Passage: Let ∫_0^ (π)/(2) (a x+b x)/((a+b)( x+ x))\,dx=k . Question: What is the value of k ?

Correct answer: A. (π)/(4) Let I = ∫_0^ (π)/(2) (a x+b x)/((a+b)( x+ x))\,dx Using the property ∫_0^a f(x)\,dx = ∫_0^a f(a-x)\,dx , we get: I = ∫_0^ (π)/(2) (a ((π)/(2)-x)+b ((π)/(2)-x))/((a+b)( ((π)/(2)-x)+ ((π)/(2)-x)))\,dx I = ∫_0^ (π)/(2) (a x+b x)/((a+b)( x+ x))\,dx Adding the two equations for I , we get: 2I = ∫_0^ (π)/(2) (a x+b x+a x+b x)/((a+b)( x+ x))\,dx 2I = ∫_0^ (π)/(2) ((a+b)( x+ x))/((a+b)( x+ x))\,dx 2I = ∫_0^ (π)/(2) 1 \,dx 2I = (π)/(2) I = (π)/(4) Thus, k = (π)/(4) Answer: (π)/(4)

NDA 2026

Exam: NDA
Year: 2026
Subject: Mathematics
Chapter: Definite Integration

NDA 2026 · Mathematics · Definite Integration

Passage: Let ∫_0^ (π)/(2) (a x+b x)/((a+b)( x+ x))\,dx=k ....

Passage: Let

\displaystyle\int_0^{\frac{\pi}{2}}\dfrac{a\sin x+b\cos x}{(a+b)(\sin x+\cos x)}\,dx=k

.

Question: What is the value of

k

Options

A $\dfrac{\pi}{4}$
B $\dfrac{\pi}{2}$
C $\pi$
D $2\pi$

Correct answer

A. $\dfrac{\pi}{4}$

Step-by-step solution

Let

I = \int_0^{\dfrac{\pi}{2}}\dfrac{a\sin x+b\cos x}{(a+b)(\sin x+\cos x)}\,dx

Using the property

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

, we get:

I = \int_0^{\dfrac{\pi}{2}}\dfrac{a\sin\left(\dfrac{\pi}{2}-x\right)+b\cos\left(\dfrac{\pi}{2}-x\right)}{(a+b)\left(\sin\left(\dfrac{\pi}{2}-x\right)+\cos\left(\dfrac{\pi}{2}-x\right)\right)}\,dx

I = \int_0^{\dfrac{\pi}{2}}\dfrac{a\cos x+b\sin x}{(a+b)(\cos x+\sin x)}\,dx

Adding the two equations for

I

, we get:

2I = \int_0^{\dfrac{\pi}{2}}\dfrac{a\sin x+b\cos x+a\cos x+b\sin x}{(a+b)(\sin x+\cos x)}\,dx

2I = \int_0^{\dfrac{\pi}{2}}\dfrac{(a+b)(\sin x+\cos x)}{(a+b)(\sin x+\cos x)}\,dx

2I = \int_0^{\dfrac{\pi}{2}} 1 \,dx

2I = \dfrac{\pi}{2}

I = \dfrac{\pi}{4}

Thus,

k = \dfrac{\pi}{4}

Answer:

\dfrac{\pi}{4}

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