 \cr & {\text{LH derivative}} = \mathop {\lim }\limits_{h \to 0} \frac{{\left| {\tan \left( {\frac{\pi }{4} - \overline {\frac{\pi }{4} - h} } \right)} \right| - \left| {\tan \,0} \right|}}{{ - h}} \cr & = \mathop {\lim }\limits_{h \to 0} \frac{{\tan \,h}}{{ - h}} \cr & = - 1 \cr} $$

Let us test each of four options.
\[\begin{array}{l} \left( {\rm{A}} \right)f\left( x \right) = \cos \,\left( {\left| x \right|} \right) + \left| x \right| = \left\{ \begin{array}{l} \cos \,x - x,\,\,x < 0\\ \cos \,x + x,\,\,x \ge 0 \end{array} \right.\\ f'\left( x \right) = \left\{ \begin{array}{l} - \sin \,x - 1,\,x < 0\\ - \sin \,x + 1,\,\,x \ge 0 \end{array} \right.\\ {\rm{At}}\,{\rm{ }}x = 0,\,\,LD = - 1,\,\,RD = 1\\ \therefore {\rm{Not\,differentiable }}\\ \left( {\rm{B}} \right)f\left( x \right) = \cos \,\left| x \right| - \left| x \right| = \left\{ \begin{array}{l} \cos \,x + x,\,\,x < 0\\ \cos \,x - x,\,\,x \ge 0 \end{array} \right.\\ \therefore {\rm{Not\,differentiable\,at }}\,x = 0\\ \left( {\rm{C}} \right)f\left( x \right) = \sin \,\left| x \right| + \left| x \right| = \left\{ \begin{array}{l} - \sin \,x - x,\,\,x < 0\\ \sin \,x - x,\,\,x \ge 0 \end{array} \right.\\ \therefore {\rm{Not\,differentiable\,at }}\,x = 0\\ \left( {\rm{D}} \right)f\left( x \right) = \sin \,\left| x \right| - \left| x \right| = \left\{ \begin{array}{l} - \sin \,x + x,\,\,x < 0\\ \sin \,x - x,\,\,x \ge 0 \end{array} \right.\\ f'\left( x \right) = \left\{ \begin{array}{l} - \cos \,x - 1,\,\,x < 0\\ \cos \,x - 1,\,\,x \ge 0 \end{array} \right.\\ {\rm{At }}x = 0,\,\,LD = 0,\,\,RD = 0\,\\ \therefore f\,{\rm{ is\,differentiable\,at }}\,x = 0 \end{array}\]

$$\eqalign{ & {I_n} = \frac{{{d^{n - 1}}}}{{d{x^{n - 1}}}}\left[ {{x^{n - 1}} + n{x^{n - 1}}\log \,x} \right] \cr & {I_n} = \left( {n - 1} \right)! + n\,{I_{n - 1}} \cr & \Rightarrow {I_n} - n\,{I_{n - 1}} = \left( {n - 1} \right)! \cr} $$