Q.Show that the function $f: \mathbf{R}_* \to \mathbf{R}_*$ defined by $f(x) = \frac{1}{x}$ is one-one and onto, where $\mathbf{R}_*$ is the set of all non-zero real numbers. Is the result true, if the domain $\mathbf{R}_*$ is replaced by $\mathbf{N}$ with co-domain being same as $\mathbf{R}_*$?

Showing the 12 most recent of 45 on this concept.

1. CBSE 2026Set CX1 markMCQ
   Q.The function $f(x)=2x$, $x\in R$ is:

   (a) one-one but not onto

   (b) one-one and onto

   (c) many-one and onto

   (d) many-one but not onto

   ›Reveal solutionSolution

   $f(x)=2x$ on $\mathbb{R}$ is a bijection — both one-one and onto — option (b).

   One-one: If $f(x_1)=f(x_2)$ then $2x_1=2x_2\Rightarrow x_1=x_2$. So $f$ is injective.

   Onto: For any $y\in\mathbb{R}$, choose $x=\dfrac{y}{2}\in\mathbb{R}$; then $f(x)=2\cdot\dfrac{y}{2}=y$. So every element of the codomain is attained — $f$ is surjective.

   Being both, $f$ is a bijection.

   ✓Final answer

   Option (b) one-one and onto.

2. CBSE 2026Set ANNUAL1 markMCQ
   Q.Let $f: R \to R$ defined as $f(x) = 3 - 4x$, then $f(x)$ is:

   (a) one-one onto

   (b) onto only

   (c) neither one-one nor onto

   (d) none of these

   ›Reveal solutionSolution

   $f(x)=3-4x$ is a linear function with non-zero slope, so it is both one-one and onto.

   Given $f:\mathbb{R}\to\mathbb{R}$, $f(x)=3-4x$.

   One-one: Let $f(x_1)=f(x_2)$. Then $3-4x_1=3-4x_2 \Rightarrow x_1=x_2$. So $f$ is injective.

   Onto: For any $y\in\mathbb{R}$, solve $3-4x=y \Rightarrow x=\dfrac{3-y}{4}$, which is a real number for every real $y$. So $f$ is surjective.

   Being both one-one and onto, $f$ is bijective.

   ✓Final answer

   Option (a): one-one onto.

3. CBSE 2026Set ANNUAL1 markMCQ
   Q.If A = {0, 1, 4, 9, 16, 25, ......} then function defined by f: Z → A, f(x) = x² is:

   (a) one-one but not onto

   (b) onto but not one-one

   (c) one-one and onto

   (d) neither one-one nor onto

   ›Reveal solutionSolution

   Two different integers with the same absolute value (like $2$ and $-2$) give the same square, so $f$ is not one-one; but every element of $A$ is a perfect square that some integer squares to, so $f$ is onto.

   $f:\mathbb{Z}\to A$ is defined by $f(x) = x^2$, where $A=\{0,1,4,9,16,25,\dots\}$ is the set of all perfect squares of non-negative integers.

   One-one check: Take $x=2$ and $x=-2$. Both are in $\mathbb{Z}$ and $f(2) = 4 = f(-2)$, but $2\ne -2$. Different inputs give the same output — $f$ is not one-one.

   Onto check: For every $a \in A$, $a$ is a perfect square, say $a = k^2$ for some non-negative integer $k$. Since $k \in \mathbb{Z}$, we have $f(k) = k^2 = a$. Every element of $A$ has a pre-image in $\mathbb Z$ — $f$ is onto.

   ✓Final answer

   $f$ is onto but not one-one (option b).

4. CBSE 2026Set ANNUAL1 markMCQ
   Q.Let f : R → R be defined by f(x) = 3x, choose the correct answer:

   (a) f is one-one onto

   (b) f is many-one onto

   (c) f is one-one but not onto

   (d) f is neither one-one nor onto

   ›Reveal solutionSolution

   $f(x)=3x$ is a straight-line map with non-zero slope, so it is both injective and surjective on $\mathbb{R}$ — a bijection.

   Checking one-one (injective):

   Let $f(x_1)=f(x_2)$. Then $3x_1=3x_2 \Rightarrow x_1=x_2$. So distinct inputs never share an output — $f$ is one-one.

   Checking onto (surjective):

   Take any $y \in \mathbb{R}$. Choosing $x=y/3 \in \mathbb{R}$ gives $f(x)=3(y/3)=y$. Every real number has a pre-image, so $f$ is onto.

   Being both one-one and onto, $f$ is a bijection.

