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UPSC NDA II 2023 Mathematics Solved Paper

Question Numbers: 49-50

Consider the following for the next items that follow:

A function is defined by f(x) = π + sin2 x.

Question : 49 of 120

Marks: +1, -0

What is the range of the function?

Solution:

Concept:

The range of

\sin^2 x

[0, 1]

, so adding a constant shifts the range.

Explanation:

The function is

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

We know

\sin x

ranges over

[-1, 1]

Squaring gives

0 \le \sin^2 x \le 1

Adding

\pi

yields

\pi \le \pi + \sin^2 x \le \pi + 1

Thus the range is

[\pi, \pi + 1]

Answer:

Option B:

[\pi, \pi + 1]

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