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)
{f}'(x)=\sqrt{2}{cos x-sin x}
For maxima,minima-
\sqrt{2}{cos x-sin x}=0
sin x=cos x
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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