GATE (TF) Textile 2019 Question Paper Solution

Table of Contents

For $x$ in [ $0, \pi$ ], is maximum value of $(\sin x+ \cos x)$ is

$F=(\sin x+ \cos x)$

Differentiate it both side with respect to x

$\frac{dF}{dx}=(\cos x-\sin x)$

For maxima minima , $\frac{dF}{dx}=0$

$(\cos x-\sin x)=0$

cos x=sin x

x= $\frac{\pi}{4}$

Again differentiate-

$\frac{d^2F}{dx^2}=(\-sin x-\cos x)$

$\frac{d^2F}{dx^2}=(\-sin\frac{\pi}{4} -\cos \frac{\pi}{4})=(-\frac{1}{\sqrt 2}-\frac{1}{\sqrt 2})=-\sqrt 2$

Hence function is minima at x= $\frac{\pi}{4}$

$F=(\sin x+ \cos x)$

By putting x= $\frac{\pi}{4}$

$F=(\sin \frac{\pi}{4}+ \cos \frac{\pi}{4})$

$F=(\frac{1}{\sqrt 2}+\frac{1}{\sqrt 2})$

$F=\frac{2}{\sqrt 2}$

$F=\sqrt 2$ (Ans)

This is minimum value of the function.