NIMCET 2015 — Mathematics PYQ

1. Find the First Derivative:

Given $f(x)=2\sin x+\cos 2x$.

Differentiating with respect to $x$:

Using the identity $\sin 2x=2\sin x\cos x$:

2. Find Critical Points:

Set $f^{\prime }(x)=0$:

This gives two cases:

• Case 1: $\cos x=0⟹x=\frac{\pi }{2}$ (Since $x\in (0,\pi )$)

• Case 2: $1-2\sin x=0⟹\sin x=\frac{1}{2}⟹x=\frac{\pi }{6},\frac{5\pi }{6}$

The critical points in the interval $(0,\pi )$ are $x=\frac{\pi }{6},\frac{\pi }{2},\text{ and }\frac{5\pi }{6}$.

3. Second Derivative Test:

Find $f^{\prime \prime }(x)$:

Now, check the sign of $f^{\prime \prime }(x)$ at the critical points:

• At $x=\frac{\pi }{6}$:

  f''\left(\frac{\pi}{6}\right) = -2\left(\frac{1}{2}\right) - 4\cos\left(\frac{\pi}{3}\right) = -1 - 4\left(\frac{1}{2}\right) = -3 < 0 \text{ (Local Maximum)}

• At $x=\frac{\pi }{2}$:

  f''\left(\frac{\pi}{2}\right) = -2\sin\left(\frac{\pi}{2}\right) - 4\cos(\pi) = -2(1) - 4(-1) = -2 + 4 = 2 > 0 \text{ (Local Minimum)}

• At $x=\frac{5\pi }{6}$:

  f''\left(\frac{5\pi}{6}\right) = -2\left(\frac{1}{2}\right) - 4\cos\left(\frac{5\pi}{3}\right) = -1 - 4\left(\frac{1}{2}\right) = -3 < 0 \text{ (Local Maximum)}

Conclusion:

The function has local maxima at $x=\frac{\pi }{6}$ and $x=\frac{5\pi }{6}$, and a local minimum at $x=\frac{\pi }{2}$. Therefore, it has both local minimum and local maximum.

Correct Option:

(c) Both local minimum and local maximum