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KCET 2026 Mathematics Solved Paper

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Question : 9 of 60

Marks: +1, -0

The maximum value of

\sin \left(x+\frac{\pi}{6}\right)+\cos \left(x+\frac{\pi}{6}\right)

x=

$\frac{\pi}{2}$
$\frac{\pi}{4}$
$\frac{\pi}{6}$
$\frac{\pi}{12}$

Solution:

Concept:

Combine sine and cosine using the identity

\sin A + \cos A = \sqrt{2}\sin\left(A + \frac{\pi}{4}\right)

.

Explanation:

Let

y = x + \frac{\pi}{6}

.

Then the expression becomes

\sin y + \cos y

.

Using the identity,

\sin y + \cos y = \sqrt{2}\sin\left(y + \frac{\pi}{4}\right)

.

The maximum value of

\sin\left(y + \frac{\pi}{4}\right)

is

1

.

So the maximum occurs when

y + \frac{\pi}{4} = \frac{\pi}{2}

.

This gives

y = \frac{\pi}{4}

.

Substitute back

y = x + \frac{\pi}{6}

:

x + \frac{\pi}{6} = \frac{\pi}{4}

x = \frac{\pi}{4} - \frac{\pi}{6} = \frac{3\pi - 2\pi}{12} = \frac{\pi}{12}

.

Answer:

x = \frac{\pi}{12}

, which is option D.

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