Question

Match the terms of column I with the terms of column II and choose the correct option from the codes given below.

Column I		Column II
A	${\sin }^{-1}\left(3x-4x^3\right),x\in \left[-\frac{1}{2},\frac{1}{2}\right]$	1	${\tan }^{-1}\frac{31}{17}$
B	${\tan }^{-1}\frac{2}{11}+{\tan }^{-1}\frac{7}{24}$	2	$3{\sin }^{-1}x$
C	${\cos }^{-1}\left(4x^3-3x\right),x\in \left[\frac{1}{2},1\right]$	3	${\tan }^{-1}\frac{1}{2}$
D	$2{\tan }^{-1}\frac{1}{2}+{\tan }^{-1}\frac{1}{7}$	4	$3{\cos }^{-1}x$

• A. A - (2), B - (1), C - (4), D - (3)
• B. A - (3), B - (1), C - (4), D - (2)
• C. A - (3), B - (4), C - (2), D - (1)
• D. A - (2), B - (3), C - (4), D - (1)

Answer 1

Answer: (D)

A. Let ${\sin }^{-1}x=\theta \Rightarrow x=\sin \theta$, then
${\sin }^{-1}\left(3x-4x^3\right)={\sin }^{-1}\left[3\sin \theta -4{\sin }^3\theta \right]$
$={\sin }^{-1}[\sin 3\theta ]=3\theta$
$\left(\because \sin 3\theta =3\sin \theta -4{\sin }^3\theta \right)$
$=3{\sin }^{-1}x$
B. ${\tan }^{-1}\left(\frac{2}{11}\right)+{\tan }^{-1}\left(\frac{7}{24}\right)={\tan }^{-1}\left(\frac{\frac{2}{11}+\frac{7}{24}}{1-\frac{2}{11}\cdot \frac{7}{24}}\right)$
$\left[\because {\tan }^{-1}x+{\tan }^{-1}y={\tan }^{-1}\left(\frac{x+y}{1-xy}\right)\right]$
$={\tan }^{-1}\left(\frac{\frac{48+77}{264}}{1-\frac{14}{264}}\right)={\tan }^{-1}\left(\frac{\frac{125}{264}}{\frac{264-14}{264}}\right)={\tan }^{-1}\frac{\frac{125}{264}}{\frac{250}{264}}$
$={\tan }^{-1}\left(\frac{125}{264}\times \frac{264}{250}\right)$
$={\tan }^{-1}\left(\frac{1}{2}\right)$
C. Let $x=\cos \theta \Rightarrow \theta ={\cos }^{-1}x$, then
${\cos }^{-1}\left(4x^3-3x\right)={\cos }^{-1}\left[4{\cos }^3\theta -3\cos \theta \right]$
$={\cos }^{-1}[\cos 3\theta ]=3\theta$
$=3{\cos }^{-1}x\left(\because \cos 3\theta =4{\cos }^3\theta -3\cos \theta \right)$
D. $2{\tan }^{-1}\left(\frac{1}{2}\right)+{\tan }^{-1}\left(\frac{1}{7}\right)={\tan }^{-1}\left(\frac{2\times \frac{1}{2}}{1-{\left(\frac{1}{2}\right)}^2}\right)+{\tan }^{-1}\left(\frac{1}{7}\right)$
$\left[\because 2{\tan }^{-1}x={\tan }^{-1}\left(\frac{2x}{1-x^2}\right)\right]$
$={\tan }^{-1}\frac{1}{1-\frac{1}{4}}+{\tan }^{-1}\frac{1}{7}={\tan }^{-1}\left(\frac{4}{3}\right)+{\tan }^{-1}\left(\frac{1}{7}\right)$
$={\tan }^{-1}\left(\frac{\frac{4}{3}+\frac{1}{7}}{1-\frac{4}{3}\times \frac{1}{7}}\right)$
$\left[\because {\tan }^{-1}x+{\tan }^{-1}y={\tan }^{-1}\left(\frac{x+y}{1-xy}\right)\right]$
$={\tan }^{-1}\left(\frac{\frac{28+3}{21}}{1-\frac{4}{21}}\right)={\tan }^{-1}\left(\frac{\frac{31}{21}}{\frac{17}{21}}\right)$
$={\tan }^{-1}\left(\frac{31}{21}\times \frac{21}{17}\right)={\tan }^{-1}\left(\frac{31}{17}\right)$