Question

The function \[ f(x)=\sin 2x+2\cos x,\qquad x\in\left(-\frac{3\pi}{4},\frac{3\pi}{4}\right) \] has:

• A. no critical point
• B. a point of local maxima and a point of local minima
• C. a point of local maxima and a point of inflection
• D. a point of local minima and a point of inflection

Hint:

A point where \(f'(x)=0\) but no sign change in monotonicity occurs is not an extremum — always check the second derivative or sign of \(f'(x)\).

Answer 1

The Correct Option is C

Step 1: Find the first derivative \[ f(x)=\sin 2x+2\cos x \] \[ f'(x)=2\cos 2x-2\sin x \] Set \(f'(x)=0\): \[ 2(\cos 2x-\sin x)=0 \Rightarrow \cos 2x=\sin x \] Using \(\cos 2x=1-2\sin^2 x\): \[ 1-2\sin^2 x=\sin x \] \[ 2\sin^2 x+\sin x-1=0 \] \[ (2\sin x-1)(\sin x+1)=0 \] \[ \sin x=\frac12 \quad \text{or} \quad \sin x=-1 \] In the interval \(\left(-\frac{3\pi}{4},\frac{3\pi}{4}\right)\), \[ \sin x=\frac12 \Rightarrow x=\frac{\pi}{6} \] (\(\sin x=-1\) gives \(x=-\frac{\pi}{2}\), which is a boundary point for critical behaviour.) 
Step 2: Second derivative test \[ f''(x)=-4\sin 2x-2\cos x \] At \(x=\frac{\pi}{6}\): \[ f''\!\left(\frac{\pi}{6}\right) =-4\sin\frac{\pi}{3}-2\cos\frac{\pi}{6} =-4\cdot\frac{\sqrt3}{2}-2\cdot\frac{\sqrt3}{2}<0 \] Hence, \(x=\frac{\pi}{6}\) is a point of local maxima. 
Step 3: Check for inflection point At \(x=-\frac{\pi}{2}\), \[ f'(x)=0 \] but \(f''(x)\) changes sign across this point, hence it is a point of inflection. Final Conclusion: The function has a point of local maxima and a point of inflection. \[ \boxed{\text{Option (3)}} \]