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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 : 94 of 120

Marks: +1, -0

What is

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

equal to ?

0
1
2
4

Solution:

Concept:

Use the substitution

x \to 4-x

and add the two forms of the integral.

Explanation:

Given

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

, so

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

.

Then

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

.

Let

I=\int_{0}^{4}\frac{f(4-x)}{g(4-x)}\,dx

.

Using

x\to 4-x

, we get

I=\int_{0}^{4}\frac{f(x)}{g(x)}\,dx

.

Adding the two expressions:

2I=\int_{0}^{4}\left(\frac{f(4-x)}{g(4-x)}+\frac{f(x)}{g(x)}\right)dx

Since

g(4-x)=g(x)

, the numerator sum is

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

.

Thus,

2I=\int_{0}^{4}1\,dx=4

So

I=2

.

Answer:

2

, i.e. option C.

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