NIMCET 2008 — Mathematics PYQ

Step 1: Find the derivative $f^{\prime }(x)$

Differentiating with respect to $x$:

Step 2: Find critical points by setting $f^{\prime }(x)=0$

Using the identity $\cos 2x=2{\cos }^{2}x-1$:

This is a quadratic equation in terms of $\cos x$. Let $y=\cos x$:

So, $y=\frac{1}{2}$ or $y=-1$.

• If $\cos x=\frac{1}{2}$, then $x=\frac{\pi }{3}$ or $x=\frac{5\pi }{3}$ (within $[0,2\pi ]$).

• If $\cos x=-1$, then $x=\pi$.

Step 3: Evaluate $f(x)$ at critical points and endpoints

The points to check are $\{0,\frac{\pi }{3},\pi ,\frac{5\pi }{3},2\pi \}$.

1. At $x=0$ and $x=2\pi$:

   $f(0)=2\sin (0)+\sin (0)=0$

   $f(2\pi )=2\sin (2\pi )+\sin (4\pi )=0$

2. At $x=\frac{\pi }{3}$:

   $f(\frac{\pi }{3})=2\sin (\frac{\pi }{3})+\sin (\frac{2\pi }{3})=2(\frac{\sqrt{3}}{2})+\frac{\sqrt{3}}{2}=\frac{3\sqrt{3}}{2}\approx 2.598$ (Absolute Maximum)

3. At $x=\pi$:

   $f(\pi )=2\sin (\pi )+\sin (2\pi )=0+0=0$

4. At $x=\frac{5\pi }{3}$:

   $f(\frac{5\pi }{3})=2\sin (\frac{5\pi }{3})+\sin (\frac{10\pi }{3})$

   $f(\frac{5\pi }{3})=2(-\frac{\sqrt{3}}{2})+(-\frac{\sqrt{3}}{2})=-\frac{3\sqrt{3}}{2}\approx -2.598$ (Absolute Minimum)

Final Answer:

The absolute maximum occurs at $x=\frac{\pi }{3}$ and the absolute minimum occurs at $x=\frac{5\pi }{3}$.

The correct option is (a).