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UPSC NDA I 2026 Mathematics Paper

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Question : 30 of 120

Marks: +1, -0

\cos x+\sqrt{3} \sin x

is maximum when

x

is equal to

$\frac{\pi}{2}$
$\frac{\pi}{3}$
$\frac{\pi}{4}$
$\frac{\pi}{6}$

Solution:

Concept:

Express

\cos x+\sqrt{3}\sin x

in the form

2\cos\left(x-\frac{\pi}{3}\right)

.

Explanation:

We have

\cos x+\sqrt{3}\sin x = 2\left(\frac{1}{2}\cos x+\frac{\sqrt{3}}{2}\sin x\right)

Since

\cos\frac{\pi}{3}=\frac{1}{2}

and

\sin\frac{\pi}{3}=\frac{\sqrt{3}}{2}

,

\cos x+\sqrt{3}\sin x = 2\cos\left(x-\frac{\pi}{3}\right)

The maximum value of

\cos

is

1

, which occurs when its angle is

0

.

So,

x-\frac{\pi}{3}=0 \Rightarrow x=\frac{\pi}{3}

.

Answer:

x=\frac{\pi}{3}

, i.e., option B.

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