NDA 2026 — Mathematics PYQ

Step 1: Simplify the expression

We have the expression $f(x)=\cos x+\sqrt{3}\sin x$.

Multiply and divide by $\sqrt{a^2+b^2}$, where $a=1$ and $b=\sqrt{3}$:

$$\sqrt{a^2+b^2}=\sqrt{1^2+(\sqrt{3}{)}^2}=\sqrt{1+3}=\sqrt{4}=2$$

Now rewrite the expression:

$$f(x)=2\left(\frac{1}{2}\cos x+\frac{\sqrt{3}}{2}\sin x\right)$$

Step 2: Use the angle addition formula

Recall that $\cos (\frac{\pi }{3})=\frac{1}{2}$ and $\sin (\frac{\pi }{3})=\frac{\sqrt{3}}{2}$. Substituting these:

$$f(x)=2\left(\cos \left(\frac{\pi }{3}\right)\cos x+\sin \left(\frac{\pi }{3}\right)\sin x\right)$$

Using the identity $\cos (A-B)=\cos A\cos B+\sin A\sin B$:

$$f(x)=2\cos \left(x-\frac{\pi }{3}\right)$$

Step 3: Determine the maximum value

The function $f(x)=2\cos \left(x-\frac{\pi }{3}\right)$ reaches its maximum value when $\cos \left(x-\frac{\pi }{3}\right)=1$.

This occurs when the angle is $0$:

$$x-\frac{\pi }{3}=0$$

$$x=\frac{\pi }{3}$$

Conclusion: The expression reaches its maximum value when $x=\frac{\pi }{3}$. Therefore, the correct option is (b).