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UPSC NDA I 2017 Mathematics Solved Paper

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Question : 103 of 120

Marks: +1, -0

What is the maximum value of the function

f(x) = 4 \sin^{2} x + 1

?

5
3
2
1

Solution:

Concept:

The maximum of

f(x) = 4 \sin^{2}x + 1

is found from the range of

\sin^{2}x

, which is

0

to

1

.

Explanation:

Given

f(x) = 4 \sin^{2}x + 1

.

The range of

\sin x

is

[-1, 1]

.

Squaring gives

\sin^{2}x \text{ in } [0, 1]

.

Multiply by

4

:

4 \sin^{2}x \text{ in } [0, 4]

.

Add

1

:

4 \sin^{2}x + 1 \text{ in } [1, 5]

.

The largest value in this interval is

5

.

Thus the maximum value of

f(x)

is

5

.

Answer:

The maximum value is

5

. Therefore, the correct option is A. 5.

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