Application of Derivatives MCQ Questions & Answers in Calculus | Maths

$$f'\left( x \right) = 4\left( {{x^2} - 1} \right).$$    So, for $$0 \leqslant x < 1,$$   we get $$f'\left( x \right) > 0$$   i.e., $$f\left( x \right)$$  is $$m.d.$$  and for $$1 < x \leqslant 2$$   we get $$f'\left( x \right) > 0,$$   i.e., $$f\left( x \right)$$  is $$m.i.$$
$$\therefore \,\min f\left( x \right) = f\left( 1 \right) = \frac{4}{3} - 4 = - \frac{8}{3}$$

$$f'\left( x \right) = \frac{1}{{1 + {x^2}}} - 1 = \frac{{ - {x^2}}}{{1 + {x^2}}} < 0{\text{ for all }}x \ne 0$$

$$\eqalign{ & {\text{Given}}\,{\text{that}}\,g\left( u \right) = 2{\tan ^{ - 1}}\left( {{e^u}} \right) - \frac{\pi }{2} \cr & \therefore g\left( { - u} \right) = 2{\tan ^{ - 1}}\left( {{e^{ - u}}} \right) - \frac{\pi }{2} = 2{\tan ^{ - 1}}\left( {\frac{1}{{{e^u}}}} \right) - \frac{\pi }{2} \cr & = 2{\cot ^{ - 1}}\left( {{e^u}} \right) - \frac{\pi }{2} = 2\left[ {\frac{\pi }{2} - {{\tan }^{ - 1}}\left( {{e^u}} \right)} \right] - \frac{\pi }{2} \cr & = \pi - 2{\tan ^{ - 1}}\left( {{e^u}} \right) - \frac{\pi }{2} = \frac{\pi }{2} - 2{\tan ^{ - 1}}\left( {{e^u}} \right) \cr & = - g\left( u \right)\,\therefore g\,{\text{is}}\,{\text{an}}\,{\text{odd}}\,{\text{function}}{\text{.}} \cr & {\text{Also}}\,g'\left( u \right) = \frac{{2{e^u}}}{{1 + {e^{2u}}}} > 0,\forall u \in \left( { - \infty ,\infty } \right) \cr & \therefore g\,{\text{is}}\,{\text{strictly}}\,{\text{increasing}}\,{\text{on}}\,\left( { - \infty ,\infty } \right) \cr} $$

$$\eqalign{ & {\text{Let }}{z^2} = {\left( {x - 0} \right)^2} + {\left( {y - 3} \right)^2} \cr & = {x^2} + {y^2} - 6y + 9 = {y^2} - 4y + 9\,\,\,\left( {\because {x^2} = 2y} \right) \cr & \therefore \,\,\frac{{d\left( {{z^2}} \right)}}{{dy}} = 2y - 4 \cr & \therefore \,\,\frac{{d\left( {{z^2}} \right)}}{{dy}} = 0\,\,\,\, \Rightarrow y = 2{\text{ and then }}x = 2,\, - 2 \cr} $$
As the question of a point being farthest does not arise (being $$\infty $$), we find $$\left( {2,\,2} \right),$$  as well as $$\left( { - 2,\,2} \right)$$  are the nearest to the point $$\left( {0,\,3} \right).$$

Since tangent is parallel to $$x$$-axis,
$$\eqalign{ & \therefore \,\,\frac{{dy}}{{dx}} = 0 \Rightarrow 1 - \frac{8}{{{x^3}}} = 0 \Rightarrow x = 2 \Rightarrow y = 3 \cr & {\text{Equation of tangent is }}y - 3 = 0\left( {x - 2} \right){\text{ }} \Rightarrow y = 3 \cr} $$

