Question

$f(x) = \log|\sin x|$, where $x \in (0, \pi)$ is strictly increasing on

• A. $\left(\frac{\pi}{2}, \pi\right)$ only
• B. $(0, \pi)$ only
• C. $\left(0, \frac{\pi}{2}\right)$ only
• D. $\left(\frac{\pi}{4}, \frac{3\pi}{4}\right)$ only

Hint:

You can solve this conceptually without calculus! On the interval $\left(0, \frac{\pi}{2}\right)$, as $x$ increases, $\sin x$ increases from $0$ to $1$. Since $\log(x)$ is a continuously increasing function, $\log(\sin x)$ must also increase over this exact same range!

Answer 1

The Correct Option is C

Step 1: Differentiate the function.
For $f(x)=\log|\sin x|$ on $(0,\pi)$, $\sin x>0$ there, so $f(x)=\log(\sin x)$ and $f'(x)=\dfrac{\cos x}{\sin x}=\cot x$.
Step 2: State the increasing test.
$f$ is strictly increasing wherever $f'(x)>0$, i.e. where $\cot x>0$.
Step 3: Analyse $\cot x$ on $(0,\pi)$.
In the first quadrant $\left(0,\dfrac{\pi}{2}\right)$ both $\sin x$ and $\cos x$ are positive, so $\cot x>0$.
Step 4: Check the second quadrant.
On $\left(\dfrac{\pi}{2},\pi\right)$, $\cos x<0$ while $\sin x>0$, so $\cot x<0$ and $f$ decreases.
Step 5: Examine the midpoint.
At $x=\dfrac{\pi}{2}$, $\cot x=0$, marking the turning point of $f$.
Step 6: Conclude the interval.
Hence $f$ is strictly increasing only on $\left(0,\dfrac{\pi}{2}\right)$. \[ \boxed{\left(0,\tfrac{\pi}{2}\right)} \]