Differentiability and Differentiation MCQ Questions & Answers in Calculus | Maths

$$\eqalign{ & \left( {{a^2} - 2a - 15} \right){e^{ax}} + \left( {{b^2} - 2b - 15} \right){e^{bx}} = 0 \cr & {\text{or, }}\left( {{a^2} - 2a - 15} \right) = 0{\text{ and }}\left( {{b^2} - 2b - 15} \right) = 0 \cr & {\text{or, }}\left( {a - 5} \right)\left( {a + 3} \right) = 0{\text{ and }}\left( {b - 5} \right)\left( {b + 3} \right) = 0 \cr & {\text{i}}{\text{.e}}{\text{., }}a = 5{\text{ or }} - 3{\text{ and }}b = 5{\text{ or }} - 3 \cr & \therefore \,a \ne b \cr & {\text{Hence, }}a = 5{\text{ and }}b = - 3{\text{ or }}a = - 3{\text{ and }}b = 5 \cr & {\text{or }}ab = - 15 \cr} $$

\[\begin{array}{l} f\left( x \right) = \left| {x - 2} \right| = \left\{ \begin{array}{l} x - 2\,\,,\,\,x - 2 \ge 0\\ 2 - x\,\,,\,\,x - 2 \le 0 \end{array} \right.\\ = \left\{ \begin{array}{l} x - 2\,\,,\,\,x \ge 2\\ 2 - x\,\,,\,\,x \le 2 \end{array} \right. \end{array}\]
Similarly, \[f\left( x \right) = \left| {x - 5} \right| = \left\{ \begin{array}{l} x - 5\,\,,\,\,x \ge 5\\ 5 - x\,\,,\,\,x \le 5 \end{array} \right.\]
$$\eqalign{ & \therefore f\left( x \right) = \left| {x - 2} \right| + \left| {x - 5} \right| \cr & = \left\{ {x - 2 + 5 - x = 3,\,\,2 \leqslant x \leqslant 5} \right\} \cr & {\text{Thus}} \cr & f\left( x \right) = 3,\,2 \leqslant x \leqslant 5 \cr & f'\left( x \right) = 0,\,2 < x < 5 \cr & f'\left( 4 \right) = 0 \cr} $$
$$\therefore $$ statement 1 is true.

$$\because f\left( 2 \right) = 0 + \left| {2 - 5} \right| = 3$$     and $$f\left( 5 \right) = \left| {5 - 2} \right| + 0 = 3$$
$$\therefore $$ statement-2 is also true and a correct explanation for statement 1.

$$\eqalign{ & y = {\cot ^{ - 1}}\left[ {\frac{{\sqrt {1 + \sin \,x} + \sqrt {1 - \sin \,x} }}{{\sqrt {1 + \sin \,x} - \sqrt {1 - \sin \,x} }}} \right] \cr & y = {\cot ^{ - 1}}\left[ {\frac{{\sqrt {{{\cos }^2}\frac{x}{2} + {{\sin }^2}\frac{x}{2} + 2\,\sin \frac{x}{2}\cos \frac{x}{2}} + \sqrt {{{\cos }^2}\frac{x}{2} + {{\sin }^2}\frac{x}{2} - 2\,\sin \frac{x}{2}\cos \frac{x}{2}} }}{{\sqrt {{{\cos }^2}\frac{x}{2} + {{\sin }^2}\frac{x}{2} + 2\,\sin \frac{x}{2}\cos \frac{x}{2}} - \sqrt {{{\cos }^2}\frac{x}{2} + {{\sin }^2}\frac{x}{2} - 2\,\sin \frac{x}{2}\cos \frac{x}{2}} }}} \right] \cr & y = {\cot ^{ - 1}}\left[ {\frac{{\sqrt {{{\left( {\cos \frac{x}{2} + \sin \frac{x}{2}} \right)}^2}} + \sqrt {{{\left( {\cos \frac{x}{2} - \sin \frac{x}{2}} \right)}^2}} }}{{\sqrt {{{\left( {\cos \frac{x}{2} + \sin \frac{x}{2}} \right)}^2}} - \sqrt {{{\left( {\cos \frac{x}{2} - \sin \frac{x}{2}} \right)}^2}} }}} \right] \cr & y = {\cot ^{ - 1}}\left[ {\frac{{\cos \frac{x}{2} + \sin \frac{x}{2} + \cos \frac{x}{2} - \sin \frac{x}{2}}}{{\cos \frac{x}{2} + \sin \frac{x}{2} - \cos \frac{x}{2} + \sin \frac{x}{2}}}} \right] \cr & y = {\cot ^{ - 1}}\left[ {\frac{{2\,\cos \frac{x}{2}}}{{2\,\sin \frac{x}{2}}}} \right] \cr & y = {\cot ^{ - 1}}\left( {\cot \frac{x}{2}} \right) \cr & y = \frac{x}{2} \cr & \frac{{dy}}{{dx}} = \frac{1}{2} \cr} $$

