Application of Derivatives MCQ Questions & Answers in Calculus | Maths

$$\eqalign{ & S = {b^2}x + \frac{{{c^2}{a^2}}}{x} \cr & \therefore \frac{{dS}}{{dx}} = {b^2} - \frac{{{c^2}{a^2}}}{{{x^2}}} = 0 \cr & \Rightarrow x = \pm \frac{{ca}}{b} \cr & \frac{{{d^2}S}}{{d{x^2}}} = 2{c^2}{a^2}.\frac{1}{{{x^3}}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\therefore {\left. {\frac{{{d^2}S}}{{d{x^2}}}} \right)_{x = \frac{{ca}}{b}}} > 0 \cr & \therefore \,\,\min \,S = {b^2}.\frac{{ca}}{b} + \frac{{{c^2}{a^2}}}{{\frac{{ca}}{b}}} = 2abc \cr} $$

Given the curve is $${x^{\frac{3}{2}}} + {y^{\frac{3}{2}}} = 2{a^{\frac{3}{2}}}.....(1)$$
$$\therefore \,\frac{3}{2}\sqrt x + \frac{3}{2}\sqrt y \frac{{dy}}{{dx}} = 0\,\,{\text{or, }}\frac{{dy}}{{dx}} = - \frac{{\sqrt x }}{{\sqrt y }}$$
Since the tangent is equally inclined to the axes,
$$\eqalign{ & \frac{{dy}}{{dx}} = \pm 1 \cr & \therefore \, - \frac{{\sqrt x }}{{\sqrt y }} = \pm 1{\text{ or }} - \frac{{\sqrt x }}{{\sqrt y }} = - 1 \cr & \therefore \,\sqrt x = \sqrt y \,\,\,\,\,\left[ {\because \,\sqrt x > 0,\,\sqrt y > 0} \right] \cr & {\text{Putting }}\sqrt y = \sqrt x {\text{ in equation}}\left( 1 \right){\text{,we get}} \cr & 2{x^{\frac{3}{2}}} = 2{a^{\frac{3}{2}}}{\text{ or }}{x^3} = {a^3} \cr & {\text{Therefore, }}x = a{\text{ and, so }}y = a \cr} $$

$$f'\left( x \right) = \frac{{\phi '\left( x \right)}}{{1 + {{\left\{ {\phi \left( x \right)} \right\}}^2}}} > 0{\text{ for }}0 < x < \frac{\pi }{2}$$         because $$\phi '\left( x \right) > 0,\,\phi \left( x \right)$$    being m.i.

For $$0 < x \leqslant \frac{\pi }{2}\,;\,\left[ {\cos \,x} \right] = 0$$
Hence, $$f\left( x \right) = 1$$   for all $$\left( {0,\,\frac{\pi }{2}} \right]$$
Trivially $$f\left( x \right)$$  is continuous on $$\left( {0,\,\frac{\pi }{2}} \right)$$
This function is neither strictly increasing nor strictly decreasing and its global maximum is $$1.$$

Let $$BP = x.$$
From similar triangle property, we get

$$\eqalign{ & \frac{{AO}}{{{l_1}}} = \frac{{{l_2}}}{x} \cr & {\text{or }}AO = \frac{{{l_1}{l_2}}}{x} \cr & {\text{or }}\frac{{d\left( {AO} \right)}}{{dt}} = \frac{{ - {l_1}{l_2}}}{{{x^2}}}\frac{{dx}}{{dt}} \cr & {\text{When, }}x = \frac{{{l_1}}}{2},\,\frac{{d\left( {AO} \right)}}{{dt}} = - \frac{{2{l_2}}}{5}\,m/s. \cr} $$

Two roots of the first equation are 0 and $$\alpha .$$ The second equation is the derived equation of the first. So, it has root in $$\left( {0,\,\alpha } \right).$$ .

$$\eqalign{ & f'\left( x \right) > 0 \cr & \Rightarrow 4{\sin ^3}x.\cos \,x - 4{\cos ^3}x.\sin \,x > 0 \cr & \Rightarrow \sin \,x.\cos \,x\left( {{{\sin }^2}x - {{\cos }^2}x} \right) > 0 \cr & {\text{or }}\sin \,2x.\cos \,2x < 0\,\,\,\,\,\,{\text{or }}\sin \,4x < 0 \cr & \therefore \pi < 4x < 2\pi \,\,\,\,{\text{or }}3\pi < 4x < 4\pi \cr & \therefore \frac{\pi }{4} < x < \frac{\pi }{2}\,\,\,{\text{or }}\frac{{3\pi }}{4} < x < \pi \cr} $$

$$y = {\log _e}x\,\,\,\, \Rightarrow \frac{{dy}}{{dx}} = \frac{1}{x}\,\,\,\,\,\,\therefore {\left. {\frac{{dy}}{{dx}}} \right)_{1,\,0}} = 1$$
$$\therefore $$ the equation of the normal at $$\left( {1,\,0} \right)$$  is
$$\eqalign{ & y - 0 = - \frac{1}{1}\left( {x - 1} \right){\text{ or }}x + y = 1 \cr & \therefore {\text{ area}} = \frac{1}{2}.1.1 = \frac{1}{2} \cr} $$

$$\eqalign{ & f''\left( x \right) = 6\left( {x - 1} \right).\,{\text{Inegrating,we}}\,{\text{get}} \cr & f'\left( x \right) = 3{x^2} - 6x + c \cr & {\text{Slope}}\,{\text{at}}\,\left( {2,1} \right) = f'\left( 2 \right) = c = 3\,\,\left[ {\because \,{\text{slope}}\,{\text{of}}\,{\text{tangent}}\,{\text{at}}\,\left( {2,1} \right)\,{\text{is}}\,3} \right] \cr & \therefore f'\left( x \right) = 3{x^2} - 6x + 3 = 3{\left( {x - 1} \right)^2} \cr & {\text{Inegrating}}\,{\text{again,we}}\,{\text{get}}\,f\left( x \right) = {\left( {x - 1} \right)^3} + D \cr & {\text{The}}\,{\text{curve}}\,{\text{passes}}\,{\text{through}}\,\left( {2,1} \right) \cr & \Rightarrow 1 = {\left( {2 - 1} \right)^3} + D \Rightarrow D = 0 \cr & \therefore f\left( x \right) = {\left( {x - 1} \right)^3} \cr} $$

We have $$e < \pi $$  and
$$\eqalign{ & f'\left( x \right) = \frac{{\frac{1}{{\pi + x}}\log \left( {e + x} \right) - \frac{1}{{e + x}}\log \left( {\pi + x} \right)}}{{{{\left\{ {\log \left( {e + x} \right)} \right\}}^2}}} \cr & = \frac{{\left( {e + x} \right)\log \left( {e + x} \right) - \left( {\pi + x} \right)\log \left( {\pi + x} \right)}}{{\left( {\pi + x} \right)\left( {e + x} \right){{\left\{ {\log \left( {e + x} \right)} \right\}}^2}}} \cr} $$
$${\text{In}}\left[ {0,\,\infty } \right),$$   denominator $$ > 0$$  and numerator $$ < 0,$$
Since, $$e + x = \pi + x.$$
Hence, $$f\left( x \right)$$  is decreasing in $$\left[ {0,\,\infty } \right).$$