Application of Derivatives MCQ Questions & Answers in Calculus | Maths

At $$x = 0,\,y = 1$$
Hence, the point at which normal is drawn is $$P\left( {0,\,1} \right).$$
Differentiating the given equation w.r.t. $$x,$$ we have
$$\eqalign{ & {\left( {1 + x} \right)^y}\left\{ {\log \left( {1 + x} \right)\frac{{dy}}{{dx}} + \frac{y}{{1 + x}}} \right\} - \frac{{dy}}{{dx}} + \frac{1}{{\sqrt {1 - {{\sin }^4}x} }}2\,\sin \,x\,\cos \,x = 0 \cr & {\text{or }}{\left( {\frac{{dy}}{{dx}}} \right)_{\left( {0,\, 1} \right)}} = \frac{{{{\left( {1 + 0} \right)}^1} \times \frac{1}{{1 + 0}} - \frac{{2\,\sin \,0}}{{\sqrt {1 - {{\sin }^2}0} }}}}{{1 - {{\left( {1 + 0} \right)}^1}\log \,1}} = 1 \cr} $$
$$\therefore $$  Slope of the normal $$=\, –1$$
Therefore, equation of the normal having slope $$–1$$ at point $$P\left( {0,\,1} \right)$$  is given by
$$y - 1 = - \left( {x - 0} \right){\text{ or }}x + y = 1$$

$$\eqalign{ & {\text{Since,}}\,\,y = \int\limits_0^x {\left| t \right|dt} ,x \in {\text{R}} \cr & {\text{therefore }}\frac{{dy}}{{dx}} = |x| \cr & {\text{But from }}y = 2x,\frac{{dy}}{{dx}} = 2 \cr & \Rightarrow |x| = 2 \Rightarrow x = \pm 2 \cr & {\text{Points }}y = \int\limits_0^{ \pm 2} {\left| t \right|dt} = \pm 2 \cr & \therefore {\text{ equation of tangent is}}\,y - 2 = 2\left( {x - 2} \right)\,{\text{or }}y + 2 = 2\left( {x + 2} \right) \cr & \Rightarrow x{\text{ - intercept }} = \pm 1. \cr} $$

$$f\left( x \right) = \left( {1 + {b^2}} \right){x^2} + 2bx + 1$$
It is a quadratic expression with coeff. of $${x^2} = 1 + {b^2} > 0.$$
$$\therefore f\left( x \right)$$   represents an upward parabola whose min value is $$\frac{{ - D}}{{4a}},D$$   being the discreminant.
$$\eqalign{ & \therefore m\left( b \right) = - \frac{{4{b^2} - 4\left( {1 + {b^2}} \right)}}{{4\left( {1 + {b^2}} \right)}} \Rightarrow m\left( b \right) = \frac{1}{{1 + {b^2}}} \cr & {\text{For range of}}\,m\left( b \right): \cr & \frac{1}{{1 + {b^2}}} > 0\,{\text{also}}\,{b^2} \geqslant 0 \Rightarrow 1 + {b^2} \geqslant 1 \cr & \Rightarrow \frac{1}{{1 + {b^2}}} \leqslant 1.\,{\text{Thus}}\,m\left( b \right) = \left( {0,1} \right] \cr} $$

$$\eqalign{ & f\left( x \right) = \int {{e^x}\left( {x - 1} \right)\left( {x - 2} \right)dx} \cr & {\text{For decreasing function, }}f'(x) < 0 \cr & \Rightarrow {e^x}(x - 1)(x - 2) < 0 \Rightarrow \,\,\,(x - 1)(x - 2) < 0 \cr & \Rightarrow 1 < x < 2,\quad \because {e^x} > 0\forall x \in R \cr} $$

