Exercise 4.6 — Book-Back Answer Key
Exercise 4.6 — Book-Back Answer Key
Twenty multiple-choice questions, each with the correct option and a one-line justification. Every answer below was independently re-worked and verified.
| Q | Answer | Why |
|---|---|---|
| 1 | (c) \(\frac{\pi}{2}-x\) | \(\cos x=\sin(\frac{\pi}{2}-x)\), and \(\frac{\pi}{2}-x\) stays in arcsine's range for \(x\in[0,\pi]\). |
| 2 | (b) \(\frac{\pi}{3}\) | Adding \(\sin^{-1}+\cos^{-1}=\frac{\pi}{2}\) twice gives total \(\pi\); subtract \(\frac{2\pi}{3}\). |
| 3 | (c) \(0\) | \(\sec^{-1}\frac{5}{3}=\cos^{-1}\frac{3}{5}\), \(\operatorname{cosec}^{-1}\frac{13}{12}=\sin^{-1}\frac{12}{13}\); the two pairs cancel. |
| 4 | (a) \(|\alpha|\le\frac{1}{\sqrt{2}}\) | Need \(2\sin^{-1}\alpha\in[-\frac{\pi}{2},\frac{\pi}{2}]\), so \(\sin^{-1}\alpha\in[-\frac{\pi}{4},\frac{\pi}{4}]\). |
| 5 | (b) \(0\le x\le\pi\) | The formula holds only while \(\frac{\pi}{2}-x\) lies in arcsine's range. |
| 6 | (a) \(0\) | Sum \(\frac{3\pi}{2}\) forces \(x=y=z=1\); then \(3-\frac{9}{3}=0\). |
| 7 | (c) \(\frac{\pi}{10}\) | \(\tan^{-1}x=\frac{\pi}{2}-\cot^{-1}x=\frac{\pi}{2}-\frac{2\pi}{5}\). |
| 8 | (a) \([1,2]\) | Need \(0\le\sqrt{x-1}\le 1\), so \(1\le x\le 2\). |
| 9 | (d) \(-\frac{1}{5}\) | Angle \(=\frac{\pi}{2}+\sin^{-1}x\); its cosine is \(-x\). |
| 10 | (d) \(\tan^{-1}\frac{1}{2}\) | Arctangent addition gives \(\frac{a+b}{1-ab}=\frac{1}{2}\). |
| 11 | (c) \([-2,-\sqrt{2}]\cup[\sqrt{2},2]\) | \(-1\le x^{2}-3\le 1\Rightarrow 2\le x^{2}\le 4\). |
| 12 | (b) \(\frac{3\pi}{4}\) | \(\cot^{-1}2+\cot^{-1}3=\frac{\pi}{4}\), so third angle \(=\pi-\frac{\pi}{4}\). |
| 13 | (b) \(x^{2}-x-12=0\) | The equation yields \(x=4\), which is a root of this quadratic. |
| 14 | (a) \(\frac{\pi}{2}\) | Both arguments equal \(\cos 2x\); then \(\sin^{-1}t+\cos^{-1}t=\frac{\pi}{2}\). |
| 15 | (c) \(-1\) | \(\cot^{-1}t+\tan^{-1}t=\frac{\pi}{2}\Rightarrow u=\frac{\pi}{2}\), \(\cos\pi=-1\). |
| 16 | (c) \(0\) | \(\sin^{-1}\frac{2x}{1+x^{2}}=2\tan^{-1}x\) for \(|x|\le 1\); the terms cancel. |
| 17 | (b) unique solution | Combined with \(\tan^{-1}x+\cot^{-1}x=\frac{\pi}{2}\) it gives \(x=\sqrt{3}\) only. |
| 18 | (b) \(\frac{1}{\sqrt{5}}\) | \(\sin^{-1}x=\tan^{-1}\frac{1}{2}\); the \(1\text{–}2\text{–}\sqrt{5}\) triangle gives \(\sin=\frac{1}{\sqrt{5}}\). |
| 19 | (d) \(3\) | \(\operatorname{cosec}^{-1}\frac{5}{4}=\sin^{-1}\frac{4}{5}\), then \(\sin^{-1}\frac{x}{5}=\sin^{-1}\frac{3}{5}\). |
| 20 | (d) \(\frac{x}{\sqrt{1+x^{2}}}\) | Triangle with legs \(x,1\) and hypotenuse \(\sqrt{1+x^{2}}\). |
All twenty supplied answers were confirmed correct; no corrections were needed.