Differentiability and Differentiation MCQ Questions & Answers in Calculus | Maths

$$\eqalign{ & \mathop {\lim }\limits_{x \to a} \frac{{f\left( a \right)g'\left( x \right) - g\left( a \right)f'\left( x \right)}}{{g'\left( x \right) - f'\left( x \right)}} = 4\,\,\left( {{\text{By L'Hospital rule}}} \right) \cr & \mathop {\lim }\limits_{x \to a} \frac{{k\,g'\left( x \right) - k\,f'\left( x \right)}}{{g'\left( x \right) - f'\left( x \right)}} = 4 \cr & \therefore k = 4 \cr} $$

$$\eqalign{ & {\text{Let }}x = \cos \,\theta \cr & {\text{Then }}y = {\text{cose}}{{\text{c}}^{ - 1}}\frac{1}{{2{x^2} - 1}} \cr & = {\text{cose}}{{\text{c}}^{ - 1}}\sec \,2\theta \cr & = {\text{cose}}{{\text{c}}^{ - 1}}\left\{ {{\text{cosec}}\left( {\frac{\pi }{2} - 2\theta } \right)} \right\} \cr & = \frac{\pi }{2} - 2\theta \cr & z = \sqrt {1 - {{\cos }^2}\theta } = \sin \,\theta \cr & \therefore \,\frac{{dy}}{{dz}} = \frac{{\frac{{dy}}{{d\theta }}}}{{\frac{{dz}}{{d\theta }}}} = \frac{{ - 2}}{{\cos \,\theta }} = \frac{{ - 2}}{x} \cr} $$

$$\mathop {\lim }\limits_{x \to 0} \frac{1}{x} = \frac{1}{0} = \infty $$    which does not exist.
Hence, statements-1 is incorrect.
Now, $$\mathop {\lim }\limits_{x \to 0} \,{e^{\frac{1}{x}}} = {e^\infty }$$    which also does not exist.
Hence, statement-2 is correct.

We have,
$$\eqalign{ & {\text{L}}f'\left( 0 \right) = \mathop {\lim }\limits_{h \to 0} \frac{{f\left( {0 - h} \right) - f\left( 0 \right)}}{{ - h}} \cr & = \mathop {\lim }\limits_{h \to 0} \frac{{ - h\,\log \,\cos \,h}}{{ - h\,\log \left( {1 + {h^2}} \right)}} \cr & = \mathop {\lim }\limits_{h \to 0} \frac{{\log \,\cos \,h}}{{\log \left( {1 + {h^2}} \right)}}\,\,\,\,\,\,\,\,\left( {\frac{0}{0}{\text{form}}} \right) \cr & = \mathop {\lim }\limits_{h \to 0} \frac{{ - \tan \,h}}{{\frac{{2h}}{{\left( {1 + {h^2}} \right)}}}} \cr & = - \frac{1}{2} \cr & {\text{R}}f'\left( 0 \right) = \mathop {\lim }\limits_{h \to 0} \frac{{f\left( {0 + h} \right) - f\left( 0 \right)}}{h} \cr & = \mathop {\lim }\limits_{h \to 0} \frac{{h\,\log \,\cos \,h}}{{h\,\log \left( {1 + {h^2}} \right)}} \cr & = \mathop {\lim }\limits_{h \to 0} \frac{{\log \,\cos \,h}}{{\log \left( {1 + {h^2}} \right)}}\,\,\,\,\,\,\,\,\left( {\frac{0}{0}{\text{form}}} \right) \cr & = \mathop {\lim }\limits_{h \to 0} \frac{{ - \tan \,h}}{{\frac{{2h}}{{\left( {1 + {h^2}} \right)}}}} \cr & = \frac{{ - 1}}{2} \cr} $$
Since $${\text{L}}f'\left( 0 \right){\text{ = R}}f'\left( 0 \right),$$    therefore $$f\left( x \right)$$  is differentiable at $$x = 0.$$
Since differentiability $$ \Rightarrow $$ continutity, therefore $$f\left( x \right)$$  iscontinuous at $$x = 0.$$

$$\eqalign{ & {\text{Let }}f\left( x \right) = \lambda {x^2} + \mu x + v \cr & {\text{Then }}f'\left( x \right) = 2\lambda x + \mu \cr & {\text{Also,}}\,\,f\left( 1 \right) = f\left( { - 1} \right) \cr & \Rightarrow \lambda + \mu + v = \lambda - \mu + v \cr & \Rightarrow \mu = 0\,\,\,\,\,\,\,\,\,\,\,\,\,\,\therefore f'\left( x \right) = 2\lambda x \cr & \therefore f'\left( {{a_1}} \right) = 2\lambda {a_1}\,\,\,\,f'\left( {{a_2}} \right) = 2\lambda {a_2}\,\,\,\,f'\left( {{a_3}} \right) = 2\lambda {a_3} \cr & {\text{As }}{a_1},\,{a_2},\,{a_3}{\text{ are in AP, }}f'\left( {{a_1}} \right),\,f'\left( {{a_2}} \right),\,f'\left( {{a_3}} \right){\text{ are in AP}}{\text{.}} \cr} $$

