Question

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

• A. Domain of $f\left(x\right)$ is $R-\left(2n+1\right)\frac{\pi }{2},n\in I$
• 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)

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