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UPSC NDA I 2022 Mathematics Solved Paper

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Question : 66 of 120

Marks: +1, -0

What is the maximum value of the functions

f(x) = \frac{1}{\tan x+\cot x},

0 < x < \frac{\pi}{2}?

$\frac{1}{4}$
$\frac{1}{2}$
1
2

Solution:

Concept:

Use trigonometric identities to simplify the function and find its maximum value by analyzing the range of sine.

Explanation:

Start with

f(x) = \frac{1}{\tan x + \cot x}

for

0 < x < \frac{\pi}{2}

.

Rewrite

\tan x = \frac{\sin x}{\cos x}

and

\cot x = \frac{\cos x}{\sin x}

.

Then

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

.

So

f(x) = \sin x \cos x

.

Use the identity

2\sin x \cos x = \sin 2x

, thus

\sin x \cos x = \frac{1}{2} \sin 2x

.

Therefore

f(x) = \frac{1}{2} \sin 2x

.

In the interval

(0,\frac{\pi}{2})

,

2x

lies in

(0,\pi)

.

The sine function has a maximum value of

1

(at

2x = \frac{\pi}{2}

).

Hence the maximum of

f(x)

is

\frac{1}{2} \times 1 = \frac{1}{2}

.

Answer:

\frac{1}{2}

(Option B).

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