Question

Let $y=f(x)$ be a function satisfying the equation ${\tan }^{-1}y={\tan }^{-1}x+C$ where $y=1$ when $x=0$. Range of the function $f(x)$, is

• A. $R-\{-1\}$
• B. $(-1,\infty )$
• C. $[1,\infty )$
• D. $(-\infty ,1]$

Answer 1

Answer: (B)

${\tan }^{-1}y={\tan }^{-1}x+C$
$x=0;y=1\Rightarrow C=\frac{\pi }{4}$
$\Rightarrow {\tan }^{-1}y={\tan }^{-1}x+\frac{\pi }{4}\Rightarrow <br/>$
note: even $\frac{-\pi }{2}<ta{n}^{-1}x+\frac{\pi }{4}<\frac{\pi }{2};\frac{-\pi }{2}<{\tan }^{-1}x<\frac{\pi }{4};-\infty <x<1\Rightarrow$ (A) $x<1$
Hence $y=\tan \left({\tan }^{-1}x+\frac{\pi }{4}\right)=\left(\frac{x+1}{1-x}\right)\Rightarrow x=\frac{y-1}{y+1}<1$
$\Rightarrow \frac{2}{y+1}>0\Rightarrow y\in (-1,\infty )$