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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

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

Solution: 👈: Video Solution

Concept:

Definite Integrals:

\int\limits_{a}^{b} f(x) dx = \int\limits_{a}^{b} f(a+b-x) dx

If

f(x)=f(2a-x)

, then

\int\limits_{0}^{2a} f(x) dx = 2 \int\limits_{0}^{a} f(x) dx

A function

f(x)

is:

- Even, if

f(-x)=f(x)

. And

\int\limits_{-a}^{a} f(x) dx = 2 \int\limits_{0}^{a} f(x) dx

- Odd,if

f(-x) = -f(x)

. And

\int\limits_{-a}^{a} f(x) dx = 0

.

- Periodic, if

f(np \pm x) = f(x)

, for some number

p

and

n \in Z

.

Calculation:

We know that

\int\limits_{a}^{b} f(x) dx = \int\limits_{a}^{b} f(a+b-x) dx

\therefore I = \int\limits_{0}^{a} \frac{f(a-x)}{f(x)+f(a-x)} dx

\; = \int\limits_{0}^{a} \frac{f[(a+0)-(a-x)]}{f[(a+0)-x] + f[(a+0)-(a-x)]} dx

\; = \int\limits_{0}^{a} \frac{f(x)}{f(a-x)+f(x)} dx

And,

2I = \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

= \int\limits_{0}^{a} 1 dx

= a

.

\Rightarrow I = \frac{a}{2}

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