Application of Derivatives MCQ Questions & Answers in Calculus | Maths

$$\eqalign{ & {\text{We have, }}A + B = \frac{\pi }{3} \cr & \therefore \,B = \frac{\pi }{3} - A \Rightarrow \tan \,B = \frac{{\sqrt 3 - \tan \,A}}{{1 + \sqrt 3 \tan \,A}} \cr & {\text{Let }}Z = \tan \,A.\frac{{\sqrt 3 - \tan \,A}}{{1 + \sqrt 3 \tan \,A}} \cr & \Rightarrow Z = \frac{{\sqrt 3 \tan \,A - {{\tan }^2}A}}{{1 + \sqrt 3 \tan \,A}} \cr & \Rightarrow Z = \frac{{\sqrt 3 x - {x^2}}}{{1 + \sqrt 3 x}},\,\,\,\,\,\,{\text{where}}\,x = \tan \,A \cr & \Rightarrow \frac{{dZ}}{{dx}} = - \frac{{\left( {x + \sqrt 3 } \right)\left( {\sqrt 3 x - 1} \right)}}{{{{\left( {1 + \sqrt 3 x} \right)}^2}}} \cr & {\text{For max }}Z,\,\frac{{dZ}}{{dx}} = 0 \Rightarrow x = \frac{1}{{\sqrt 3 }},\, - \sqrt 3 \cr} $$
$$x \ne - \sqrt 3 $$   because $$A + B = \frac{\pi }{3}$$   which implies that $$x = \tan \,A > 0.$$
It can be easily checked that $$\frac{{{d^2}Z}}{{d{x^2}}} < 0{\text{ for }}x = \frac{1}{{\sqrt 3 }}.$$
Hence, $$Z$$ is maximum for $$x = \frac{1}{{\sqrt 3 }}$$  i.e., $$\tan \,A = \frac{1}{{\sqrt 3 }}{\text{ or }}A = \frac{\pi }{6}$$
For this value of $$x,\,Z = \frac{1}{3}.$$

Shortest distance between two curve occurred along the common normal
Slope of normal to $${y^2} = x$$   at point $$P\left( {{t^2},t} \right)$$   is $$ - 2t$$  and slope of line $$y - x = 1$$   is 1.
As they are perpendicular to each other
$$\eqalign{ & \therefore \,\left( { - 2t} \right) = - 1 \Rightarrow t = \frac{1}{2} \cr & \therefore \,P\left( {\frac{1}{4},\frac{1}{2}} \right) \cr} $$
and shortest distance $$ = \left| {\frac{{\frac{1}{2} - \frac{1}{4} - 1}}{{\sqrt 2 }}} \right|$$

So shortest distance between them is $$\frac{{3\sqrt 2 }}{8}$$

Product of two increasing function is always an increasing function.
$$\therefore \,fog$$   is always an increasing function.

$$\eqalign{ & f\left( x \right) = \sqrt {9 - {x^2}} \cr & f'\left( x \right) = \frac{1}{{2\sqrt {9 - {x^2}} }} \times \left( { - 2x} \right) = - \frac{x}{{\sqrt {9 - {x^2}} }} \cr} $$
For function to be increasing
$$ - \frac{x}{{\sqrt {9 - {x^2}} }} > 0{\text{ or }} - x > 0{\text{ or }}x < 0$$
but $$\sqrt {9 - {x^2}} $$   is defined only when
$$\eqalign{ & 9 - {x^2} > 0{\text{ or }}{x^2} - 9 < 0 \cr & \left( {x + 3} \right)\left( {x - 3} \right) < 0 \cr & {\text{i}}{\text{.e}}{\text{., }} - 3 < x < 3 \cr & - 3 < x < 3 \cap x < 0\, \Rightarrow - 3 < x < 0 \cr} $$

$$\eqalign{ & {\text{Here, }}y = \sin \,\theta .\sin \,\theta = \sin \,\theta .\sin \left( {\frac{\pi }{3} - \theta } \right) \cr & = \frac{1}{2}\left\{ {\cos \left( {2\theta - \frac{\pi }{3}} \right) - \cos \frac{\pi }{3}} \right\} \cr & = \frac{1}{2}\cos \left( {2\theta - \frac{\pi }{3}} \right) - \frac{1}{4} \cr & {\text{This has a maximum at }}2\theta - \frac{\pi }{3} = 0 \cr} $$

$$\frac{{dy}}{{dx}} = \frac{{\frac{{dy}}{{dt}}}}{{\frac{{dx}}{{dt}}}} = \frac{{3a\,{{\sin }^2}t\,\cos \,t}}{{ - 3a\,{{\cos }^2}t\,\sin \,t}} = - \tan \,t$$
$$\therefore $$  equation of the tangent at $$‘t’$$ is
$$\eqalign{ & y - a\,{\sin ^3}t = - \tan \,t\left( {x - a\,{{\cos }^3}t} \right) \cr & \Rightarrow x\,\tan \,t + y - a\left( {{{\sin }^3}t + \sin \,t.{{\cos }^2}t} \right) = 0 \cr & \Rightarrow x\,\tan \,t + y - a\,\sin \,t = 0 \cr} $$
$$\therefore $$   distance from the origin to this tangent
$$ = \frac{{\left| { - a\,\sin \,t} \right|}}{{\sqrt {{{\tan }^2}t + 1} }} = \frac{{a\,\sin \,t}}{{\sec \,t}} = \frac{a}{2}\sin \,2t$$

