Question

The range of the function $y = \log(\sin x)$ where $\sin x > 0$ is:

• A. $[0, \infty)$
• B. $(-\infty, \infty)$
• C. $(-\infty, 0]$
• D. $[-1, 1]$

Hint:

The log of a value between 0 and 1 is always negative. Since $\sin x$ never exceeds 1, its log can never be positive.

Answer 1

The Correct Option is C

Step 1: Concept
We determine the range by analyzing the composite function $f(g(x))$ where $g(x) = \sin x$ and $f(u) = \log u$.

Step 2: Meaning
Since the domain is restricted to $\sin x > 0$, we know that for any real $x$, the value of the sine function is bounded such that $0 < \sin x \leq 1$.

Step 3: Analysis
The natural logarithm function $y = \log u$ is monotonically increasing. As the argument $u$ ranges from values approaching $0$ up to $1$, $\log u$ ranges from $-\infty$ to $\log(1) = 0$.

Step 4: Conclusion
Therefore, the outputs of the function cover all values from negative infinity up to and including zero, making the range $(-\infty, 0]$. Final Answer: (C)