Let $$f\left( x \right) = p\,\sin \,x + \frac{{\sin \,3x}}{3}$$
Differentiate both side w.r.t $$\left( x \right)$$
$$ \Rightarrow f'\left( x \right) = p\,\cos \,x + \frac{{3\,\cos \,3x}}{3} = p\,\cos \,x + \cos \,3x$$
It is given that $$f\left( x \right)$$  has extremum value at $$x = \frac{\pi }{3}$$
$$\eqalign{ & \therefore f'\left( {\frac{\pi }{3}} \right) = 0 \cr & \Rightarrow p\,\cos \,\frac{\pi }{3} + \cos \,\pi = 0 \cr & \Rightarrow \frac{p}{2} - 1 = 0 \cr & \Rightarrow p = 2 \cr} $$

$$\eqalign{ & {\text{we}}\,{\text{have}}\,f:R \to \left[ { - \frac{1}{2},\frac{1}{2}} \right] \cr & f\left( x \right) = \frac{x}{{1 + {x^2}}},\forall x \in R \cr & \Rightarrow \,f'\left( x \right) = \frac{{\left( {1 + {x^2}} \right).1 - x.2x}}{{{{\left( {1 + {x^2}} \right)}^2}}} = \frac{{ - \left( {x + 1} \right)\left( {x - 1} \right)}}{{{{\left( {1 + {x^2}} \right)}^2}}} \cr} $$

⇒ $$f'\left( x \right)$$  changes sign in different intervals.
$$\therefore $$ Not injective
$$\eqalign{ & {\text{Now}}\,y = \frac{x}{{1 + {x^2}}} \cr & \Rightarrow y + y{x^2} = x \cr & \Rightarrow y{x^2} - x + y = 0 \cr & {\text{For}}\,y \ne 0,\,{\text{D}} = 1 - 4{y^2} \geqslant 0 \cr & \Rightarrow y \in \left[ {\frac{{ - 1}}{2},\frac{1}{2}} \right] - \left\{ 0 \right\} \cr & {\text{For}}\,y = 0 \Rightarrow x = 0 \cr & \therefore \,\,{\text{Range}}\,{\text{is}}\,\left[ {\frac{{ - 1}}{2},\frac{1}{2}} \right] \cr & \Rightarrow \,{\text{Surjective}}\,{\text{but}}\,{\text{not}}\,{\text{injective}} \cr} $$

$$\eqalign{ & \frac{{dV}}{{dt}} = \frac{{d\left( {\frac{4}{3}\pi {r^3}} \right)}}{{dt}} = 4\pi {r^2}\frac{{dr}}{{dt}} = a\,\,\,\left( {{\text{given}}} \right) \cr & \therefore \frac{{dr}}{{dt}} = \frac{a}{{4\pi {r^2}}} \cr & \frac{{dS}}{{dt}} = \frac{{d\left( {4\pi {r^2}} \right)}}{{dt}} = 8\pi r.\frac{{dr}}{{dt}} = 8\pi r.\frac{a}{{4\pi {r^2}}} = \frac{{2a}}{r} \cr & \therefore \,{\left. {\frac{{dS}}{{dt}}} \right)_{r = b}} = \frac{{2a}}{b} \cr} $$

$$\eqalign{ & {\text{Give}}\,{\text{that}}\,{p^2} + {q^2} = 1\,\,\,\,\therefore p = \cos \theta \,{\text{and}}\,q = \sin \theta \cr & {\text{Then}}\,p + q = \cos \theta + \sin \theta \cr & {\text{we}}\,{\text{know}}\,{\text{that}} \cr & - \sqrt {{a^2} + {b^2}} \leqslant a\cos \theta + b\sin \theta \leqslant \sqrt {{a^2} + {b^2}} \cr & \therefore - \sqrt 2 \leqslant \cos \theta + \sin \theta \leqslant \sqrt 2 \cr & {\text{Hence}}\,{\text{max}}{\text{.value}}\,{\text{of}}\,\,p + q\,\,{\text{is}}\,\sqrt 2 \cr} $$

$$\eqalign{ & {\text{Let }}f\left( x \right) = 4\alpha {x^2} + \frac{1}{x} \cr & {\text{For }}x > 0,{f_{\min }} = 1 \cr & f'\left( x \right) = 8\alpha x - \frac{1}{{{x^2}}} = 0 \Rightarrow x = \frac{1}{{2{\alpha ^{\frac{1}{3}}}}} \cr & {f_{\min }} = 1 \Rightarrow 4\alpha {\left( {\frac{1}{{2{\alpha ^{\frac{1}{3}}}}}} \right)^2} + 2{\alpha ^{\frac{1}{3}}} = 1 \cr & \Rightarrow 3{\alpha ^{\frac{1}{3}}} = 1{\text{ or }}\alpha = \frac{1}{{27}} \cr} $$