$$\eqalign{ & {\text{Given}}\,{\text{that}}\,f\left( x \right) = x\left| x \right|\,{\text{and}}\,g\left( x \right) = \sin x \cr & {\text{So}}\,{\text{that}}\,gof\left( x \right) = g\left( {f\left( x \right)} \right) = g\left( {x\left| x \right|} \right) = \sin x\left| x \right| \cr & = \left\{ {_{\sin \left( {{x^2}} \right),{\text{ if }}x\, \geqslant \,0}^{\sin \left( { - {x^2}} \right),{\text{ if }}x\, < \,0}} \right. \cr & = \left\{ {_{\sin {x^2},{\text{ if }}x\, \geqslant \,0}^{ - \sin {x^2},{\text{ if }}x\, < \,0}} \right. \cr & \therefore \left( {gof} \right)'\left( x \right) = \left\{ {_{2x\,\cos {x^2},{\text{ if }}x \geqslant 0}^{ - 2x\cos {x^2},{\text{ if }}x < 0}} \right. \cr & {\text{Here we observe}} \cr & L\left( {{\text{go}}f} \right)'{\text{ }}\left( 0 \right) = 0 = R\left( {gof} \right)'\left( 0 \right){\text{ }} \cr & \Rightarrow gof\,{\text{is differentiable at}}\,x = 0 \cr & {\text{and}}\,{\left( {gof} \right)^\prime }\,{\text{is}}\,{\text{continuous at }}x{\text{ }} = {\text{ }}0 \cr & {\text{Now}}\,{\left( {gof} \right)^{\prime \prime }}\left( x \right) = \left\{ {_{2\cos {x^2} - 4{x^2}\sin {x^2},\,x\, \geqslant \,0}^{ - 2\cos {x^2} + 4{x^2}\sin {x^2},\,x\, < \,0}} \right. \cr & {\text{Here }}L{\left( {gof} \right)^{\prime \prime }}\left( 0 \right){\text{ }} = - 2{\text{ and }}R{\left( {gof} \right)^{\prime \prime }}\left( 0 \right) = 2 \cr & \because \,\,L{\left( {gof} \right)^{\prime \prime }}\left( 0 \right) \ne R{\left( {gof} \right)^{\prime \prime }}\left( 0 \right) \cr & \Rightarrow gof\left( x \right)\,{\text{is not twice differentiable at}}\,x = 0. \cr} $$
∴ Statement - 1 is true but statement -2 is false.

$$\eqalign{ & {y^2} = 2x\,\,\,\,\, \Rightarrow 2y\frac{{dy}}{{dx}} = 2\,\,\,\, \Rightarrow \frac{{dy}}{{dx}} = \frac{1}{y} \cr & {\text{The slope of }}PQ = \frac{{2 + 1}}{{2 - \frac{1}{2}}}{\text{ = 2}} \cr & \therefore \frac{1}{y} = 2\,\, \Rightarrow y = \frac{1}{2} \cr & \therefore x = \frac{{{y^2}}}{2} = \frac{{{{\left( {\frac{1}{2}} \right)}^2}}}{2} = \frac{1}{8} \cr} $$

Let $$a,\,b,\,c$$   be the roots. Then $$ab + bc + ca = \mu ,\,abc = 6$$
$$\eqalign{ & {\text{Dividing, }}\,\,\frac{1}{a} + \frac{1}{b} + \frac{1}{c} = \frac{\mu }{6} \cr & {\text{AM}} > {\text{GM}}\,\,\,\,\,\,\, \Rightarrow \frac{{\frac{1}{a} + \frac{1}{b} + \frac{1}{c}}}{3} \geqslant \root 3 \of {\frac{1}{{abc}}} \cr & {\text{or }}\mu \geqslant 18.\,\root 3 \of {\frac{1}{{abc}}} = {3.6^{\frac{2}{3}}} \cr} $$
$$\therefore $$ the minimum value of $$\mu = {3.6^{\frac{2}{3}}}$$

Let cost $$C = av + \frac{b}{v}$$
According to given question,
$$\eqalign{ & 30a + \frac{b}{{30}} = 75.....\left( {\text{i}} \right) \cr & 40a + \frac{b}{{40}} = 65.....\left( {{\text{ii}}} \right) \cr & {\text{On solving }}\left( {\text{i}} \right){\text{and}}\left( {{\text{ii}}} \right){\text{, we get}}\,a = \frac{1}{2}{\text{and}}\,b = 1800 \cr & {\text{Now, }}C = av + \frac{b}{v} \Rightarrow \frac{{dC}}{{dv}} = a - \frac{b}{{{v^2}}} \cr & \frac{{dC}}{{dv}} = 0 \cr & \Rightarrow a - \frac{b}{{{v^2}}} = 0 \cr & \Rightarrow v = \sqrt {\frac{b}{a}} = \sqrt {3600} \cr & \Rightarrow v = 60\,{\text{kmph}} \cr} $$