Question

If $f:R\to R$ defined by $f\left(x\right)=\{\begin{array}{c}2x+5,x>0\\ 3x-2,x\le 0\end{array}$ then $f$ is

• A. many-one function
• B. one-one function
• C. onto function
• D. into function

Answer 1

Answer: (B), (D)

into function

$f:R\to R$
$f\left(x\right)=\{\begin{array}{c}2x+5,x>0\\ 3x-2,x\le 0\end{array}$
$\Rightarrow f^{\prime }\left(x\right)=\{\begin{array}{c}2,x>0\\ 3,x\le 0\end{array}$
$\because f^{\prime }\left(x\right)>0\forall x\in R$
So, $f$ is strictly increasing function
Therefore, $f$ is one-one function.
Now,
When $f\left(x\right)=2x+5$ for $x>0$ its range lies from $(5,\infty ).$
Similarly when $f\left(x\right)=3x-2$ for $x\le 0$ its range lies from $(-\infty ,-2].$
So,Range of $f\left(x\right)=(-\infty ,-2]\cup (5,\infty )$
$\therefore f\left(x\right)$ is not onto function because the function range is not equal to its co-domain.