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: Understanding the Question:
We are given a logarithmic trigonometric function $f(x) = \log|\sin x|$ on the open interval $(0, \pi)$. We need to determine the specific sub-interval where this function is strictly increasing.

Step 2: Key Formula or Approach:
A continuous function $f(x)$ is strictly increasing on an interval if its first derivative with respect to $x$ is strictly positive for all points in that interval: $$f'(x) > 0$$ We will use the chain rule to differentiate the function: $\frac{d}{dx}(\log|g(x)|) = \frac{g'(x)}{g(x)}$.

Step 3: Detailed Explanation:
Differentiate the given function $f(x) = \log|\sin x|$: $$f'(x) = \frac{1}{\sin x} \cdot \frac{d}{dx}(\sin x) = \frac{\cos x}{\sin x} = \cot x$$ Now, analyze where $f'(x) > 0$, which means finding where: $$\cot x > 0$$ We look within the total domain boundary given as $x \in (0, \pi)$:

• In the first quadrant, $x \in \left(0, \frac{\pi}{2}\right)$, the cotangent function is strictly positive ($\cot x > 0$). Thus, $f'(x) > 0$.

• In the second quadrant, $x \in \left(\frac{\pi}{2}, \pi\right)$, the cotangent function is strictly negative ($\cot x < 0$). Thus, $f'(x) < 0$.
Since $f'(x) > 0$ exclusively when $x$ is in the first quadrant, the function is strictly increasing on the sub-interval $\left(0, \frac{\pi}{2}\right)$.

Step 4: Final Answer:
The function is strictly increasing on $\left(0, \frac{\pi}{2}\right)$ only, which corresponds to option (C).