If $y = cos x + sin x,$ then maximum value of y is

2 \sqrt{2}

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\frac{1}{\sqrt{2}}

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2

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\sqrt{2}

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Solution

The correct option is D $\sqrt{2}$
$y = sin x + cos x$
$\frac{d y}{d x} = cos x - sin x$
$\frac{d y}{d x} = 0$
$cos x = sin x$
Maxima occurs at $x = π / 4$
$y (π / 4) = \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}} = \sqrt{2}$

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