Application of Derivatives MCQ Questions & Answers in Calculus | Maths

$$\eqalign{ & f'\left( x \right) = \frac{{\sin \,x - x\,\cos \,x}}{{{{\sin }^2}x}},\,\,g'\left( x \right) = \frac{{\tan \,x - x\,{{\sec }^{2\,}}x}}{{{{\tan }^2}x}} \cr & {\text{Now, }}\frac{d}{{dx}}\left( {\sin \,x - x\,\cos \,x} \right) = x\sin \,x > 0,\,{\text{when }}0 < x \leqslant 1 \cr} $$
$$\therefore \,\sin \,x - x\cos \,x$$    is an increasing function. But at $$x = 0,\,x\sin \,x{\text{ is }}0$$
$$\therefore \,\,{\text{in }}0 < x \leqslant 1,\,\sin \,x - x\cos \,x > 0$$
$$\therefore \,f'\left( x \right) > 0{\text{ for }}0 < x \leqslant 1$$
So, $$f\left( x \right)$$  is increasing in the interval.
Again, $$\frac{d}{{dx}}\left( {\tan \,x - x{{\sec }^2}x} \right) = - x.2{\sec ^2}x.\tan \,x < 0{\text{ for }}0 < x \leqslant 1$$
$$\therefore \,\,g\left( x \right)$$  is decreasing in $$0 < x \leqslant 1$$

$$\eqalign{ & {\text{Equation}}\,{\text{of}}\,{\text{tangent}}\,{\text{to}}\,{\text{the}}\,{\text{ellipse}}\,\frac{{{x^2}}}{{27}} + {y^2} = 1\,{\text{at}} \cr & \left( {3\sqrt 3 \cos \theta ,\sin \theta } \right),\theta \in \left( {0,\frac{\pi }{2}} \right)\,{\text{is}}\,\frac{{\sqrt 3 x\cos \theta }}{9} + y.\sin \theta = 1 \cr & \therefore {\text{Intercept}}\,{\text{on}}\,x - {\text{axis}}\, = \frac{9}{{\sqrt 3 \cos \theta }}; \cr & {\text{Intercept}}\,{\text{on}}\,y - {\text{axis}} = \frac{1}{{\sin \theta }} \cr & \therefore {\text{Sum}}\,{\text{of}}\,{\text{intercepts}} = S = 3\sqrt 3 \sec \theta + \operatorname{cosec} \theta \cr & {\text{For}}\,{\text{min}}{\text{.}}\,{\text{value}}\,{\text{of}}\,S,\frac{{dS}}{{d\theta }} = 0 \cr & \Rightarrow 3\sqrt 3 \sec \theta \tan \theta - \operatorname{cosec} \theta \cot \theta = 0 \cr & \Rightarrow \frac{{3\sqrt 3 \sin \theta }}{{{{\cos }^2}\theta }} - \frac{{\cos \theta }}{{{{\sin }^2}\theta }} = 0 \Rightarrow 3\sqrt 3 {\sin ^3}\theta - {\cos ^3}\theta = 0 \cr & \Rightarrow {\tan ^3}\theta = \frac{1}{{3\sqrt 3 }} = {\left( {\frac{1}{{\sqrt 3 }}} \right)^3} \cr & \Rightarrow \tan \theta = \frac{1}{{\sqrt 3 }} = \tan \frac{\pi }{6} \Rightarrow \theta = \frac{\pi }{6} \cr} $$

$$\eqalign{ & {h^2} + {r^2} = { \approx ^2} = 9\,......\left( 1 \right) \cr & {\text{Volume}}\,{\text{of}}\,{\text{cone}} \cr & V = \frac{1}{3}\pi {r^2}h\,......\left( 2 \right) \cr & {\text{From }}\left( {\text{1}} \right){\text{ and }}\left( {\text{2}} \right){\text{,}} \cr & \Rightarrow V = \frac{1}{3}\pi \left( {9 - {h^2}} \right)h \Rightarrow V = \frac{1}{3}\pi \left( {9h - {h^3}} \right) \cr & \Rightarrow \frac{{dv}}{{dh}} = \frac{1}{3}\pi \left( {9 - 3{h^2}} \right) \cr & {\text{For}}\,{\text{maxima/minima,}} \cr & \frac{{dV}}{{dh}} = 0 \Rightarrow \frac{1}{3}\pi \left( {9 - 3{h^2}} \right) = 0 \cr & \Rightarrow h = \pm \sqrt 3 \Rightarrow h = \sqrt 3 \cr & {\text{Now}}\,;\frac{{{d^2}V}}{{d{h^2}}} = \frac{1}{3}\pi \left( { - 6h} \right) \cr & {\text{Here}},\,{\left( {\frac{{{d^2}V}}{{d{h^2}}}} \right)_{{\text{at}}\,h = \sqrt 3 }} < 0 \cr & {\text{Then,}}\,h = \sqrt 3 \,{\text{is point of maxima}} \cr & {\text{Hence, the required maximum volume is,}} \cr & V = \frac{1}{3}\pi \left( {9 - 3} \right)\sqrt 3 = 2\sqrt 3 \pi \cr} $$

