Question

Let $\text{sgn}\left(y\right)$ and $\left\{y\right\}$ denote signum function of $y$ and fractional part function of $y$ respectively.
Which of the following functions is (are) bijective?

• A. $f:(-\infty ,0]\to (0,\frac{\pi }{2}]$defined by $f\left(x\right)={\sin }^{-1}\left(e^x\right)$
• B. $f:[-1,1]\to \{-1,0,1\}$ defined by $f\left(x\right)=\text{sgn}({\sin }^{-1}|x|-{\cos }^{-1}|x|)$
• C. $f:[-3,0]\to [\cos 3,1]$ defined by $f\left(x\right)=\cos x$
• D. $f:R-Z\to R$ defined by $f\left(x\right)=\ln \left\{x\right\}$

Answer 1

Answer: (A), (C)

$f:[-3,0]\to [\cos 3,1]$ defined by $f\left(x\right)=\cos x$

We have, $-\infty <x\le 0$
$\Rightarrow 0<e^x\le 1$
$\Rightarrow {\sin }^{-1}\left(e^x\right)\in (0,\frac{\pi }{2}]$
Since $e^x$ and ${\sin }^{-1}x$ are one-one, therefore ${\sin }^{-1}\left(e^x\right)$ is also one-one.
$\therefore f\left(x\right)={\sin }^{-1}\left(e^x\right)$ is one-one and onto.
Hence, $f\left(x\right)$ is bijective.


$f\left(x\right)=\text{sgn}({\sin }^{-1}|x|-{\cos }^{-1}|x|)$
Since $f(-x)=f\left(x\right),$ therefore $f\left(x\right)$ is many-one.
And range of $f$ is $\{-1,0,1\}.$
$\therefore f\left(x\right)$ is onto.
Hence, $f\left(x\right)$ is not bijective.


$f\left(x\right)=\cos x$

From the graph, it is clear that $f\left(x\right)$ is one-one and onto.
Hence, $f\left(x\right)$ is bijective.


$f\left(x\right)=\ln \left\{x\right\}$
$f\left(1.2\right)=\ln \left(0.2\right)$ and $f\left(2.2\right)=\ln \left(0.2\right),$ therefore $f\left(x\right)$ is many-one.
Range of $f$ is $(-\infty ,0)$
$\therefore f$ is many-one and into.
Hence, $f\left(x\right)$ is not bijective.