Trigonometry & Inverse Trigonometry Question - None

Question

Range of $f(x)=\cot^{-1}(\log_e(1-x^2))$ is:

Accepted Answer

Step 1: Need $1-x^2>0\implies x\in(-1,1)$ .

Step 2: $1-x^2\in(0,1]$ , so $\log_e(1-x^2)\in(-\infty,0]$ .

Step 3: $\cot^{-1}$ is decreasing: $\cot^{-1}(0)=\pi/2$ and $\lim_{u o-\infty}\cot^{-1}(u)=\pi$ .

Answer: (C) $[\pi/2,\pi)$

Trap: Forgetting that

\cot^{-1}

is decreasing — the range flips direction.

Key Concept: Decreasing functions reverse inequality direction in range calculations

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