NDA 2026 — Mathematics PYQ

1. Using the Property of Definite Integrals

We use the property ${\int }_0^{a}f(x)dx={\int }_0^{a}f(a-x)dx$. Let the given integral be $I$:

$$I=\frac{1}{a+b}{\int }_0^{\frac{\pi }{2}}\frac{a\sin x+b\cos x}{\sin x+\cos x}dx\dots \text{(Eq. 1)}$$

Applying the property $x\to \frac{\pi }{2}-x$:

$$I=\frac{1}{a+b}{\int }_0^{\frac{\pi }{2}}\frac{a\sin (\frac{\pi }{2}-x)+b\cos (\frac{\pi }{2}-x)}{\sin (\frac{\pi }{2}-x)+\cos (\frac{\pi }{2}-x)}dx$$

$$I=\frac{1}{a+b}{\int }_0^{\frac{\pi }{2}}\frac{a\cos x+b\sin x}{\cos x+\sin x}dx\dots \text{(Eq. 2)}$$

2. Adding the Two Equations

Adding (Eq. 1) and (Eq. 2):

$$2I=\frac{1}{a+b}{\int }_0^{\frac{\pi }{2}}\frac{(a\sin x+b\cos x)+(a\cos x+b\sin x)}{\sin x+\cos x}dx$$

$$2I=\frac{1}{a+b}{\int }_0^{\frac{\pi }{2}}\frac{(a+b)\sin x+(a+b)\cos x}{\sin x+\cos x}dx$$

$$2I=\frac{a+b}{a+b}{\int }_0^{\frac{\pi }{2}}\frac{\sin x+\cos x}{\sin x+\cos x}dx$$

$$2I={\int }_0^{\frac{\pi }{2}}1dx$$

3. Evaluating the Result

$$2I=[x{]}_0^{\frac{\pi }{2}}=\frac{\pi }{2}-0=\frac{\pi }{2}$$

$$I=\frac{\pi }{4}$$

Conclusion: The value of $k$ is $\frac{\pi }{4}$.

Correct Option: (a)