Application of Derivatives MCQ Questions & Answers in Calculus | Maths

Given velocity is $$v = s + 1$$
Since, velocity $$ = \frac{{ds}}{{dt}}$$
$$\therefore \,\frac{{ds}}{{dt}} = s + 1 \Rightarrow \frac{{ds}}{{s + 1}} = dt$$
Integrate both side we get
$$\eqalign{ & \log \left( {s + 1} \right) = t \cr & {\text{At }}s = 9m, \cr & t = \log \left( {10} \right){\text{second}} \cr} $$

$$\eqalign{ & {\text{Tangent}}\,{\text{to}}\,y = {x^2} + bx - b\,{\text{at}}\,\left( {1,1} \right)\,{\text{is}}\,y - 1 = \left( {2 + b} \right)\left( {x - 1} \right) \Rightarrow \left( {b + 2} \right)x - y = b + 1 \cr & x - {\text{intercept}} = \frac{{b + 1}}{{b + 2}}\,{\text{and}}\,y - {\text{intercept}} = - \left( {b + 1} \right) \cr & {\text{Given}}\,Ar\left( \Delta \right) = 2 \Rightarrow \frac{1}{2}\left( {\frac{{b + 1}}{{b + 2}}} \right)\left[ { - \left( {b + 1} \right)} \right] = 2 \cr & \Rightarrow {b^2} + 2b + 1 = - 4\left( {b + 2} \right) \Rightarrow {b^2} + 6b + 9 = 0 \cr & \Rightarrow {\left( {b + 3} \right)^2} = 0 \Rightarrow b = - 3 \cr} $$

$$\eqalign{ & {\text{The}}\,{\text{given}}\,{\text{curve}}\,{\text{is}}\,{y^3} + 3{x^2} = 12y \cr & \Rightarrow 3{y^2}\frac{{dy}}{{dx}} + 6x = 12\frac{{dy}}{{dx}} \Rightarrow \quad \frac{{dy}}{{dx}} = \frac{{2x}}{{4 - {y^2}}} \cr & {\text{For}}\,{\text{vertical}}\,{\text{tangents}}\,\frac{{dy}}{{dx}} = \frac{1}{0} \Rightarrow 4 - {y^2} = 0 \Rightarrow y = \pm 2 \cr & {\text{For}}\,y = 2,{x^2} = \frac{{24 - 8}}{3} = \frac{{16}}{3} \Rightarrow x = \pm \frac{4}{{\sqrt 3 }} \cr & {\text{For}}\,y = - 2,{x^2} = \frac{{ - 24 + 8}}{3} = - ve\,\left( {{\text{not}}\,{\text{possible}}} \right) \cr & \therefore {\text{Req}}{\text{.}}\,{\text{points}}\,{\text{are}}\,\left( { \pm \frac{4}{{\sqrt 3 }},2} \right) \cr} $$

$$\eqalign{ & g'\left( x \right) = xf'\left( {2{x^2} - 1} \right) - xf'\left( {1 - {x^2}} \right) \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = x\left( {f'\left( {2{x^2} - 1} \right) - f'\left( {1 - {x^2}} \right)} \right) \cr & g'\left( x \right) > 0 \cr} $$
$${\text{If }}x > 0,\,2{x^2} - 1 > 1 - {x^2}$$       (as $$f'$$ is an increasing function)
$$\eqalign{ & {\text{or }}3{x^2} > 2 \cr & {\text{or }}x\, \in \left( { - \infty ,\, - \sqrt {\frac{2}{3}} } \right) \cup \left( {\sqrt {\frac{2}{3}} ,\,\infty } \right) \cr & {\text{or }}x\, \in \left( {\sqrt {\frac{2}{3}} ,\,\infty } \right) \cr & {\text{If }}x < 0,\,2{x^2} - 1 < 1 - {x^2} \cr & {\text{or }}3{x^2} < 2 \cr & {\text{or }}x\, \in \left( { - \sqrt {\frac{2}{3}} ,\,\sqrt {\frac{2}{3}} } \right) \cr & {\text{or }}x\, \in \left( { - \sqrt {\frac{2}{3}} ,\,0} \right) \cr} $$

$$\eqalign{ & f'\left( x \right) = - \left( {x + 2} \right){e^{ - x}} + {e^{ - x}} = - \left( {x + 1} \right){e^{ - x}} = 0 \Rightarrow x = - 1 \cr & {\text{For}}\,x \in \left( { - \infty , - 1} \right),f'\left( x \right) > 0\,{\text{and}}\,{\text{for}}\,x \in \left( { - 1,\infty } \right),f'\left( x \right) < 0 \cr & \therefore f\left( x \right){\text{ is increasing on}}\,\left( { - \infty , - 1} \right)\,{\text{and}}\,{\text{decreasing on}}\,\left( { - 1, - \infty } \right) \cr} .$$

