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UPSC CDS I 2024 Math Paper

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Question : 34 of 100

Marks: +1, -0

What is the minimum value of

\frac{sin^{2}A+5sinA+1}{sinA}

\frac{\pi }{2}

?

3
5
7
9

Solution:

Concept:

The expression can be simplified to

x + \frac{1}{x} + 5

, where

x = \sin A

.

For

x > 0

, the minimum of

x + \frac{1}{x}

is always

2

.

Explanation:

First, simplify the given expression:

\frac{\sin^2 A + 5\sin A + 1}{\sin A} = \sin A + 5 + \frac{1}{\sin A}

.

Let

x = \sin A

. Since

0 < A \le \frac{\pi}{2}

, we have

0 < x \le 1

.

The function

x + \frac{1}{x}

attains its minimum value when

x = 1

, giving

1 + \frac{1}{1} = 2

.

Thus, the minimum value of

\sin A + \frac{1}{\sin A}

is

2

.

Adding the constant

5

, the minimum of the whole expression is

2 + 5 = 7

.

Answer:

7

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