Application of Derivatives MCQ Questions & Answers in Calculus | Maths

$$\eqalign{ & f\left( x \right) = a\,{\log _e}\left| x \right| + b{x^2} + x \cr & f'\left( x \right) = \frac{a}{x} + 2bx + 1 \cr & {\text{Given, }}x = 1,\,3{\text{ are extremum of }}f\left( x \right) \cr & \Rightarrow f'\left( 1 \right) = 0 = f'\left( 3 \right) \cr & \Rightarrow a + 2b + 1 = 0......\left( 1 \right) \cr & {\text{and }}\frac{a}{3} + 6b + 1 = 0......\left( 2 \right) \cr & {\text{solving equation }}\left( 1 \right){\text{ and }}\left( 2 \right){\text{ we get,}} \cr & a = - \frac{3}{4},\,b = - \frac{1}{8} \cr} $$

$${\text{Area}}\,{\text{of}}\,\Delta ABC,\,A = \frac{1}{2} \times d\cos \alpha \times d\sin \alpha = \frac{{{d^2}}}{4}\sin 2\alpha $$

$$\eqalign{ & {\text{which is max}}{\text{. when}}\,\sin 2\alpha = 1 \cr & {\text{i}}{\text{.e}}{\text{.}}\,\alpha = {45^ \circ } \cr & \therefore \Delta ABC\,{\text{is an isosceles triangle}}{\text{.}} \cr} $$

It is clear from figure that at $$x = 0,f\left( x \right)$$   is not differentiable.

⇒ $$f\left( x \right)$$  has neither maximum nor minimum at $$x = 0.$$

Here, $$b{y^2} = {\left( {x + a} \right)^3}$$
Differentiating both the sides, we get
$$2by\frac{{dy}}{{dx}} = 3{\left( {x + a} \right)^2} \Rightarrow \frac{{dy}}{{dx}} = \frac{{3{{\left( {x + a} \right)}^2}}}{{2by}}$$
$$\therefore $$  length of subnormal $$SN = y.\frac{{dy}}{{dx}} = \frac{3}{2}\frac{{{{\left( {x + a} \right)}^2}}}{b}$$
$$\therefore $$  length of subtangent $$ST = y.\frac{{dx}}{{dy}} = \frac{{2b{y^2}}}{{3{{\left( {x + a} \right)}^2}}}$$
$$\eqalign{ & \therefore \,p\left( {SN} \right) = q{\left( {ST} \right)^2} \cr & \Rightarrow \frac{p}{q} = \frac{{{{\left( {ST} \right)}^2}}}{{\left( {SN} \right)}} \cr & \Rightarrow \frac{p}{q} = \frac{8}{{27}}\frac{{{b^3}{y^4}}}{{{{\left( {x + a} \right)}^6}}} \cr & \Rightarrow \frac{p}{q} = \frac{{8b}}{{27}}\,\,\,\,\,\,\,\,\left( {\because \,\frac{{{b^2}{y^4}}}{{{{\left( {x + a} \right)}^6}}} = 1} \right) \cr} $$

The point of intersection is $$\left( {1,\,\frac{1}{e}} \right)$$
Around $$x = 1,\,\,\left| x \right| = x$$
$$\eqalign{ & {\text{So, }}y = {e^{ - x}} \cr & \therefore \frac{{dy}}{{dx}} = - {e^{ - x}} \cr & \therefore {\left. {\frac{{dy}}{{dx}}} \right)_{x = 1}} = - {e^{ - 1}} \cr} $$
$$\therefore $$ the equation of the tangent at $$\left( {1,\,\frac{1}{e}} \right)$$  is
$$y - \frac{1}{e} = - \frac{1}{e}\left( {x - 1} \right){\text{ or }}x + ey = 2$$

$$\eqalign{ & {\text{Given}}\,f\left( x \right) = {\tan ^{ - 1}}\left( {\sin x + \cos x} \right) \cr & f'\left( x \right) = \frac{1}{{1 + {{\left( {\sin x + \cos x} \right)}^2}}} \cdot \left( {\cos x - \sin x} \right) \cr & = \frac{{\sqrt 2 \cdot \left( {\frac{1}{{\sqrt 2 }}\cos x - \frac{1}{{\sqrt 2 }}\sin x} \right)}}{{1 + {{\left( {\sin x + \cos x} \right)}^2}}} \cr & = \frac{{\left( {\cos \frac{\pi }{4} \cdot \cos x - \sin \frac{\pi }{4} \cdot \sin x} \right)}}{{1 + {{\left( {\sin x + \cos x} \right)}^2}}} \cr & \therefore f'\left( x \right) = \frac{{\sqrt 2 \cos \left( {x + \frac{\pi }{4}} \right)}}{{1 + {{\left( {\sin x + \cos x} \right)}^2}}} \cr & {\text{if}}\,f'\left( x \right) > O\,{\text{then}}\,f\left( x \right)\,{\text{is increasing function}}{\text{.}} \cr & {\text{Hence}}\,f\left( x \right)\,{\text{is increasing,}}\,{\text{if}}\, - \frac{\pi }{2} < x + \frac{\pi }{4} < \frac{\pi }{2} \cr & \Rightarrow - \frac{{3\pi }}{4} < x < \frac{\pi }{4} \cr & {\text{Hence, }}f\left( x \right)\,{\text{is increasing when}}\,n \in \left( { - \frac{\pi }{2},\frac{\pi }{4}} \right) \cr} $$

