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UPSC NDA I 2023 Mathematics Solved Paper

Question Numbers: 37-38

Consider the following for the next two (02) items that follow :

Let

x=\frac{sin^{2}A+sinA+1}{sinA}

where 0 < A ≤

\frac{\pi }{2}

Question : 38 of 120

Marks: +1, -0

At what value of A does x attain the minimum value ?

Solution:

Concept:

For

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

\sin A > 0

The expression

t + \frac{1}{t}

attains its minimum value

2

when

t = 1

Explanation:

We rewrite

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

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

Since

\sin A > 0

, we can apply the inequality:

\sin A + \frac{1}{\sin A} \ge 2

, with equality when

\sin A = 1

Thus the minimum value of

x

2 + 1 = 3

, achieved when

\sin A = 1

Within the interval

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

\sin A = 1

implies

A = \frac{\pi}{2}

Answer:

A = \frac{\pi}{2}

(Option D).

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