GATE (TF) Textile 2015 Question Paper Solution | GATE/2015/TF/36

Question 36 (Textile Engineering & Fibre Science)

The maximum value of $f(x)=\sqrt{2}(\sin x + \cos x)$ for $x$ in $\left [ 0, \pi \right ]$ is __2__.

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$f(x)=\sqrt{2}(\sin x + \cos x)$
Differentiate-
${f}'(x)=\sqrt{2}{cos x-sin x}$
For maxima,minima-
${f}'(x)=0$
$\sqrt{2}{cos x-sin x}$ =0
sin x=cos x
x= $\frac{\pi}{4}$
Again differentiate-
${f}''(x)=\sqrt{2}{-sin x-cos x}$
Put x= $\frac{\pi}{4}$
${f}''(x)=\sqrt{2}{-sin \frac{\pi}{4}-cos \frac{\pi}{4}}$
${f}''(x)=-2$ The value is negative,it means this is the case of maxima.
Now maximum value of the function at x= $\frac{\pi}{4}$
$f(x)=\sqrt{2}(\sin\frac{\pi}{4}+ \cos\frac{\pi}{4})$
$f(x)=\sqrt{2}(\frac{1}{\sqrt 2}+\frac{1}{\sqrt 2} )$
f(x)=2 (Ans)

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