Application of Derivatives MCQ Questions & Answers in Calculus | Maths

Differentiating w.r.t.$$x,\,\frac{{n{x^{n - 1}}}}{{{a^n}}} + \frac{{n{y^{n - 1}}}}{{{b^n}}}.\frac{{dy}}{{dx}} = 0$$
$${\text{or }}\frac{{dy}}{{dx}} = - \frac{{{b^n}{x^{n - 1}}}}{{{a^n}{y^{n - 1}}}}$$
$$\therefore $$ The equation of the tangent at $$\left( {\alpha ,\,\beta } \right)$$  is
$$\eqalign{ & y - \beta = - \frac{{{b^n}{\alpha ^{n - 1}}}}{{{a^n}{\beta ^{n - 1}}}}\left( {x - \alpha } \right) \cr & {\text{or }}{a^n}{\beta ^{n - 1}}y + {b^n}{\alpha ^{n - 1}}x = {a^n}{\beta ^n} + {b^n}{\alpha ^n} \cr} $$
But $$\left( {\alpha ,\,\beta } \right)$$  is on the curve.
$$\therefore \,{a^n}{\beta ^{n - 1}}y + {b^n}{\alpha ^{n - 1}}x = {a^n}{b^n}.2......(1)$$
The line touches the curve at $$\left( {\alpha ,\,\beta } \right)$$  if (1) and $$\frac{x}{a} + \frac{y}{b} = 2$$   are identical.
$$\eqalign{ & \therefore \frac{{{b^n}{\alpha ^{n - 1}}}}{{\frac{1}{a}}} = \frac{{{a^n}{\beta ^{n - 1}}}}{{\frac{1}{b}}} = \frac{{2{a^n}{b^n}}}{2} \cr & \Rightarrow {\alpha ^{n - 1}} = {a^{n - 1}},\,{\beta ^{n - 1}} = {b^{n - 1}} \cr & \therefore \,\left( {\alpha ,\,\beta } \right) = \left( {a,\,b} \right) \cr} $$

$$\eqalign{ & \frac{{dx}}{{d\theta }} = - a\sin \theta \,{\text{and}}\,\frac{{dy}}{{d\theta }} = a\cos \theta \cr & \therefore \frac{{dy}}{{dx}} = - \cot \theta \cr & \therefore {\text{The slope of the normal}}\,{\text{at}}\,\theta = \tan \theta \cr & \therefore {\text{The equation of the normal at}}\,\theta \,{\text{is}} \cr & y - a\sin \theta = \tan \theta \left( {x - a - a\cos \theta } \right) \cr & \Rightarrow y\cos \theta - a\sin \theta \cos \theta = x\sin \theta - a\sin \theta - a\sin \theta \cos \theta \cr & \Rightarrow x\sin \theta - y\cos \theta = a\sin \theta \cr & \Rightarrow y = \left( {x - a} \right)\tan \theta \cr & {\text{which always passes through}}\,\left( {a,0} \right) \cr} $$

For $${y^2} = 4ax,\,y $$    - axis is tangent at (0, 0), while for $${x^2} = 4ay,\,x $$    - axis is tangent at (0, 0). Thus the two curves cut each other at right angles.

$$\eqalign{ & \frac{{dy}}{{dx}} = \frac{{{x^2}}}{{\sqrt {{x^2} + 1} }} \cr & \therefore \,\,{\left( {\frac{{dy}}{{dx}}} \right)_{x = 1}} = \frac{1}{{\sqrt {1 + 1} }} = \frac{1}{{\sqrt 2 }} \cr} $$

$$\eqalign{ & y = \frac{2}{3}{x^3} + \frac{1}{2}{x^2} \cr & \therefore \,\frac{{dy}}{{dx}} = \frac{2}{3}3{x^2} + \frac{1}{2}2x \cr & = 2{x^2} + x \cr} $$
Since the tangent makes equal angles with the axes.
$$\eqalign{ & \therefore \,\frac{{dy}}{{dx}} = \pm 1 \cr & {\text{or 2}}{x^2} + x = \pm 1 \cr & {\text{or 2}}{x^2} + x - 1 = 0\,\,\,\,\,\left( {{\text{2}}{x^2} + x + 1 = 0{\text{ has no real roots}}} \right) \cr & {\text{or }}\left( {2x - 1} \right)\left( {x + 1} \right) = 0 \cr & {\text{i}}{\text{.e}}{\text{., }}x = \frac{1}{2}{\text{ or }}x = - 1 \cr} $$

Let the speed of the train be $$v$$ and distance to be covered be $$s$$ so that total time taken is $$\frac{s}{v}$$ hours. Cost of fuel per hour $$ = k{v^2}$$   ($$k$$ is constant)
Also $$48 = k.\,{16^2}$$   by given condition
$$\therefore \,k = \frac{3}{{16}}$$
$$\therefore $$  Cost to fuel per hour $$\frac{3}{{16}}{v^2}$$
Other charges per hour are $$300$$
Total running cost,
$$\eqalign{ & C = \left( {\frac{3}{{16}}{v^2} + 300} \right)\frac{s}{v} = \frac{{3s}}{{16}}v + \frac{{300s}}{v} \cr & \frac{{dC}}{{dv}} = \frac{{3s}}{{16}} - \frac{{300s}}{{{v^2}}} = 0 \Rightarrow v = 40 \cr & \frac{{{d^2}C}}{{d{v^2}}} = \frac{{600s}}{{{v^3}}} > 0 \cr} $$
$$\therefore \,v = 40$$   results in minimum running cost.

