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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 : 37 of 120

Marks: +1, -0

What is the minimum value of x ?

Solution:

Concept:

For a positive number

t

, the minimum value of

t + \frac{1}{t}

2

, achieved when

t = 1

Explanation:

Given

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

for

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

Simplify the expression:

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

Since

\sin A > 0

in this interval, let

t = \sin A

Then

x = t + \frac{1}{t} + 1

The minimum of

t + \frac{1}{t}

2

when

t = 1

, i.e.,

\sin A = 1

(which gives

A = \frac{\pi}{2}

Therefore,

x_{\text{min}} = 2 + 1 = 3

Answer:

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