$f(x)=\sin x+\frac{1}{2} \cos 2 x$ 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:
Answer & explanation
Correct answer: option 1
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