Q233 / 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 the value of $k$ ?
Question: What is the value of $k$ ?
- Correct answerA$\dfrac{\pi}{4}$
- B$\dfrac{\pi}{2}$
- C$\pi$
- D$2\pi$
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: $\dfrac{\pi}{4}$
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: $\dfrac{\pi}{4}$
Hence, the correct answer is option: 'A'.