Question

The minimum and maximum value of $f\left( x \right)=\sin \left( \cos x \right)+\cos \left( \sin x \right)\forall -\dfrac{\pi }{2}\le x\le \dfrac{\pi }{2}$ are respective,
A. $\cos 1\ and\ 1+\sin 1$
B. $\sin 1\ and\ 1+\cos 1$
C. $\cos 1\ and\ \cos \left( \dfrac{1}{\sqrt{2}} \right)+\sin \left( \dfrac{1}{\sqrt{2}} \right)$
D. None of these

Answer 1

Hint: We will start by using the fact that at the point of minima and maxima the derivative of the function is zero. Then we will use the fact that $\dfrac{d}{dx}\left( \sin x \right)=\cos x\ and\ \dfrac{d}{dx}\left( \cos x \right)=-\sin x$ and chain rule to find the derivative of give function.

Complete step-by-step answer:
Now, we have to find the minimum and maximum value of $f\left( x \right)=\sin \left( \cos x \right)+\cos \left( \sin x \right)\forall -\dfrac{\pi }{2}\le x\le \dfrac{\pi }{2}$.
Now, we will differentiate the function and equate it to zero to find the critical points we will use the chain rule for finding derivative of the function as we know that according to chain rule $\dfrac{d}{dx}f\left( g\left( x \right) \right)=\dfrac{d}{dx}f\left( g\left( x \right) \right)\times \dfrac{d}{dx}\left( g\left( x \right) \right)$
Now, we have,
$f\left( x \right)=\sin \left( \cos x \right)+\cos \left( \sin x \right)$
We know that,
$\begin{aligned} & \dfrac{d}{dx}\left( \sin x \right)=\cos x \\\ & \dfrac{d}{dx}\left( \cos x \right)=\sin x \\\ & \dfrac{d}{dx}\left( f\left( x \right) \right)=\dfrac{d}{dx}\left( \sin \left( \cos x \right) \right)+\dfrac{d}{dx}\left( \cos \left( \sin x \right) \right) \\\ & =\cos \left( \cos x \right)\left( -\sin x \right)-\sin \left( \sin x \right)\left( \cos x \right) \\\ \end{aligned}$
Since, we know that $\sin \left( -x \right)=-\sin x$ we get,
$=-\cos \left( \cos x \right)\sin x-\sin \left( \sin x \right)\cos x$
Now, we know that critical points or points of minima and maxima are the points at which $f'\left( x \right)=0$. So, we have,
$-\cos \left( \cos x \right)\sin x-\sin \left( \sin x \right)\cos x=0.........\left( 1 \right)$
Now, we know that $\sin \left( 0 \right)=0,\cos \left( \dfrac{\pi }{2} \right)=0$. So, we can see that equation (1) is satisfied for $x=0\ and\ x=\dfrac{\pi }{2}$ as $\sin \left( 0 \right)=0\ and\ \cos \left( \dfrac{\pi }{2} \right)=0$.
Now, we have critical points of the function $\forall -\dfrac{\pi }{2}\le x\le \dfrac{\pi }{2}$ as 0 only. So, now the value of function is,
$f\left( 0 \right)=\sin \left( \cos 0 \right)+\cos \left( \sin 0 \right)$
Now, we know that
$\sin 0=0\ and\ \cos 0=1$
So, we have,
$\begin{aligned} & f\left( 0 \right)=\sin \left( 1 \right)+\cos \left( 0 \right) \\\ & =\sin \left( 1 \right)+1 \\\ & =1+\sin 1 \\\ \end{aligned}$
Now, we will find the value at end point i.e.
$f\left( \dfrac{\pi }{2} \right)=\sin \left( \cos \left( \dfrac{\pi }{2} \right) \right)+\cos \left( \sin \dfrac{\pi }{2} \right)$
Now, we know that,
$\cos \left( \dfrac{\pi }{2} \right)=0\ and\ \sin \left( \dfrac{\pi }{2} \right)=1$
So, we have,
$\begin{aligned} & f\left( \dfrac{\pi }{2} \right)=\sin \left( 0 \right)+\cos \left( 1 \right) \\\ & =0+\cos \left( 1 \right) \\\ & =\cos \left( 1 \right) \\\ \end{aligned}$
Now, we know that $1>\dfrac{\pi }{4}$ and after $\dfrac{\pi }{4}\cos \theta <\sin \theta$. So, we have, $1+\sin 1>\cos 1$.
Hence, the minimum and maximum value of function is cos1 and 1 + sin1 respectively. So, the correct option is (A).

Note: It is important to note that we have found the solution of the derivative of the given function analytically seeing that both terms will be zero when $\sin x=0$ for all x belongs to $\left[ -\dfrac{\pi }{2},\dfrac{\pi }{2} \right]$. Also, we have then find the maximum and minimum value by checking which value is smaller and finding the values at the end points of the range given i.e. $-\dfrac{\pi }{2}\le x\le \dfrac{\pi }{2}$. So, we have checked for $x=\dfrac{\pi }{2}$.