$$\eqalign{ & {\text{Here,}}\,\,f'\left( 1 \right) = \tan \frac{\pi }{6} = \frac{1}{{\sqrt 3 }} \cr & f'\left( 2 \right) = \tan \frac{\pi }{3} = \sqrt 3 \cr & f'\left( 3 \right) = \tan \frac{\pi }{4} = 1 \cr & {\text{Now, }}\int_2^3 {f'\left( x \right).f''\left( x \right)dx = \left[ {\frac{1}{2}{{\left\{ {f'\left( x \right)} \right\}}^2}} \right]_2^3 = \frac{1}{2}\left[ {{{\left\{ {f'\left( 3 \right)} \right\}}^2} - {{\left\{ {f'\left( 2 \right)} \right\}}^2}} \right]} \cr & \int_1^3 {f''\left( x \right)dx} = \left[ {f'\left( x \right)} \right]_1^3 = f'\left( 3 \right) - f'\left( 1 \right) \cr & \therefore {\text{ value}}\, = \frac{1}{2}\left\{ {{1^2} - {{\left( {\sqrt 3 } \right)}^2}} \right\} + 1 - \frac{1}{{\sqrt 3 }} = - \frac{1}{{\sqrt 3 }} \cr & \cr} $$

$$\eqalign{ & {\text{Slope of tangent }}y = f\left( x \right){\text{ is }}\frac{{dy}}{{dx}} = f'{\left( x \right)_{\left( {3,4} \right)}} \cr & {\text{Therefore, slope of normal}} = - \frac{1}{{f'{{\left( x \right)}_{\left( {3,4} \right)}}}} = - \frac{1}{{f'\left( 3 \right)}} \cr & {\text{but }} - \frac{1}{{f'\left( 3 \right)}} = \tan \left( {\frac{{3\pi }}{4}} \right){\text{ }}\left( {{\text{given}}} \right) \cr & {\text{or}}\, - \frac{1}{{f'\left( 3 \right)}} = \tan \left( {\frac{\pi }{2} + \frac{\pi }{4}} \right) = - 1 \Rightarrow f'\left( 3 \right) = 1 \cr} $$

$$\eqalign{ & f\left( \theta \right) = a\sec \,\theta - b\tan \,\theta \cr & \therefore f'\left( \theta \right) = a\sec \,\theta .\tan \,\theta - b{\sec ^2}\theta \cr & \therefore f'\left( \theta \right) = 0 \Rightarrow \sec \,\theta \left( {a\tan \,\theta - b\sec \,\theta } \right) = 0 \cr & \Rightarrow \sec \,\theta = 0{\text{ or }}a\tan \,\theta = b\sec \,\theta \cr & {\text{But }}\sec \,\theta \ne 0 \cr & \therefore f'\left( \theta \right) = 0 \cr & \Rightarrow a\tan \,\theta = b\sec \,\theta \,\,\,\,\,\,\,\, \Rightarrow b = \frac{{a\tan \,\theta }}{{\sec \,\theta }} = a\sin \,\theta \cr & \therefore \sin \,\theta = \frac{b}{a} \cr & f''\left( \theta \right) = a\sec \,\theta .{\tan ^2}\theta + a{\sec ^3}\theta - 2b{\sec ^2}\theta .\tan \,\theta \cr & = a.\frac{a}{{\sqrt {{a^2} - {b^2}} }}.\frac{{{b^2}}}{{{{\left( {\sqrt {{a^2} - {b^2}} } \right)}^2}}} + a.{\left( {\frac{a}{{\sqrt {{a^2} - {b^2}} }}} \right)^3} - 2b{\left( {\frac{a}{{\sqrt {{a^2} - {b^2}} }}} \right)^2}.\frac{b}{{\sqrt {{a^2} - {b^2}} }} \cr & = \frac{1}{{{{\left( {{a^2} - {b^2}} \right)}^{\frac{3}{2}}}}}.\left\{ {{a^2}{b^2} + {a^4} - 2{a^2}{b^2}} \right\} \cr & = \frac{{{a^2}\left( {{a^2} - {b^2}} \right)}}{{{{\left( {{a^2} - {b^2}} \right)}^{\frac{3}{2}}}}} \cr & = \frac{{{a^2}}}{{\sqrt {{a^2} - {b^2}} }} > 0 \cr & \therefore \,\min f\left( \theta \right) = a.\frac{a}{{\sqrt {{a^2} - {b^2}} }} - b.\frac{b}{{\sqrt {{a^2} - {b^2}} }} \cr & = \frac{{{a^2} - {b^2}}}{{\sqrt {{a^2} - {b^2}} }} \cr & = \sqrt {{a^2} - {b^2}} \cr} $$