$$\eqalign{ & f'\left( x \right) = ab - {b^2} - 2 + {\cos ^4}x + {\sin ^4}x < 0 \cr & \Rightarrow ab - {b^2} - 2 + {\left( {{{\cos }^2}x + {{\sin }^2}x} \right)^2} - 2{\sin ^2}x.\,{\cos ^2}x < 0 \cr & {\text{or, }}ab - {b^2} - 1 < \frac{1}{2}{\sin ^2}2x < \frac{1}{2} \cr & {\text{or, }}2ab - 2{b^2} - 2 < 1 \cr & {\text{or, }}2{b^2} - 2ab + 3 > 0 \cr} $$
This is true for any $$b\, \in \,R$$   if $$D < 0,$$  that is, $$4{a^2} - 4.2.3 < 0$$
$${\text{or, }}{a^2} < 6\,\,\,\,\,{\text{or, }} - \sqrt 6 < a < \sqrt 6 $$

$$\eqalign{ & f'\left( x \right) = 3{x^2} + 2\lambda x + 5 + 2\cos \,2x \leqslant 3{x^2} + 2\lambda x + 7\,\,\,\,\left( {\because \max \,\cos \,2x = 1} \right) \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \geqslant 3{x^2} + 2\lambda x + 3\,\,\,\left( {\because \min \,\cos \,2x = - 1} \right) \cr & {\text{But }}3{x^2} + 2\lambda x + 7 < 0{\text{ is not true for all}}\;x\, \in \,R \cr & \therefore \,f'\left( x \right) \geqslant 3{x^2} + 2\lambda x + 3 > 0{\text{ for all }}x{\text{ if }}D < 0,{\text{ i}}{\text{.e}}{\text{., }}4{\lambda ^2} - 4.3.3 < 0 \cr & {\text{or, }}{\lambda ^2} - 9 < 0\,\,\,\, \Rightarrow - 3 < \lambda < 3 \cr} $$
$$\therefore $$ if $$ - 3 < \lambda < 3,\,f\left( x \right)$$     is strictly $$m.i.$$  and so $$f\left( x \right)$$  is invertible.

At time $$t,$$ ball drops $$16{t^2}\,ft$$   distance. Therefore, $$y = 50 - 16{t^2}.....\left( 1 \right)$$
Point $$A$$ is the position of the falling ball at some time $$t.$$ So, $$\frac{{dy}}{{dt}} = - 32t$$
From the figure, $$\tan \,\theta = \frac{y}{x} = \frac{{50}}{{30 + x}}{\text{ or }}y = \left( {\frac{{50}}{{30 + x}}} \right).x$$
$$\eqalign{ & \therefore \,\frac{{dy}}{{dt}} = \frac{d}{{dt}}\left( {\frac{{50}}{{30 + x}}} \right) = \frac{{1500}}{{{{\left( {30 + x} \right)}^2}}}.\frac{{dx}}{{dt}} \cr & {\text{or }}\frac{{dx}}{{dt}} = \frac{{\left( {30 + {x^2}} \right)}}{{1500}}\left( { - 32t} \right) \cr & {\text{When }}f = \frac{1}{2},\,y = 46\,\,\,\,\,\,\,\left[ {{\text{using}}\left( 1 \right)} \right] \cr & {\text{and }}x = 345\,\,\,\,\,\,\,\left[ {{\text{using}}\left( 2 \right)} \right] \cr & \therefore \,\frac{{dx}}{{dt}} = - 16\frac{{{{\left( {375} \right)}^2}}}{{1500}} = - 1500\,{\text{ft/s}} \cr} $$

$$\eqalign{ & f'\left( x \right) > 0{\text{ if }}x \geqslant 0{\text{ and }}g'\left( x \right) < 0{\text{ if }}x \geqslant 0 \cr & {\text{Let }}h\left( x \right) = f\left( {g\left( x \right)} \right){\text{ then}} \cr & h'\left( x \right) = f'\left( {g\left( x \right)} \right).g'\left( x \right) < 0{\text{ if }}x \geqslant 0 \cr & \therefore \,h\left( x \right){\text{is decreasing function}} \cr & \therefore \,h\left( x \right) \leqslant h\left( 0 \right){\text{ if }}x \geqslant 0 \cr & \therefore \,f\left( {g\left( x \right)} \right) \leqslant f\left( {g\left( 0 \right)} \right) = 0 \cr & {\text{But codomain of each function is }}\left[ {0,\,\infty } \right) \cr & \therefore \,f\left( {g\left( x \right)} \right) = 0{\text{ for all }}x \geqslant 0 \cr & \therefore \,f\left( {g\left( x \right)} \right) = 0 \cr & {\text{Also }}g\left( {f\left( x \right)} \right) \leqslant g\left( {f\left( 0 \right)} \right){\text{ }}\left[ {{\text{as above}}} \right] \cr} $$