$$\eqalign{ & \frac{{dy}}{{dx}} = 2\cos \,2x = 2\sin \left( {2x + \frac{\pi }{2}} \right) \cr & \therefore \frac{{{d^2}y}}{{d{x^2}}} = {2^2}\cos \left( {2x + \frac{\pi }{2}} \right)\,\,\, = {2^2}\sin \left( {2x + 2.\frac{\pi }{2}} \right) \cr & \frac{{{d^6}y}}{{d{x^6}}} = {2^6}.\sin \left( {2x + 6.\frac{\pi }{2}} \right) = {2^6}\sin \left( {2x + 3\pi } \right) \cr} $$

Let $$y = {\left( {1 + \frac{1}{x}} \right)^x}$$
Taking logarithm of both sides, we get
$$\eqalign{ & \log \,y = x\left[ {\log \left( {1 + \frac{1}{x}} \right)} \right] \cr & \Rightarrow \frac{1}{y}{y_1}\left( x \right) = \frac{{{x^2}}}{{x + 1}}\left( { - \frac{1}{{{x^2}}}} \right) + \log \left( {1 + \frac{1}{x}} \right) \cr & = - \frac{1}{{x + 1}} + \log \left( {1 + \frac{1}{x}} \right).....(1) \cr & {\text{Since,}}\,y\left( 2 \right) = {\left( {1 + \frac{1}{2}} \right)^2} = \frac{9}{4} \cr & {\text{So, }}{y_1}\left( 2 \right) = \left( {\frac{9}{4}} \right)\left( { - \frac{1}{3} + \log \frac{3}{2}} \right) \cr} $$
Again differentiate equation (1) w.r.t. $$\left( x \right),$$ we get
$$\frac{{y\left( x \right){y_2}\left( x \right) - {{\left[ {{y_1}\left( x \right)} \right]}^2}}}{{{{\left( {y\left( x \right)} \right)}^2}}} = \frac{1}{{{{\left( {1 + x} \right)}^2}}} - \frac{1}{{x\left( {x + 1} \right)}}$$
By putting $$x = 2,$$  we get
$$\frac{{y\left( 2 \right){y_2}\left( 2 \right) - {{\left( {{y_1}\left( 2 \right)} \right)}^2}}}{{{{\left( {y\left( 2 \right)} \right)}^2}}} = \frac{{ - 1}}{{18}}$$
Now, put value of $$y\left( 2 \right)$$  and $${y_1}\left( 2 \right)$$
$$\eqalign{ & \Rightarrow {y_2}\left( 2 \right) = \left( {\frac{9}{4}} \right){\left( { - \frac{1}{3} + \log \frac{3}{2}} \right)^2} - \frac{1}{8} \cr & \Rightarrow {\left( {{y_2}\left( 2 \right) + \frac{1}{8}} \right)^4} = 9{\left( {\log \frac{3}{2} - \frac{1}{3}} \right)^2} \cr & \Rightarrow {\text{ Required expression }} = 3 \cr} $$

$$\eqalign{ & F'\left( x \right) = \left[ {f\left( {\frac{x}{2}} \right).f'\left( {\frac{x}{2}} \right) + g\left( {\frac{x}{2}} \right).g'\left( {\frac{x}{2}} \right)} \right] \cr & {\text{Here, }}g\left( x \right) = f'\left( x \right){\text{ and }}g'\left( x \right) = f''\left( x \right) = - f\left( x \right) \cr & {\text{So, }}F'\left( x \right) = f\left( {\frac{x}{2}} \right).g\left( {\frac{x}{2}} \right) - f\left( {\frac{x}{2}} \right).g\left( {\frac{x}{2}} \right) = 0 \cr & \Rightarrow \,F\left( x \right){\text{ is constant function}} \cr & {\text{So, }}F\left( {10} \right) = 5 \cr} $$

