Application of Derivatives MCQ Questions & Answers in Calculus | Maths

$$y = x - {x^2}$$   where $$y$$ is the greatest difference.
$$\eqalign{ & \frac{{dy}}{{dx}} = 1 - 2x \cr & \therefore \frac{{dy}}{{dx}} = 0 \Rightarrow x = \frac{1}{2} \cr} $$
Again, $$\frac{{{d^2}y}}{{d{x^2}}} = - 2$$
$$\therefore \,y$$   is the maximum at $$x = \frac{1}{2}$$

$$\eqalign{ & f\left( x \right) = 3{x^2} - 2\left( {p + q + r} \right)x + {p^2} + {q^2} + {r^2} \cr & \therefore \,f'\left( x \right) = 6x - 2\left( {p + q + r} \right) \cr & \therefore \,f'\left( x \right) = 0 \cr & \Rightarrow x = \frac{{p + q + r}}{3} = {\text{AM of }}p,\,q,\,r \cr & {\text{Also, }}f''\left( x \right) = 6 > 0 \cr} $$

Here $$y > 0.$$  Putting $$y=x$$  in $$y = \sqrt {4 - {x^2}} ,$$   we get $$x = \sqrt 2 ,\, - \sqrt 2 $$
So, the point is $$\left( {\sqrt 2 ,\,\sqrt 2 } \right)$$
Differentiating $${y^2} + {x^2} = 4$$   w.r.t. $$x,$$
$$\eqalign{ & 2y\frac{{dy}}{{dx}} + 2x = 0\,\,\,\,\,\,\,{\text{or, }}\frac{{dy}}{{dx}} = - \frac{x}{y} \cr & \therefore \,{\text{ at }}\left( {\sqrt 2 ,\,\sqrt 2 } \right),\,\frac{{dy}}{{dx}} = - 1 \cr} $$

$$\eqalign{ & f\left( x \right) = \cos \,\pi x + 10x + 3{x^2} + {x^3} \cr & f'\left( x \right) = - x\,\sin \,\pi x + 10 + 6x + 3{x^2} \cr & f'\left( x \right) = 3{x^2} + 6x = 10 - x\,\sin \,\pi x \cr & f'\left( x \right) = 3{\left( {{x^2} + 1} \right)^2} + 7 - \pi \,\sin \,\pi x \cr & f'\left( x \right) > 0 \cr & f\left( x \right){\text{ is an }} \uparrow {\text{ function}} \cr & f\left( { - 2} \right) = \cos \left( { - 2,\,\pi } \right) + \left( { - 20} \right) + 12 + \left( { - 8} \right) \cr & = 1 - 20 + 12 - 8 \cr & = - 15 \cr} $$

$$\eqalign{ & \frac{{dy}}{{dx}} = 3{x^2} - 2ax + 1 \cr & {\text{Given that }}\frac{{dy}}{{dx}} \geqslant 0 \cr & {\text{or }}3{x^2} - 2ax + 1 \geqslant 0{\text{ for all }}x \cr & {\text{or }}D \leqslant 0 \cr & {\text{or }}4{a^2} - 12 \leqslant 0 \cr & {\text{or }} - \sqrt 3 \leqslant a \leqslant \sqrt 3 \cr} $$

We have, $${y^2} = 2ax......\left( {\text{i}} \right)$$
Put $$x = \frac{a}{2}\,;\,{y^2} = 2a\left( {\frac{a}{2}} \right) \Rightarrow y = \pm a$$
$$\therefore $$  The points are $$\left( {\frac{a}{2},\,a} \right)$$  and $$\left( {\frac{a}{2},\, - a} \right)$$
Differentiating (i) with respect to $$x,$$ we get
$$2y\frac{{dy}}{{dx}} = 2a \Rightarrow \frac{{dy}}{{dx}} = \frac{a}{y}$$
At $$\left( {\frac{a}{2},\,a} \right)\,;\,\frac{{dy}}{{dx}} = \frac{a}{y} = \frac{a}{a} = 1 = {m_1}\,\,\left( {{\text{say}}} \right)$$
At $$\left( {\frac{a}{2},\, - a} \right)\,;\,\frac{{dy}}{{dx}} = \frac{a}{y} = \frac{a}{{ - a}} = - 1 = {m_2}\,\,\left( {{\text{say}}} \right)$$
Since $${m_1}{m_2} = - 1,$$   the two tangents are at right angles.

