Q234 / 270 · NDA 2026 (Phase 1) · Mathematics
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 $\displaystyle\int_0^{\frac{\pi}{2}}\dfrac{a\cos x+b\sin x}{\sin x+\cos x}\,dx$ equal to ?
Question: What is $\displaystyle\int_0^{\frac{\pi}{2}}\dfrac{a\cos x+b\sin x}{\sin x+\cos x}\,dx$ equal to ?
- A$k$
- B$2k$
- Correct answerC$k(a+b)$
- D$\dfrac{k}{a+b}$
Solution
Given $\displaystyle k = \int_0^{\dfrac{\pi}{2}}\dfrac{a\sin x+b\cos x}{(a+b)(\sin x+\cos x)}\,dx$
Multiplying both sides by $(a+b)$:
$\displaystyle k(a+b) = \int_0^{\dfrac{\pi}{2}}\dfrac{a\sin x+b\cos x}{\sin x+\cos x}\,dx$
Using the definite integral property $\displaystyle \int_0^a f(x)\,dx = \int_0^a f(a-x)\,dx$, replacing $x$ with $\dfrac{\pi}{2}-x$:
$\displaystyle k(a+b) = \int_0^{\dfrac{\pi}{2}}\dfrac{a\sin\left(\dfrac{\pi}{2}-x\right)+b\cos\left(\dfrac{\pi}{2}-x\right)}{\sin\left(\dfrac{\pi}{2}-x\right)+\cos\left(\dfrac{\pi}{2}-x\right)}\,dx$
$\displaystyle k(a+b) = \int_0^{\dfrac{\pi}{2}}\dfrac{a\cos x+b\sin x}{\cos x+\sin x}\,dx$
This matches the required integral. Thus, the value is $k(a+b)$.
Answer: $k(a+b)$
Multiplying both sides by $(a+b)$:
$\displaystyle k(a+b) = \int_0^{\dfrac{\pi}{2}}\dfrac{a\sin x+b\cos x}{\sin x+\cos x}\,dx$
Using the definite integral property $\displaystyle \int_0^a f(x)\,dx = \int_0^a f(a-x)\,dx$, replacing $x$ with $\dfrac{\pi}{2}-x$:
$\displaystyle k(a+b) = \int_0^{\dfrac{\pi}{2}}\dfrac{a\sin\left(\dfrac{\pi}{2}-x\right)+b\cos\left(\dfrac{\pi}{2}-x\right)}{\sin\left(\dfrac{\pi}{2}-x\right)+\cos\left(\dfrac{\pi}{2}-x\right)}\,dx$
$\displaystyle k(a+b) = \int_0^{\dfrac{\pi}{2}}\dfrac{a\cos x+b\sin x}{\cos x+\sin x}\,dx$
This matches the required integral. Thus, the value is $k(a+b)$.
Answer: $k(a+b)$
Hence, the correct answer is option: 'C'.