Trigonometry & Inverse Trigonometry
General
Grade 12
Question:
<p>Which are correct?<br>(A) \(\cot^{-1}x=\tan^{-1}(1/x)\ \forall x\in\mathbb{R}\setminus\{0\}\)<br>(B) \(f(x)=\text{sgn}(e^x)\) is into<br>(C) \(f:\mathbb{R}^+\to\mathbb{R},\,f(x)=\sin x+x\) is odd<br>(D) \(f(x)=e^x/e^{[x]}\) is periodic</p>
cot⁻¹x=tan⁻¹(1/x) ∀x∈ R{0}
f(x)=sgn(eˣ) is into
f:R⁺→ R, f(x)=sinx+x is odd
f(x)=eˣ/e^[x] is periodic
Step-by-Step Solution
<div class="solution"><p>(A) False: for \(x<0\), \(\cot^{-1}x=\pi+\tan^{-1}(1/x)\).</p><p>(B) \(e^x>0\) always, so \(\text{sgn}(e^x)=1\), range={1}≠ℝ \to into ✓</p><p>(C) Domain is \(\mathbb{R}^+\) only -- no symmetry about 0 \to cannot be odd.</p><p>(D) \(f(x)=e^{x-[x]}=e^{\{x\}}\), period 1 ✓</p><p><strong>Answer: (B),(D)</strong></p><div class="trap-box"><strong>Trap:</strong> The cot⁻^1 vs tan⁻^1 identity is one of the most tested piecewise identities in ITF.<div class="key-concept"><strong>Key Concept:</strong> \(\cot^{-1}x=\tan^{-1}(1/x)\) only for \(x>0\); add \pi for \(x<0\)
Correct Answer: B,D