\[\begin{array}{l} {\rm{We\,\, have }}\left| x \right| = \left\{ \begin{array}{l} - x\,\,\,\,\,{\rm{if}}\,\,x < 0\\ x\,\,\,\,\,\,\,\,{\rm{if}}\,\,x \ge 0 \end{array} \right.\\ {\rm{Also }}\left| {{x^2} - 3x + 2} \right| = \left| {\left( {x - 1} \right)\left( {x - 2} \right)} \right|\\ = \left\{ \begin{array}{l} \left( {1 - x} \right)\left( {2 - x} \right)\,\,{\rm{if}}\,\,x < 1\\ \left( {x - 1} \right)\left( {2 - x} \right)\,\,{\rm{if}}\,\,1 \le x < 2\\ \left( {x - 1} \right)\left( {x - 2} \right)\,\,{\rm{if}}\,\,x \ge 2 \end{array} \right. \end{array}\]
As $$\cos \left( { - \theta } \right) = \cos \,\theta \,\,\,\, \Rightarrow \cos \,\left| x \right| = \cos \,x$$
$$\therefore $$ Given function can be written as
\[f\left( x \right) = \left\{ \begin{array}{l} \left( {{x^2} - 1} \right)\left( {x - 1} \right)\left( {x - 2} \right) + \cos \,x\,\,{\rm{if}}\,\,x \le 1\\ - \left( {{x^2} - 1} \right)\left( {x - 1} \right)\left( {x - 2} \right) + \cos \,x\,\,{\rm{if}}\,\,1 \le x < 2\\ \left( {{x^2} - 1} \right)\left( {x - 1} \right)\left( {x - 2} \right) + \cos \,x\,\,{\rm{if}}\,\,x \ge 2 \end{array} \right.\]
This function is differentiable at all points except possibly at $$x = 1$$  and $$x=2.$$
$$\eqalign{ & {\text{L}}f'(1) = {\left\{ {\frac{d}{{dx}}\left[ {\left( {{x^2} - 1} \right)\left( {x - 1} \right)\left( {x - 2} \right) + \cos \,x} \right]} \right\}_{x\, = \,1}} \cr & = - \sin \,1 \cr & {\text{R}}f'(1) = {\left\{ {\frac{d}{{dx}}\left[ { - \left( {{x^2} - 1} \right)\left( {x - 1} \right)\left( {x - 2} \right) + \cos \,x} \right]} \right\}_{x\, = \,1}} \cr & = - \sin \,1 \cr & \because {\text{L}}f'\left( 1 \right) = {\text{R}}f'\left( 1 \right) \cr & \therefore \,f\,\,{\text{is differentiable at }}x = 1 \cr & {\text{L}}f'(2) = {\left\{ {\frac{d}{{dx}}\left[ { - \left( {{x^2} - 1} \right)\left( {x - 1} \right)\left( {x - 2} \right) + \cos \,x} \right]} \right\}_{x\, = \,2}} \cr & = - 3 - \sin \,2 \cr & {\text{R}}f'(2) = {\left\{ {\frac{d}{{dx}}\left[ {\left( {{x^2} - 1} \right)\left( {x - 1} \right)\left( {x - 2} \right) + \cos \,x} \right]} \right\}_{x\, = \,2}} \cr & = 3 - \sin \,2 \cr & {\text{L}}f'\left( 2 \right) \ne {\text{R}}f'\left( 2 \right) \cr} $$
$$\therefore \,f$$ is not differentiable at $$x =2.$$

$$\eqalign{ & y'\left( x \right) = f'\left( {f\left( {f\left( {f\left( x \right)} \right)} \right)} \right)f'\left( {f\left( {f\left( x \right)} \right)} \right)f'\left( {f\left( x \right)} \right)f'\left( x \right) \cr & \Rightarrow y'\left( 0 \right) = f'\left( {f\left( {f\left( {f\left( 0 \right)} \right)} \right)} \right)f'\left( {f\left( {f\left( 0 \right)} \right)} \right)f'\left( {f\left( 0 \right)} \right)f'\left( 0 \right) \cr & = f'\left( {f\left( {f\left( 0 \right)} \right)} \right)f'\left( {f\left( 0 \right)} \right)f'\left( 0 \right)f'\left( 0 \right) \cr & = f'\left( {f\left( 0 \right)} \right)f'\left( 0 \right)f'\left( 0 \right)f'\left( 0 \right) \cr & = f'\left( 0 \right)f'\left( 0 \right)f'\left( 0 \right)f'\left( 0 \right) \cr & = {\left( {f'\left( 0 \right)} \right)^4} \cr & = {2^4} \cr & = 16 \cr} $$

$$f'\left( 1 \right) = \mathop {\lim }\limits_{h \to 0} \frac{{f\left( {1 + h} \right) - f\left( 1 \right)}}{h}\,;$$
As function is differentiable so it is continuous as it is given that $$\mathop {\lim }\limits_{h \to 0} \frac{{f\left( {1 + h} \right)}}{h} = 5$$     and hence $$f\left( 1 \right) = 0$$
Hence, $$f'\left( 1 \right) = \mathop {\lim }\limits_{h \to 0} \frac{{f\left( {1 + h} \right)}}{h} = 5$$

$$\because P\left( {{x^2} + 1} \right) = {\left\{ {P\left( x \right)} \right\}^2} + 1$$      and $$P\left( x \right)$$  is a polynomial, we have $$P\left( x \right) = x$$   which also satisfies $$P\left( 0 \right) = 0.$$   Therefore, $$P'\left( x \right) = 1.$$   So, $$P'\left( 0 \right) = 1.$$

$$\eqalign{ & f\left( {{{\log }_e}\,x} \right) = {\log _e}\left( {{{\log }_e}\,x} \right) \cr & \therefore \,\frac{{df\left( {{{\log }_e}\,x} \right)}}{{dx}} = \frac{1}{{{{\log }_e}\,x}}.\frac{1}{x} \cr} $$