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UPSC NDA I 2026 Mathematics Paper

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Question Numbers: 83-84

For the next two (02) items that follow:

\text { Let } \int_{0}^{\frac{\pi}{2}} \frac{a \sin x+b \cos x}{(a+b)(\sin x+\cos x)} d x=k

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Question : 83 of 120

Marks: +1, -0

What is the value of

k

?

$\frac{\pi}{4}$
$\frac{\pi}{2}$
$\pi$
$2 \pi$

Solution:

Concept:

Use symmetry of

\sin x

and

\cos x

over

0

to

\frac{\pi}{2}

.

Explanation:

Given:

k = \int_{0}^{\frac{\pi}{2}} \frac{a \sin x + b \cos x}{(a+b)(\sin x + \cos x)}\, dx

Split the integral:

k = \frac{a}{a+b} \int_{0}^{\frac{\pi}{2}} \frac{\sin x}{\sin x + \cos x}\, dx + \frac{b}{a+b} \int_{0}^{\frac{\pi}{2}} \frac{\cos x}{\sin x + \cos x}\, dx

Let:

I_1 = \int_{0}^{\frac{\pi}{2}} \frac{\sin x}{\sin x + \cos x}\, dx, \quad I_2 = \int_{0}^{\frac{\pi}{2}} \frac{\cos x}{\sin x + \cos x}\, dx

By symmetry,

I_1 = I_2

. Also:

I_1 + I_2 = \int_{0}^{\frac{\pi}{2}} 1\, dx = \frac{\pi}{2}

So:

I_1 = I_2 = \frac{\pi}{4}

Hence:

k = \frac{a}{a+b} \cdot \frac{\pi}{4} + \frac{b}{a+b} \cdot \frac{\pi}{4} = \frac{\pi}{4}

Answer:

k = \frac{\pi}{4}

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