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	$\sin ^{-1}\left(3 x-4 x^3\right), x \in\left[-\frac{1}{2}, \frac{1}{2}\right]$	1	$\tan ^{-1} \frac{31}{17}$
B	$\tan ^{-1} \frac{2}{11}+\tan ^{-1} \frac{7}{24}$	2	$3 \sin ^{-1} x$
C	$\cos ^{-1}\left(4 x^3-3 x\right), x \in\left[\frac{1}{2}, 1\right]$	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(3 x-4 x^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-x y}\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
$\cos ^{-1}\left(4 x^3-3 x\right)=\cos ^{-1}\left[4 \cos ^3 \theta-3 \cos \theta\right]$
$=\cos ^{-1}[\cos 3 \theta]=3 \theta$
$=3 \cos ^{-1} x \left(\because \cos 3 \theta=4 \cos ^3 \theta-3 \cos \theta\right)$
D. $2 \tan ^{-1}\left(\frac{1}{2}\right)+\tan ^{-1}\left(\frac{1}{7}\right)=\tan ^{-1}\left(\frac{2 \times \frac{1}{2}}{1-\left(\frac{1}{2}\right)^2}\right)+\tan ^{-1}\left(\frac{1}{7}\right)$
$\left[\because 2\tan^{-1} x = \tan^{-1}\left(\frac{2x}{1-x^2}\right)\right]$
$=\tan ^{-1} \frac{1}{1-\frac{1}{4}}+\tan ^{-1} \frac{1}{7}=\tan ^{-1}\left(\frac{4}{3}\right)+\tan ^{-1}\left(\frac{1}{7}\right)$
$=\tan ^{-1}\left(\frac{\frac{4}{3}+\frac{1}{7}}{1-\frac{4}{3} \times \frac{1}{7}}\right)$
$\left[\because \tan ^{-1} x+\tan ^{-1} y=\tan ^{-1}\left(\frac{x+y}{1-x y}\right)\right]$
$=\tan ^{-1}\left(\frac{\frac{28+3}{21}}{1-\frac{4}{21}}\right)=\tan ^{-1}\left(\frac{\frac{31}{21}}{\frac{17}{21}}\right)$
$=\tan ^{-1}\left(\frac{31}{21} \times \frac{21}{17}\right)=\tan ^{-1}\left(\frac{31}{17}\right)$