Given that $$f\left( x \right) = \left| {1 - x} \right|$$
\[\therefore \,f\left( {\left| x \right|} \right) = \left\{ \begin{array}{l} \,\,\,\,x - 1,\,\,\,\,\,\,\,\,\,\,\,\,x > 1\\ \,\,\,\,1 - x,\,\,\,\,\,\,\,\,0 < x \le 1\\ \,\,\,\,1 + x,\,\,\,\,\, - 1 \le x \le 0\\ - x - 1,\,\,\,\,\,\,\,\,\,\,x < - 1 \end{array} \right.\]
Clearly, the domain of $${\sin ^{ - 1}}\left( {f\left| x \right|} \right)$$   is $$\left[ { - 2,\,2} \right].$$
Therefore, it is non-differentiable at the points $$\left\{ { - 1,\,0,\,1} \right\}.$$

$$y = {\cos ^{ - 1}}\cos \left( {\pi + \overline {x - \pi } } \right) = {\cos ^{ - 1}}\cos \left( {\pi - \overline {x - \pi } } \right).$$          When $$x$$ is around $$\frac{{5\pi }}{4},\,\pi - \overline {x - \pi } $$     is in the second quadrant. So, $$y = \pi - \left( {x - \pi } \right)\,;\,\,\,\,\,\,\therefore \,\,\frac{{dy}}{{dx}} = - 1$$

$$\eqalign{ & {y^2} = P\left( x \right) \cr & \Rightarrow 2y\frac{{dy}}{{dx}} = P'\left( x \right) \cr & \Rightarrow 2{\left( {\frac{{dy}}{{dx}}} \right)^2} + 2y\frac{{{d^2}y}}{{d{x^2}}} = P''\left( x \right) \cr & \therefore 2y = \frac{{{d^2}y}}{{d{x^2}}} = P''\left( x \right) - 2{\left( {\frac{{dy}}{{dx}}} \right)^2} \cr & {\text{or,}}\,\,2{y^3}\frac{{{d^2}y}}{{d{x^2}}} = {y^2}P''\left( x \right) - 2{y^2}{\left( {\frac{{dy}}{{dx}}} \right)^2} \cr & {\text{or,}}\,\,2{y^3}\frac{{{d^2}y}}{{d{x^2}}} = P\left( x \right).P''\left( x \right) - \frac{1}{2}{\left\{ {P'\left( x \right)} \right\}^2} \cr & {\text{Differentiating w}}{\text{.r}}{\text{.t}}{\text{. }}x, \cr & \,\,2\frac{d}{{dx}}\left( {{y^3}\frac{{{d^2}y}}{{d{x^2}}}} \right) \cr & = P'\left( x \right).P''\left( x \right) + P\left( x \right).P'''\left( x \right) - P'\left( x \right).P''\left( x \right) \cr & = P\left( x \right).P'''\left( x \right) \cr} $$

$$\eqalign{ & {\text{Putting }}x = 0,\,y = 0,\,\,{\text{we get }}2f\left( 0 \right){\text{ + }}{\left\{ {f\left( 0 \right)} \right\}^2}{\text{ = 1}} \cr & \Rightarrow f\left( 0 \right) = \sqrt 2 - 1\,\,\,\,\left\{ {\because \,f\left( 0 \right) > 0} \right\} \cr & {\text{Putting }}y = x,\,\,\,2f\left( x \right) + {\left\{ {f\left( x \right)} \right\}^2} = 1 \cr & {\text{Differentiating w}}{\text{.r}}{\text{.t}}{\text{. }}x,{\text{ }} \cr & {\text{2}}f'\left( x \right) + 2f\left( x \right).f'\left( x \right) = 0\,\,\,\,\,{\text{or}},\,\,f'\left( x \right)\left\{ {1 + f\left( x \right)} \right\} = 0 \cr & \Rightarrow f'\left( x \right) = 0,\,\,{\text{because }}f\left( x \right) > 0 \cr} $$