The given polynomial is
$$P\left( x \right) = {a_0} + {a_1}{x^2} + {a_2}{x^4} + ..... + {a_n}{x^{2n}},\,x\, \in \,R{\text{ and }}\,0 < {a_0} < {a_1} < {a_2} < ..... < {a_n}.$$
Here, we observe that all coefficients of different powers of $$x,$$ i.e., $${a_0},\,{a_1},\,{a_2},\,.....,{a_n},$$     are positive.
Also, only even powers of $$x$$ are involved.
Therefore, $$P\left( x \right)$$  cannot have any maximum value.
Moreover, $$P\left( x \right)$$  is minimum, when $$x = 0,$$  i.e., $${a_0}.$$
. Therefore, $$P\left( x \right)$$  has only one minimum.

$$\eqalign{ & {\text{For Rolle's theorem in}}\,\left[ {a,b} \right] \cr & f\left( a \right) = f\left( b \right),\ln \left[ {0,1} \right] \Rightarrow f\left( 0 \right) = f\left( 1 \right) = 0 \cr & \because {\text{The function has to be continuous in }}\left[ {0,1} \right] \cr & \Rightarrow f\left( 0 \right) = \mathop {\lim }\limits_{x \to {0^ + }} f\left( x \right) = 0 \Rightarrow \mathop {\lim }\limits_{x \to 0} {x^\alpha }\log x = 0 \cr & \Rightarrow \mathop {\lim }\limits_{x \to 0} \frac{{\log x}}{{{x^{ - \alpha }}}} = 0 \cr & {\text{Applying L' Hospital's Rule}} \cr & \mathop {\lim }\limits_{x \to 0} \frac{{\frac{1}{x}}}{{ - a{x^{ - \alpha - 1}}}} = 0 \Rightarrow \mathop {\lim }\limits_{x \to 0} \frac{{ - {x^\alpha }}}{\alpha } = 0 \Rightarrow \alpha > 0 \cr} $$

Let $$y = {e^{\left( {2{x^2} - 2x - 1} \right){{\sin }^2}x}}$$     and $$u = \left( {2{x^2} - 2x - 1} \right){\sin ^2}x$$
$$\eqalign{ & {\text{Now, }}\frac{{du}}{{dx}} = \left( {2{x^2} - 2x - 1} \right)2\,\sin \,x\,\cos \,x + \left( {4x - 2} \right){\sin ^2}x \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \sin \,x\left[ {2\left( {2{x^2} - 2x} \right)\cos \,x + \left( {4x - 2} \right)\sin \,x} \right] \cr & \frac{{du}}{{dx}} = 0 \Rightarrow \sin \,x = 0 \Rightarrow x = n\pi \cr & \frac{{{d^2}u}}{{d{x^2}}} = \sin \,x\frac{d}{{dx}}\left[ {2\left( {2{x^2} - 2x - 1} \right)\cos \,x + \left( {4x - 2} \right)\sin \,x} \right] + \cos \,x\left[ {2\,\cos \,x\left( {2{x^2} - 2x - 1} \right) + \left( {4x - 2} \right)\sin \,x} \right] \cr & {\text{At }}x = n\pi ,\,\frac{{{d^2}u}}{{d{x^2}}} = 0 + 2\,{\cos ^2}n\pi \left( {2{n^2}{\pi ^2} - 1} \right) > 0 \cr} $$
Hence at $$x = n\pi ,$$  the value of $$u$$ and so its corresponding the value of $$y$$ is minimum and minimum value $$ = {e^0} = 1$$

$$\eqalign{ & {\text{Let}}\,y = {x^{25}}{\left( {1 - x} \right)^{75}} \Rightarrow \frac{{dy}}{{dx}} = 25{x^{24}}{\left( {1 - x} \right)^{75}} - 75{x^{25}}{\left( {1 - x} \right)^{74}} \cr & = 25{x^{24}}{\left( {1 - x} \right)^{74}}\left( {1 - x - 3x} \right) = 25{x^{24}}{\left( {1 - x} \right)^{74}}\left( {1 - 4x} \right) \cr & {\text{For}}\,{\text{maximum}}\,{\text{value}}\,{\text{of}}\,y,\frac{{dy}}{{dx}} = 0 \cr & \Rightarrow \quad x = 0,1,\frac{1}{4},x = \frac{1}{4} \in \left( {0,1} \right) \cr & {\text{Also}}\,{\text{at}}\,x = 0,y = 0,\,{\text{at}}\,x = 1,y = 0,\,{\text{and}}\,{\text{at}}\,x = \frac{1}{4},y > 0 \cr & \therefore {\text{Max}}{\text{.}}\,{\text{value}}\,{\text{of}}\,y\,{\text{occurs}}\,{\text{at}}\,x = \frac{1}{4} \cr} $$

$$\eqalign{ & y = {x^4} + 3{x^2} + 2x \cr & \therefore \,\frac{{dy}}{{dx}} = 4{x^3} + 6x + 2 \cr} $$
Point on curve which is nearest to the line $$y = 2x – 1$$   is the point where tangent to curve is parallel to given line. Therefore,
$$\eqalign{ & 4{x^3} + 6x + 2 = 2 \cr & {\text{or, }}2{x^3} + 3x = 0 \cr & {\text{or, }}x = 0,\,\,y = 0 \cr} $$
Therefore, point on the curve at the least distance from the line $$y = 2x – 1$$   is $$\left( {0,\,0} \right).$$
Distance of this point from line is $$\frac{1}{{\sqrt 5 }}.$$