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UPSC NDA Math Model Paper 9

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Question : 108 of 120

Marks: +1, -0

Maximum value of function f (x) = sin x + cos x

2
$\quad \sqrt{2}$
$\quad \frac{1}{\sqrt{2}}$
1

Solution:

Given: f (x) = sin x + cos x

f'(x) = \frac{d}{dx}(f(x))

= \frac{d}{dx}(\sin x + \cos x)

= \cos x - \sin x

Now,

f''(x) = \frac{d}{dx}(f'(x))

= \frac{d}{dx}(\cos x - \sin x)

= -\sin x - \cos x

= -(\sin x + \cos x)

Here, f’’(x) < 0, so given function is maxima

∴ f’(x) = 0

\Rightarrow \cos x - \sin x = 0

\Rightarrow \cos x = \sin x

We know at x = π/4, sin x = cos x

f (x) = sin x + cos x

= sin x + sin x (∵ sin x = cos x)

= 2 sin x

F(π/4) = 2 sin(π/4)

= 2\left(\frac{1}{\sqrt{2}}\right)

Divide and multiply by

\sqrt{2}

, we get

= \sqrt{2}

Hence, option (2) is correct.

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