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UPSC NDA 2 2021 Mathematics Solved Paper

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Question : 23 of 120

Marks: +1, -0

What is the period of the function

f(x) = \ln(2 + \sin^{2}x)

?

$\text{rm} \frac{\pi}{2}$
π
$2\pi$
$3\pi$

Solution:

Concept:

The period of

f(x) = \ln(2 + \sin^2 x)

is the same as the period of

\sin^2 x

because

\ln

is a one‑to‑one function.

Explanation:

Use the identity

\cos 2x = 1 - 2\sin^2 x

to rewrite

\sin^2 x

.

So

\sin^2 x = \frac{1 - \cos 2x}{2}

.

Then

f(x) = \ln\left(2 + \frac{1 - \cos 2x}{2}\right) = \ln\left(\frac{5 - \cos 2x}{2}\right) = \ln(5 - \cos 2x) - \ln 2

.

The constant

-\ln 2

does not affect the period.

The function

\cos 2x

has period

\frac{2\pi}{2} = \pi

.

Hence

\ln(5 - \cos 2x)

also has period

\pi

, so

f(x)

has period

\pi

.

Answer:

\pi

(Option B).

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