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UPSC NDA I 2026 Mathematics Paper

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Question Numbers: 93-94

For the next two (02) items that follow:

Let

f(x)=\sin x \text { and } g(x)-f(x)=f(4-x) \text {. }

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Question : 93 of 120

Marks: +1, -0

What is

\int_{0}^{4} \frac{f(x)}{g(x)} d x

equal to ?

0
1
2
4

Solution:

Concept:

Use the definite integral property

\int_0^a h(x)\,dx = \int_0^a h(a-x)\,dx

.

Explanation:

Given

g(x) - f(x) = f(4-x)

and

f(x) = \sin x

, we get:

g(x) = f(x) + f(4-x) = \sin x + \sin(4-x)

So the integral becomes:

I = \int_0^4 \frac{f(x)}{g(x)}\,dx = \int_0^4 \frac{\sin x}{\sin x + \sin(4-x)}\,dx

Using the property, replace

x

by

4-x

:

I = \int_0^4 \frac{\sin(4-x)}{\sin(4-x) + \sin x}\,dx

Adding the two expressions for

I

:

2I = \int_0^4 \frac{\sin x + \sin(4-x)}{\sin x + \sin(4-x)}\,dx = \int_0^4 1\,dx = 4

Hence,

I = 2

Answer:

2

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