0 votes

1 vote

__ 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.