Which of the following are injections or surjections or bijections ? Justify your answers.

$\text{ i) }f:R\to R\text{ defined by }f\left(x\right)=\frac{2x+1}{3}$

$\text{ ii) }f:R\to [0,\mathrm{\infty })\text{ defined by }f\left(x\right)=x^2$

$\text{ iii) }f:R\to (0,\mathrm{\infty })\text{ defined by }f\left(x\right)=2^x$

$\text{ iv) }f:(0,\mathrm{\infty })\to R\text{ defined by }f\left(x\right)=\log x$

$\text{ i) }{x}_1,x_2\in R\text{ and }f\left(x_1\right)=f\left(x_2\right)$

$\Rightarrow \frac{2x_1+1}{3}=\frac{2x_2+1}{3}\Rightarrow x_1=x_2$

$\therefore f:R\to R\text{ is an injection }$

Let $y\in R,f\left(x\right)=y\Rightarrow \frac{2x+1}{3}=y\Rightarrow 2x+1=3y$

$\Rightarrow x=\frac{3y-1}{2}\Rightarrow f\left(x\right)=\frac{2\left(\frac{3y-1}{2}\right)+1}{3}=y$

Let $y\in R,f\left(x\right)=y\Rightarrow \frac{2x+1}{3}=y\Rightarrow 2x+1=3y$

$\Rightarrow x=\frac{3y-1}{2}\Rightarrow f\left(x\right)=\frac{2\left(\frac{3y-1}{2}\right)+1}{3}=y$

$\therefore f:R\to R\text{ is a surjection }$

$\therefore f:R\to R\text{ is a bijection. }$

$\text{ ii) }f(-2)=4,f\left(2\right)=4$

Different elements in R have same

f -image in $[0,\mathrm{\infty });f:R\to [0,\mathrm{\infty })$ is not one-one

Let $y\in [0,\mathrm{\infty })\text{ thus }y>0$

Let $x=\sqrt{y}\in R,\text{ now }f\left(x\right)=x^2$

$(\sqrt{y}{)}^2=y$

$\therefore f:R\to [0,\mathrm{\infty })$ is onto

$\therefore f:R\to [0,\mathrm{\infty })$ is a sujection but not injection

$\text{ iii) }{x}_1,x_2\in R,f\left(x_1\right)=f\left(x_2\right)$

$2^{x_1}=2^{x_2}\Rightarrow x_1=x_2$

$\therefore f:R\to (0,\mathrm{\infty })$ is an injection

If $y\in (0,\mathrm{\infty })$ thus $x={\log }_2^y\in R$

$f\left(x\right)=2^x\Rightarrow 2^{{\log }_2^y}=y$

$\therefore f:R\to (0,\mathrm{\infty })$ is a surjection'

$\therefore f:R\to (0,\mathrm{\infty })$ is a bijection

$\text{ iv) }{x}_1,x_2\in (0,\mathrm{\infty }),f\left(x_1\right)=f\left(x_2\right)$

$\Rightarrow {\log }_e{x}_1={\log }_e{x}_2;x_1=x_2$

$\therefore f:(0,\mathrm{\infty })\to R$ is an injection.

$y\in R,x=e^y\in (0,\mathrm{\infty })$

$\Rightarrow f\left(x\right)={\log }_{e}x\Rightarrow {\log }_e{e}^y=y$

$\therefore f:(0,\mathrm{\infty })\to R$ is a surjection

$\therefore f:(0,\mathrm{\infty })\to R$ is bijection