CUET UG 2025 — Mathematics Algebra

To determine where $f(x) = \log_e(\sin x)$f(x)=loge​(sinx) is increasing or decreasing, find the derivative.

Using the chain rule:

$$f(x) = \log_e(\sin x)$$f(x)=loge​(sinx)

$$f'(x) = \frac{1}{\sin x} \times \cos x$$f′(x)=sinx1​×cosx

$$f'(x) = \frac{\cos x}{\sin x}$$f′(x)=sinxcosx​

$$f'(x) = \cot x$$f′(x)=cotx

---

For interval $(0, \frac{\pi}{2})$(0,2π​):

In this interval, $\sin x > 0$sinx>0 and $\cos x > 0$cosx>0.

Therefore $\cot x = \frac{\cos x}{\sin x} > 0$cotx=sinxcosx​>0

Since $f'(x) > 0$f′(x)>0, the function is strictly increasing on $(0, \frac{\pi}{2})$(0,2π​).

Statement (A) is correct.

---

For interval $(\frac{\pi}{2}, \pi)$(2π​,π):

In this interval, $\sin x > 0$sinx>0 and $\cos x < 0$cosx<0.

Therefore $\cot x = \frac{\cos x}{\sin x} < 0$cotx=sinxcosx​<0

Since $f'(x) < 0$f′(x)<0, the function is strictly decreasing on $(\frac{\pi}{2}, \pi)$(2π​,π).

Statement (D) is correct.

---

Statements (B), (C), and (E) are incorrect based on the analysis above.

The correct answer is (A) and (D).