$$\eqalign{ & AD = AB\,\cos \,\theta = 2R\,\cos \,\theta ,\,AE = AD\,\cos \,\theta = 2R\,{\cos ^2}\theta \cr & {\text{or }}EF = AB - 2AE = 2R - 4R\,{\cos ^2}\theta \cr & DE = AD\,\sin \,\theta = 2R\,\sin \,\theta \,\cos \,\theta \cr} $$
Thus, area of trapezium
$$\eqalign{ & S = \frac{1}{2}\left( {AB + CD} \right) \times DE \cr & = \frac{1}{2}\left( {2R + 2R - 4R\,{{\cos }^2}\theta } \right) \times 2R\,\sin \,\theta \,\cos \,\theta \cr & = 4{R^2}{\sin ^3}\theta \,\cos \,\theta \cr & \frac{{dS}}{{d\theta }} = 12{R^2}{\sin ^2}\theta \,{\cos ^2}\theta - 4{R^2}{\sin ^4}\theta \cr & = 4{R^2}{\sin ^2}\theta \left( {3\,{{\cos }^2}\theta - {{\sin }^2}\theta } \right) \cr} $$
For maximum area,
$$\eqalign{ & \frac{{dS}}{{d\theta }} = 0 \cr & {\text{or }}{\tan ^2}\theta = 3 \cr & {\text{or }}\tan \,\theta = \sqrt 3 \left( {\theta {\text{ is acute}}} \right) \cr & {\text{or }}{S_{\max }} = \frac{{3\sqrt 3 }}{4}{R^2} \cr} $$

There is only one function in option (A) whose critical point $$\frac{1}{2} \in \left( {0,1} \right)$$   for the rest of the parts critical point $$0 \notin \left( {0,1} \right).$$   It can be easily seen that functions in options (B), (C) and (D) are continuous on [0, 1] and differentiable in (0, 1).
\[{\text{Now}}\,{\text{for}}\,f\left( x \right) = \left\{ {\begin{array}{*{20}{c}} {\left( {\frac{1}{2} - x} \right),}&{x < \frac{1}{2}} \\ {{{\left( {\frac{1}{2} - x} \right)}^2},}&{x \geqslant \frac{1}{2}} \end{array}} \right.\]
$$\eqalign{ & {\text{Here}}\,f'\left( {\frac{{{1^ - }}}{2}} \right) = - 1\,{\text{and}}\,f'\left( {\frac{{{1^ + }}}{2}} \right) = - 2\left( {\frac{1}{2} - \frac{1}{2}} \right) = 0 \cr & \therefore f'\left( {\frac{{{1^ - }}}{2}} \right) \ne f'\left( {\frac{{{1^ + }}}{2}} \right) \cr & \therefore f\,{\text{is not differentiable at }}\frac{1}{2} \in \left( {0,1} \right) \cr & \therefore LMV{\text{ is not applicable for this function in}}\,\left[ {0,1} \right] \cr} $$

$$\eqalign{ & {\text{Given, }}f\left( x \right) = {x^2} + x + 1 \cr & f'\left( x \right) = 2x + 1 \cr & f''\left( x \right) = 2 \cr & {\text{For extrema of }}f\left( x \right) \cr & f'\left( x \right) = 0 \Rightarrow x = - \frac{1}{2} \cr & {\text{also }}f''\left( x \right) > 0 \cr & {\text{Thus }}f\left( x \right){\text{ will achieve it's minima at }}x = - \frac{1}{2} \cr & {\text{which is }}\frac{3}{4} \cr} $$

Let us define a function
$$f\left( x \right) = \frac{{a{x^3}}}{3} + \frac{{b{x^2}}}{2} + cx$$
Being polynomial, it is continuous and differentiable, also,
$$\eqalign{ & f\left( 0 \right) = 0\,{\text{and}}\,f\left( 1 \right) = \frac{a}{3} + \frac{b}{2} + c \cr & \Rightarrow f\left( 1 \right) = \frac{{2a + 3b + 6c}}{6} = 0\,\left( {{\text{given}}} \right) \cr & \therefore f\left( 0 \right) = f\left( 1 \right) \cr & \therefore f\left( x \right)\,{\text{satisfies all conditions of Rolle}}\,{\text{theorem}}\,{\text{therefore}}\,f'\left( x \right) = 0\,{\text{has a root in}}\left( {{\text{0,1}}} \right) \cr & {\text{i}}{\text{.e}}{\text{.}}\,a{x^2} + bx + c = 0\,{\text{has at lease one root in }}\left( {{\text{0,1}}} \right) \cr} $$