$$\eqalign{ & {\text{Let }}y = \cos \,x\,\,\,\,\,\,\therefore \frac{{dy}}{{dx}} = - \sin \,x \cr & {\text{Now,}}\,\,\cos \,{60^ \circ }1' = \cos \,{60^ \circ } + \delta y \cr & {\left. {\delta y = \frac{{dy}}{{dt}}} \right)_{x = {{60}^ \circ }}}.\,\delta x = - \frac{{\sqrt 3 }}{2}.\,\delta x = - \frac{{\sqrt 3 }}{2}.1' = \frac{{\sqrt 3 }}{2} \times \frac{\alpha }{{60}} \cr & \therefore \,\,\,\,\cos \,{60^ \circ }1' = \frac{1}{2} - \frac{{\alpha \sqrt 3 }}{{120}} \cr} $$

$$\eqalign{ & {\text{Substituting }}y = 1{\text{ in }}\frac{{{x^2}}}{{16}} + \frac{{{y^2}}}{9} = 1 \cr & {\text{We get }}x = \pm \frac{{8\sqrt 2 }}{3} \cr & {\text{And slope of tangents}} \cr & \frac{{dy}}{{dx}} = \frac{{9x}}{{16y}} = {m_1},\,{m_2} = \pm \frac{{3\sqrt 2 }}{2} \cr & {\text{Therefore }}\tan \,\left( \theta \right) = \frac{{{m_1} - {m_2}}}{{1 + {m_1}{m_2}}} = \left| { - \frac{{6\sqrt 2 }}{7}} \right| \cr & \Rightarrow \theta = {\tan ^{ - 1}}\left( {\frac{{6\sqrt 2 }}{7}} \right) \cr} $$

Let $$CD$$  be the position of man at any time $$t.$$
Let $$BD$$  be $$x.$$
Then $$EC = x.$$
Let $$\angle ACE$$  be $$\theta $$
Given $$AB = 41.6\,m,\,CD = 1.6\,m,{\text{ and }}\frac{{dx}}{{dt}} = 2{\text{ }}m/s.$$

$$AE = AB - EB = AB - CD = 41.6 - 1.6 = 40\,m$$
We have to find $$\frac{{d\theta }}{{dt}}$$  when $$x = 30\,m$$
From $$\Delta AEC,\,\tan \,\theta = \frac{{AE}}{{EC}} = \frac{{40}}{x}$$
Differentiating w.r.t. to $$t,$$
$$\eqalign{ & {\sec ^2}\theta \frac{{d\theta }}{{dt}} = \frac{{ - 40}}{{{x^2}}}\frac{{dx}}{{dt}} \cr & {\text{or }}{\sec ^2}\theta \frac{{d\theta }}{{dt}} = \frac{{ - 40}}{{{x^2}}} \times 2 \cr & {\text{or }}\frac{{d\theta }}{{dt}} = \frac{{ - 80}}{{{x^2}}}{\cos ^2}\theta \cr & {\text{or }}\frac{{d\theta }}{{dt}} = - \frac{{80}}{{{x^2}}}\frac{{{x^2}}}{{{x^2} + {{40}^2}}} \cr & {\text{or }}\frac{{d\theta }}{{dt}} = - \frac{{80}}{{{x^2} + {{40}^2}}} \cr} $$
When $$x = 30\,m,\,\,\,\frac{{d\theta }}{{dt}} = - \frac{{80}}{{{{30}^2} + {{40}^2}}} = - \frac{4}{{125}}\,rad/s.$$

$$\eqalign{ & y = {\sin ^{ - 1}}\left( {{{\sin }^2}x} \right) \cr & \frac{{dy}}{{dx}} = \frac{{2\,\sin \,x\,\cos \,x}}{{\sqrt {1 - {{\sin }^4}x} }} \cr & \Rightarrow \frac{{dy}}{{dx}} = \frac{{\sin \,2x}}{{\sqrt {1 - {{\sin }^4}x} }} \cr & {\text{at }}x = 0,\,\,\frac{{dy}}{{dx}} = 0 \cr} $$