Answer: $\mathbf D$
Explanation:
Method 1:
Maximum and minimum values of the expression of the form
$a\sin x + b \cos x$ is given by
$\pm \sqrt{a^2+b^2}$
Here, $a = 1, \;and \; b = 1$
So, Maximum value $= \sqrt{1^2 + 1^2} = \sqrt2$
So, option $\mathbf D$ is the correct answer.
Method 2:
Let $y = \cos x + \sin x$
On differentiating with respect to x, we get:
$\frac{dy}{dx} = -\sin x + \cos x$
Put this equation equals to 0, for calculating the maxima.
Now,
$\frac{dy}{dx} = 0 \implies \sin x = \cos x \implies x = \frac{\pi}{4}$
So, we will get the maximum value at $x = \frac{\pi}{4} = \frac{1}{\sqrt 2} = \sqrt 2$
So, again by this method option $\mathbf D$ is the right answer.