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UPSC NDA I 2021 Mathematics Solved Paper

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Question : 93 of 120

Marks: +1, -0

What is

\int\limits_0^a \frac{f(a-x)}{f(x)+f(a-x)} \, dx

equal to?

a
2a
0
$\frac{a}{2}$

Solution:

Concept:

Use the property of definite integrals:

\int\limits_0^a f(x) \, dx = \int\limits_0^a f(a-x) \, dx

.

Explanation:

Let

I = \int\limits_0^a \frac{f(a-x)}{f(x)+f(a-x)} \, dx

.

Apply the property: replace

x

by

a-x

in the integral.

Then

I = \int\limits_0^a \frac{f[a-(a-x)]}{f(a-x)+f[a-(a-x)]} \, dx = \int\limits_0^a \frac{f(x)}{f(a-x)+f(x)} \, dx

.

Add the two expressions for

I

:

I + I = \int\limits_0^a \frac{f(a-x)}{f(x)+f(a-x)} \, dx + \int\limits_0^a \frac{f(x)}{f(a-x)+f(x)} \, dx

.

The denominators are the same, so sum the numerators:

2I = \int\limits_0^a \frac{f(a-x)+f(x)}{f(x)+f(a-x)} \, dx = \int\limits_0^a 1 \, dx

.

Evaluate:

\int\limits_0^a 1 \, dx = a

. Hence

2I = a

and

I = \frac{a}{2}

.

Answer:

I = \frac{a}{2}

(Option D).

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