Question

$\text{ If } f(x) = \sin x + \frac{1}{2} \cos 2x \text{ in } \left[ 0, \frac{\pi}{2} \right], \text{ then:}$
(A) $f'(x) = \cos x - \sin 2x$
(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 $\frac{3}{4}$

• A. (A), (B), and (D) only
• B. (A), (B), and (C) only
• C. (B), (C), and (D) only
• D. (A), (C), and (D) only

Answer 1

Answer: (A)

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

Answer 2

$- \text{Differentiate } f(x) = \sin x + \frac{1}{2} \cos 2x \text{ to find } f'(x) = \cos x - \sin 2x, \text{ verifying option (A).}$
$- \text{Find the critical points by setting } f'(x) = 0, \text{ confirming option (B).}$
$- \text{Evaluate the function at critical points to confirm the minimum and maximum values, verifying option (D) and showing that (C) is incorrect.}$