Surface area of sphere $$S = 4\pi {r^2}$$
Differentiate both sides w.r.t. $$t'$$
$$ \Rightarrow \frac{{dS}}{{dt}} = \frac{{8\pi rdr}}{{dt}}$$
and Volume $$ = V = \frac{4}{3}\pi {r^3}$$
$$\eqalign{ & \Rightarrow \frac{{dV}}{{dt}} = \frac{4}{3}\pi .3{r^2}\frac{{dr}}{{dt}} \cr & = 4\pi {r^2}\frac{{dr}}{{dt}} \cr & = \frac{{4\pi {r^2}}}{{8\pi r}}.\frac{{dS}}{{dt}} \cr & = \frac{1}{2}r\frac{{dS}}{{dt}} \cr} $$

$$3a{x^2} + 2bx + c = 0\,\,\,\,\, \Rightarrow a{x^3} + b{x^2} + cx = 0\,\,\left( {{\text{on integration}}} \right)$$
Clearly $$x=0,\,2$$   are the roots of $$a{x^3} + b{x^2} + cx = 0\,\,\,\,\,\left( {\because 4a + 2b + c = 0} \right)$$
$$\therefore $$ the derived equation $$3a{x^2} + 2bx + c = 0$$     will have a root lying between 0 and 2.

Eliminating $$t$$ gives $${y^2}\left( {x - 1} \right) = 1$$
Equation of the tangent at $$P\left( {2,\,1} \right)$$  is $$x + 2y = 4$$
Solving with curve $$x = 5$$  and $$y = - \frac{1}{2},$$  we get
$$Q \equiv \left( {5,\, - \frac{1}{2}} \right){\text{ or }}PQ = \frac{{3\sqrt 5 }}{2}$$

$$\eqalign{ & \frac{{dy}}{{dx}} = \frac{{\frac{{dy}}{{dt}}}}{{\frac{{dx}}{{dt}}}} = \frac{{{e^t}\left( {\cos \,t - \sin \,t} \right)}}{{{e^t}\left( {\sin \,t + \cos \,t} \right)}} \cr & \therefore {\left. {\frac{{dy}}{{dx}}} \right)_{t = \frac{\pi }{4}}} = \frac{{\frac{1}{{\sqrt 2 }} - \frac{1}{{\sqrt 2 }}}}{{\frac{1}{{\sqrt 2 }} + \frac{1}{{\sqrt 2 }}}} = 0 \cr} $$

$$\eqalign{ & {\text{Differentiating w}}{\text{.r}}{\text{.t}}{\text{. }}\,x,\,1 + \frac{{dy}}{{dx}} = {e^{xy}}\left( {y + x\frac{{dy}}{{dx}}} \right) \cr & {\text{or }}\frac{{dy}}{{dx}} = \frac{{y{e^{xy}} - 1}}{{1 - x{e^{xy}}}} \cr & \frac{{dy}}{{dx}} = \infty \,\,\,\,\,\,\,\, \Rightarrow 1 - x{e^{xy}} = 0\,\,\,\,\,\,\, \Rightarrow 1 - x\left( {x + y} \right) = 0 \cr & {\text{This holds for }}x = 1,{\text{ }}y = 0. \cr} $$

$$\eqalign{ & y = f\left( x \right) = {x^3} - {x^2} - 2x \cr & \Rightarrow \frac{{dy}}{{dx}} = 3{x^2} - 2x - 2 \cr & f\left( 1 \right) = 1 - 1 - 2 = - 2,{\text{ }}f\left( { - 1} \right) = - 1 - 1 + 2 = 0 \cr} $$
Since the tangent to the curve is parallel to the line segment joining the points (1, -2) and (-1, 0) And their slopes are equal.
$$ \Rightarrow 3{x^2} - 2x - 2 = \frac{{ - 2 - 0}}{2} \Rightarrow x = 1,\frac{{ - 1}}{3}$$
Hence, the required set $$S = \left\{ {\frac{{ - 1}}{3},1} \right\}$$

By Lagrange’s theorem, $$\frac{{f\left( 1 \right) - f\left( { - 1} \right)}}{{1 - \left( { - 1} \right)}} = f'\left( c \right)$$     i.e., $$0 = f'\left( c \right),$$   where $$ - 1 < c < 1$$
If $$x > 0,\,f\left( x \right) = x$$     and, therefore, $$f'\left( x \right) = 1,$$   and
if $$x < 0,\,f\left( x \right) = - x$$     and, therefore, $$f'\left( x \right) = - 1$$
Hence, there is no $$c \ne 0$$  for which $$0 = f'\left( c \right)$$   holds.
Also at $$x = 0,\,f\left( x \right)$$   is not differentiable. Hence, $$c$$ has no value.
Note : $$f\left( x \right)$$  is not differentiable in $$\left[ { - 1,\,1} \right].$$   So, Lagrange’s theorem fails for the interval.