Question bank/JEE Main/Mathematics

JEE MainMathematicsToughFunctions

Let f(x): (-1, 1) →\rightarrow B be a function defined by f(x)
=tan⁡−1(2x1−x2)= \tan^{- 1}\left( \frac{2x}{1 - x^{2}} \right), if f (x) is
bijective, then set B is

Show answer and solution
Let
tan⁡−1(x)=θ∈(−π4,π4)\tan^{- 1}(x) = \theta \in \left( - \frac{\pi}{4},\frac{\pi}{4} \right)
⇒tan⁡−1(2x1−x2)=tan⁡−1(2tan⁡θ1−tan⁡2θ)\Rightarrow \tan^{- 1}\left( \frac{2x}{1 - x^{2}} \right) = \tan^{- 1}\left( \frac{2\tan\theta}{1 - \tan^{2}\theta} \right)
=tan⁡−1(tan⁡2θ)= \tan^{- 1}\left( \tan{2\theta} \right)
−π4<θ<π4- \frac{\pi}{4} < \theta < \frac{\pi}{4}
−π2<2θ<π2- \frac{\pi}{2} < 2\theta < \frac{\pi}{2}
=2θ= 2\theta
⇒tan⁡−1(2x1−x2)=2tan⁡−1(x)\Rightarrow \tan^{- 1}\left( \frac{2x}{1 - x^{2}} \right) = 2\tan^{- 1}(x)
So f(x) = 2tan^-2 (x)
f′(x)=21+x2>0f^{'}(x) = \frac{2}{1 + x^{2}} > 0
So f(x) is increasing
⇒−π2<f(x)<π2\Rightarrow - \frac{\pi}{2} < f(x) < \frac{\pi}{2}

Source: Grass-G / M-18 Mathematics Paper 27 Nov Third.docx

More JEE Main questionsChapter practice

Need a video for this idea?

Scan any doubt from your notes and get a whiteboard explanation.

Scan a doubtView plans