The function $f(x) = \sin 3x,x∈ [0,\frac{\pi}{2}]$
(A) is increasing on $[0,\frac{\pi}{6}]$
(B) is decreasing on $[\frac{\pi}{6},\frac{\pi}{2}]$
(C) is increasing on $[0,\frac{\pi}{2}]$
(D) is decreasing on $[0,\frac{\pi}{2}]$
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → (A) and (B) only
(A) is increasing on $[0,\frac{\pi}{6}]$
(B) is decreasing on $[\frac{\pi}{6},\frac{\pi}{2}]$
Given $f(x)=\sin 3x,\ x\in\left[0,\frac{\pi}{2}\right]$
$f'(x)=3\cos 3x$
Sign of $f'(x)$ on the interval: $3x\in\left[0,\frac{3\pi}{2}\right]$. $\cos 3x>0$ for $3x\in\left(0,\frac{\pi}{2}\right)$ i.e. $x\in\left(0,\frac{\pi}{6}\right)$; $\cos 3x=0$ at $x=\frac{\pi}{6}$ and $x=\frac{\pi}{2}$; $\cos 3x<0$ for $3x\in\left(\frac{\pi}{2},\frac{3\pi}{2}\right)$ i.e. $x\in\left(\frac{\pi}{6},\frac{\pi}{2}\right)$.
Therefore $f$ is increasing on $\left[0,\frac{\pi}{6}\right]$ and decreasing on $\left[\frac{\pi}{6},\frac{\pi}{2}\right]$ (derivative nonnegative on the first subinterval and nonpositive on the second, with zeros at the common endpoint).
Correct statements: (A) and (B).