JEE Main 12 Jan 2019 Morning Question 5
Question: Considering only the principal values of inverse functions, the set
$ A={ x\ge 0:{{\tan }^{-1}}(2x)+ta{n^{-1}}(3x)=\frac{\pi }{4} } $ [JEE Main Online Paper Held On 12-Jan-2019 Morning]
Options:
A) contains two elements
B) contains more than two elements
C) is an empty set
D) is a singleton
Show Answer
Answer:
Correct Answer: D
Solution:
- Here, $ A={ x\ge 0:{{\tan }^{-1}}(2x)+\tan {{,}^{-1}}(3x)=\frac{\pi }{4} } $ Now, $ \tan {{,}^{-1}}(2x)+ta{n^{-3}}(3x)=\frac{\pi }{4} $
$ \Rightarrow $ $ {{\tan }^{-1}}( \frac{2x+3x}{1-2x\times 3x} )=\frac{\pi }{4} $
$ \Rightarrow $ $ {{\tan }^{-1}}( \frac{5x}{1-6x^{2}} )=\frac{\pi }{4} $
$ \Rightarrow $ $ \frac{5x}{1-6x^{2}}=\tan \frac{\pi }{4}=1 $
$ \Rightarrow $ $ 5x=1-6x^{2} $
$ \Rightarrow $ $ 6x^{2}+5x-1=0 $
$ \Rightarrow $ $ (6x-1)(x+1)=0 $
$ \Rightarrow $ $ x=\frac{1}{6} $ $ [\because x\ge 0] $
