$f(x)=\sin x+\frac{1}{2} \cos 2 x$ in $\left[0, \frac{\pi}{2}\right]$

(A) $f'(x)=\cos x-\sin 2 x$
(B) The critical points of the function are $x=\frac{\pi}{6}$ and $x=\frac{\pi}{2}$
(C) The minimum value of the function is 2
(D) The maximum value of the function is $\frac{3}{4}$

Choose the correct answer from the options given below:

(A), (B) and (D) only

The correct answer is Option (1) → (A), (B) and (D) only

$f(x)=\sin x+\frac{\cos 2 x}{2}$

$f'(x)=\cos x-\sin 2x=0$

$\cos x=2\sin x\cos x$

$\cos x=0,\sin x=\frac{1}{2}$

$x=\frac{π}{6},\frac{π}{2}$

critical points

$f(0)=\frac{1}{2}$ → minimum

$f(\frac{π}{6})=\frac{3}{4}$ → maximum

$f(\frac{π}{2})=\frac{1}{2}$

only A, B, D → correct