Question

Let $M$ and $N$ are the maximum and minimum values of the slope of the tangents to the curve $f(x) = {e^x}\sin x:x \in [0,\pi ]$ , then the value of $\log \left| {\dfrac{M}{N}} \right|$
A. $- \dfrac{\pi }{4}$
B. $- \dfrac{\pi }{2}$
C. $- \pi$
D. $2\pi$

Answer 1

Hint : In this question we are asked the maximum and minimum values of the slope of the tangents to the curve in the given range. In order to proceed with this question we need to calculate the first, second and third derivative of the function. Because the derivative provides information about the slope of the graph of a function and we can use it to locate points on a graph where the slope is zero.

Complete step by step solution:
We are given,
$f(x) = {e^x}\sin x:x \in [0,\pi ]$ s
The slope of the this curve is given as,
$\Rightarrow m = \dfrac{{dy}}{{dx}} = {e^x}\dfrac{d}{{dx}}\sin x + \sin x\dfrac{d}{{dx}}{e^x}$
$\Rightarrow y' = {e^x} \times \cos x + \sin x \times {e^x}$
$\Rightarrow y' = {e^x}(\cos x + \sin x)$
$M$ is the maximum value of the slope and $N$ is the minimum value of the slope in $x \in [0,\pi ]$
Therefore, $M$ is the maximum and $N$ is the minimum of the function $y' = {e^x}(\cos x + \sin x)$
Second derivation,
$\Rightarrow y'' = {e^x}\dfrac{d}{{dx}}(\cos x + \sin x) + (\cos x + \sin x)\dfrac{d}{{dx}}{e^x}$
$\Rightarrow y'' = {e^x}( - \sin x + \cos x) + (\cos x + \sin x){e^x}$
$\Rightarrow y'' = {e^x}( - \sin x + \cos x + \cos x + \sin x)$
$\Rightarrow y'' = {e^x}( - \sin x + \cos x + \cos x + \sin x)$
$\Rightarrow y'' = {e^x}(2\cos x)$
We know that,
$y'' = 0$ at $x = \dfrac{\pi }{2}$
The point $x = \dfrac{\pi }{2}$ can be the point where the slope takes either maximum or minimum value.
Third derivative
\Rightarrow y''' = 2[{e^x}( - \sin x) + \cos x \times {e^x}\\}
$\Rightarrow y''' = 2\\{ - {e^x}\sin x + {e^x}\cos x\\}$
$\Rightarrow y''' = 2{e^x}\\{ \cos x - \sin x\\}$
At $x = \dfrac{\pi }{2}$ ,
$y''' < 0$
Thus this is the point where the slope is maximum.
We’ll find the maximum value of the slope by putting $x = \dfrac{\pi }{2}$ in the $y'$
$\Rightarrow M = {e^{\dfrac{\pi }{2}}}(\cos \dfrac{\pi }{2} + \sin \dfrac{\pi }{2})$
$\Rightarrow M = {e^{\dfrac{\pi }{2}}}$
Since we got only one value for $y'' = 0$ , and this point was for maximum value of the slope, to find the minimum value of the slope we can check the value at the end points of the given interval.
$\Rightarrow y'(0) = {e^0}(\cos 0 + \sin 0) = 1$
And
$\Rightarrow y'(\pi ) = {e^\pi }(\cos \pi + \sin \pi ) = - {e^\pi }$
Thus the minimum value of the slope is,
$\Rightarrow N = - {e^\pi }$
$\Rightarrow \log \left| {\dfrac{M}{N}} \right| = \log \left| {\dfrac{{{e^{\dfrac{\pi }{2}}}}}{{ - {e^\pi }}}} \right|$
$\Rightarrow \log \left| { - {e^{\dfrac{\pi }{2} - \pi }}} \right|$
$\Rightarrow \log \left| { - {e^{ - \dfrac{\pi }{2}}}} \right|$
$\Rightarrow \log {e^{ - \dfrac{\pi }{2}}}$
$\Rightarrow - \dfrac{\pi }{2}$
This is the required answer.
Option(B)
So, the correct answer is “Option B”.

Note : Points to remember,
if $\dfrac{{dy{\text{ }}}}{{dx}}{\text{ }} = {\text{ }}0$ at a point, and if $\dfrac{{{d^2}y}}{{d{x^2}}} > 0$ there, then it is minimum at that point.
if $\dfrac{{dy{\text{ }}}}{{dx}}{\text{ }} = {\text{ }}0$ at a point, and if $\dfrac{{{d^2}y}}{{d{x^2}}} < 0$ there, then it is maximum at that point.