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UPSC NDA I 2019 Mathematics Solved Paper

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Question : 26 of 120

Marks: +1, -0

What is the least value of

25 \csc^{2} x + 36 \sec^{2} x

?

1
11
120
121

Solution:

Concept:

The minimum value of

m \sec^{2} x + n \csc^{2} x

is

(\sqrt{m} + \sqrt{n})^{2}

.

Explanation:

We are given

25 \csc^{2} x + 36 \sec^{2} x

.

Here,

m = 36

and

n = 25

because

\sec^{2} x

is multiplied by 36 and

\csc^{2} x

by 25.

Using the formula, the least value is

(\sqrt{36} + \sqrt{25})^{2}

.

Compute

\sqrt{36} = 6

and

\sqrt{25} = 5

.

Sum

6 + 5 = 11

.

Square

11^{2} = 121

.

Therefore, the minimum value of the expression is 121.

Answer:

121 (Option D).

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