Trigonometry & Inverse Trigonometry Question - None

Question

Which are correct?
(A) $\cot^{-1}x=\tan^{-1}(1/x)\ \forall x\in\mathbb{R}\setminus\{0\}$
(B) $f(x)=\text{sgn}(e^x)$ is into
(C) $f:\mathbb{R}^+\to\mathbb{R},\,f(x)=\sin x+x$ is odd
(D) $f(x)=e^x/e^{[x]}$ is periodic

Accepted Answer

(A) False: for $$x<0$$ , $\cot^{-1}x=\pi+\tan^{-1}(1/x)$ .

(B) $$e^x>0$$ always, so $\text{sgn}(e^x)=1$ , range={1}≠ℝ → into ✓

(C) Domain is $\mathbb{R}^+$ only — no symmetry about 0 → cannot be odd.

(D) $f(x)=e^{x-[x]}=e^{\{x\}}$ , period 1 ✓

Answer: (B),(D)

Trap: The cot⁻¹ vs tan⁻¹ identity is one of the most tested piecewise identities in ITF.

Key Concept:

\cot^{-1}x=\tan^{-1}(1/x)

only for

$x>0$

; add π for

$x<0$

Question:

Step-by-Step Solution

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