The function $sin x + cos x$ is maximum when $x$ is equal to:

\frac{π}{6}

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\frac{π}{4}

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\frac{π}{3}

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\frac{π}{2}

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Solution

The correct option is B $\frac{π}{4}$
Let $y = sin x + cos x$

$= \sqrt{2} (\frac{sin x + cos x}{\sqrt{2}})$

$= \sqrt{2} (sin (\frac{π}{4} + x))$

Here, $y$ will be maximum when $(sin (\frac{π}{4} + x)) = 1$

But, $sin \frac{π}{2} = 1$

So, $\frac{π}{4} + x = \frac{π}{2}$

Hence, $x = \frac{π}{4}$

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