Question

If $f\left(tan x\right)=sin2x:x\neq \left(2 n + 1\right)\frac{\pi }{2},n\in Ι$ , then which of the following is an incorrect statement?

• A. Domain of $f\left(x\right)$ is $R-\left(2 n + 1\right)\frac{\pi }{2},n\in Ι$
• B. Range of $f\left(x\right)$ is $\left[- 1,1\right]$
• C. $f\left(x\right)$ is an odd function
• D. $f\left(x\right)$ is many-one function

Answer 1

Answer: (A)

$f\left(tan x\right)=\frac{2 tan x}{1 + \left(tan\right)^{2} x}\Rightarrow f\left(x\right)=\frac{2 x}{1 + x^{2}}$ Now, domain of $f\left(x\right)$ is $x\in R$
For the range of
$f\left(x\right)\Rightarrow y=\frac{2 x}{1 + x^{2}}\Rightarrow yx^{2}-2x+y=0$
$\because x\in R\therefore D\geq 0$
$4-4y^{2}\geq 0\Rightarrow y\in \left[- 1,1\right]$ is the range
Also, $f\left(- x\right)=\frac{2 \left(- x\right)}{1 + \left(- x\right)^{2}}=-f\left(x\right)$
$\therefore f\left(x\right)$ is an odd function
$f^{'}\left(x\right)=\frac{\left(1 + x^{2}\right) 2 - 2 x \left(2 x\right)}{\left(1 + x^{2}\right)^{2}}=\frac{2 \left(1 - x^{2}\right)}{\left(1 + x^{2}\right)^{2}}$
$f^{'}\left(x\right)$ can be positive & negative $\forall x\in R$
$\therefore f\left(x\right)$ is many-one