MCQ
In which of the following function is onto :
  • A
    $f: Z \rightarrow Z$,$
    f(x)=|x|
    $
  • B
    $f: N \rightarrow N$,$
    f(x)=|x|
    $
  • $f: R _0 \rightarrow R ^{+}, \quad f(x)=|x|$
  • D
    $f: C \rightarrow R$,$
    f(x)=|x|
    $

Answer

Correct option: C.
$f: R _0 \rightarrow R ^{+}, \quad f(x)=|x|$
(C) $f: R _0 \rightarrow R ^{+}, \quad f(x)=|x|$
$\because$ Every positive real number has exist the pre-image in domain $R _0$.

Need a full question paper?

Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.

Start Generating Free

Similar questions

If $K \in R_0$ then det. $ |{adj (KI_n)}|$ is equal to
If  m[-3, 4] + n[4, -3] = [10, -11] then 3m + 7n = 3m + 7n =
  1. 3
  2. 5
  3. 10
  4. 1
$A$ and $B$ are two square matrix such that $A^2B = BA$ and If $(AB)^{10} = A^K B^{10}$ then $k$ is
If $\int\frac{1}{(\text{x}+2)(\text{x}^2+1)}\text{ dx}=\text{a}\log|1+\text{x}^2|+\text{b}\tan^{-1}\text{x}+\frac{1}{5}\log|\text{x}+2|+\text{C},$ then
  1. $\text{a}=-\frac{1}{10},\text{ b}=\frac{2}{5}$
  2. $\text{a}=\frac{1}{10},\text{ b}=\frac{2}{5}$
  3. $\text{a}=-\frac{1}{10},\text{ b}=\frac{2}{5}$
  4. $\text{a}=\frac{1}{10},\text{ b}=\frac{2}{5}$
Define a function $f(x)=\frac{16 x^2-96 x+153}{x-3}$ for all real $x \neq 3$. The least positive value of $f(x)$ is
If the system of equations

$ 11 x+y+\lambda z=-5 $

$ 2 x+3 y+5 z=3 $

$ 8 x-19 y-39 z=\mu$

has infinitely many solutions, then $\lambda^4-\mu$ is equal to :

If $\frac{{dy}}{{dx}} = \frac{{xy + y}}{{xy + x}}$, then the solution of the differential equation is
If points $\text{A}(60\hat{\text{i}}+3\hat{\text{j}}),\ \text{B}(40\hat{\text{i}}-8\hat{\text{j}})$ and $\text{C}(\text{a}\hat{\text{i}}+52\hat{\text{j}})$ are collinear, then a is equal to,
  1. 40
  2. -40
  3. 20
  4. -20
If f(x) = |x2 - 9x + 20|, then f(x) is equal to:
  1. -2x + 9 for all $\text{x}\in\text{R}$
  2. 2x - 9 if 4 < x < 5
  3. -2x + 9, if 4 < x < 5
  4. None of these.
Let $\vec a = 2\hat i + \hat j - 2\hat k$ and $\vec b = \hat i + \hat j$ . Let $\vec c$ be vector such that $\left| {\vec c - \vec a} \right| = 3,\;\left| {\left( {\vec a \times \vec b} \right) \times \vec c} \right| = 3$ and the angle between $\vec c$ and $\vec a \times \vec b$ be $30^\circ $ . Then $\vec a \cdot \vec c$ is equal to :