$$\eqalign{ & {\text{Differentiating w}}{\text{.r}}{\text{.t}}{\text{. }}x,{\text{ }}{e^y}{\text{.}}\frac{{dy}}{{dx}}{\text{ }} = 2x \cr & {\text{or,}}\,\,\frac{{dy}}{{dx}} = \frac{{2x}}{{1 + {x^2}}}\,\,\,\,\left( {\because {e^y} = 1 + {x^2}} \right) \cr & \therefore m = \frac{{2x}}{{1 + {x^2}}}{\text{ or }}\left| m \right| = \frac{{2\left| x \right|}}{{1 + {{\left| x \right|}^2}}} \cr & {\text{But }}1 + {\left| x \right|^2} - 2\left| x \right| = \left( {1 - {{\left| x \right|}^2}} \right) \geqslant 0 \cr & \therefore 1 + {\left| x \right|^2} \geqslant 2\left| x \right| \cr & \therefore \left| m \right| \leqslant 1 \cr} $$

Here, $$\frac{{dx}}{{dt}} = b{\text{ cm/s,}}\,\,\,{x^2} + {y^2} = {a^2}$$
Differentiating w.r.t. $$t,\,\,2x\frac{{dx}}{{dt}} + 2y\frac{{dy}}{{dt}} = 0$$
or $$2bx + 2y\frac{{dy}}{{dt}} = 0\,\,\,\,\, \Rightarrow \frac{{dy}}{{dt}} = - \frac{{bx}}{y}$$
The velocity of the middle point at time $$t$$
$$\eqalign{ & = \sqrt {{{\left\{ {\frac{{d\left( {\frac{x}{2}} \right)}}{{dt}}} \right\}}^2} + {{\left\{ {\frac{{d\left( {\frac{y}{2}} \right)}}{{dt}}} \right\}}^2}} \cr & = \frac{1}{2}\sqrt {{{\left( {\frac{{dx}}{{dt}}} \right)}^2} + {{\left( {\frac{{dy}}{{dt}}} \right)}^2}} \cr & = \frac{1}{2}\sqrt {{b^2} + \frac{{{b^2}{x^2}}}{{{y^2}}}} \cr} $$
When the stick is equally inclined to the wall and to the floor, $$x=y$$
$$\therefore $$ the required velocity $$ = \frac{1}{2}\sqrt {{b^2} + {b^2}} = \frac{b}{{\sqrt 2 }}$$

Since, function $$f\left( x \right)$$  have local extreme points at $$x = - 1,0,1.$$
Then
$$\eqalign{ & f\left( x \right) = K\left( {x + 1} \right)x\left( {x - 1} \right) = K\left( {{x^3} - x} \right) \cr & \Rightarrow f\left( x \right) = K\left( {\frac{{{x^4}}}{4} - \frac{{{x^2}}}{2}} \right) + C{\text{ (using integration)}} \cr & \Rightarrow f\left( 0 \right) = C \cr & \because f\left( x \right) = f\left( 0 \right) \Rightarrow \frac{{{x^2}}}{2}\left( {\frac{{{x^2}}}{2} - 1} \right) = 0 \cr & \Rightarrow \frac{{{x^2}}}{2}\left( {\frac{{{x^2}}}{2} - 1} \right) = 0 \Rightarrow x = 0,0,\sqrt 2 , - \sqrt 2 \cr & \therefore S = \left\{ {0, - \sqrt 2 ,\sqrt 2 } \right\} \cr} $$