   ✓Final answer

   $f$ is one-one onto. (Option a)

5. CBSE 2026Set ANNUAL1 mark
   Q.Prove that the function $f : \mathbb{R} \to \mathbb{R}$, given by $f(x)=2x$, is both one-one and onto.

   ›Reveal solutionSolution

   Prove injectivity by showing $f(x_1)=f(x_2)$ forces $x_1=x_2$, and prove surjectivity by exhibiting a pre-image for an arbitrary $y\in\mathbb{R}$.

   Given $f:\mathbb{R}\to\mathbb{R}$, $f(x)=2x$.

   Step 1: One-one (injective).

   Let $x_1,x_2\in\mathbb{R}$ such that $f(x_1)=f(x_2)$.

   $2x_1 = 2x_2 \implies x_1 = x_2$

   So $f(x_1)=f(x_2) \implies x_1=x_2$, hence $f$ is one-one.

   Step 2: Onto (surjective).

   Let $y\in\mathbb{R}$ be arbitrary (any element of the codomain). We need $x\in\mathbb{R}$ (the domain) such that $f(x)=y$.

   Take $x = \dfrac{y}{2}$. Since $y\in\mathbb{R}$, $x=\dfrac{y}{2}\in\mathbb{R}$ too, and

   $f(x) = f\left(\frac{y}{2}\right) = 2\cdot\frac{y}{2} = y$

   So every $y\in\mathbb{R}$ has a pre-image in $\mathbb{R}$, hence $f$ is onto.

   Since $f$ is both one-one and onto, $f$ is bijective.

   ✓Final answer

   $f(x)=2x$ is both one-one and onto (bijective).

6. CBSE 2026Set ANNUAL1 markMCQ
   Q.Let $f : \mathbb{R} \to \mathbb{R}$ be defined as $f(x)=x^4$. Then

   (a) $f$ is one-one and onto

   (b) $f$ is many-one and onto

   (c) $f$ is one-one but not onto

   (d) $f$ is neither one-one nor onto

   ›Reveal solutionSolution

   Find a counter-example pair with equal outputs to disprove one-one, and note the range excludes negative numbers to disprove onto.

   Given $f:\mathbb{R}\to\mathbb{R}$, $f(x)=x^4$.

   One-one? Consider $x_1=1$ and $x_2=-1$:

   $f(1) = 1^4 = 1, \qquad f(-1) = (-1)^4 = 1$

   $f(1)=f(-1)=1$ but $1\ne-1$. So $f$ is not one-one (it is many-one).

   Onto? Since $x^4\ge0$ for every real $x$, the range of $f$ is $[0,\infty)$, which is a proper subset of the codomain $\mathbb{R}$. For example, $y=-1$ has no pre-image ($x^4=-1$ has no real solution). So $f$ is not onto.

   Hence $f$ is neither one-one nor onto.

   ✓Final answer

   (d) $f$ is neither one-one nor onto

7. CBSE 2025Set 65/4/11 markMCQ
   Q.For real $x$, let $f(x) = x^3 + 5x + 1$. Then : (A) $f$ is one-one but not onto on $R$ (B) $f$ is onto on $R$ but not one-one (C) $f$ is one-one and onto on $R$ (D) $f$ is neither one-one nor onto on $R$

   ›Reveal solutionSolution

   The function $f(x)=x^3+5x+1$ is strictly increasing (since $f'(x)=3x^2+5>0$ for all real $x$), so it is one-one. As a cubic with odd degree and positive leading coefficient, its range is all real numbers, so it is onto $\mathbb{R}$. Hence the correct option is (C).

   1. Understanding one-one (injective) — why the derivative tells the story

      A function is one-one if different inputs give different outputs. For a differentiable function, a sufficient condition is that the derivative never changes sign — that is, the function is strictly monotonic (always increasing or always decreasing).

      Here, $f'(x) = 3x^2 + 5$. Since $x^2 \ge 0$ for all real $x$, we have $3x^2 \ge 0$, so $3x^2 + 5 \ge 5 > 0$. The derivative is always positive.

      Therefore $f$ is strictly increasing on $\mathbb{R}$. A strictly increasing function is automatically one-one: if $x_1 < x_2$, then $f(x_1) < f(x_2)$, so no two distinct $x$'s can map to the same $y$.

   2. Understanding onto (surjective) — why the range is all reals

      A function $f: \mathbb{R} \to \mathbb{R}$ is onto if every real number appears as an output. For a polynomial of odd degree with a positive leading coefficient, the end behaviour guarantees this:

      • As $x \to -\infty$, $x^3 \to -\infty$, so $f(x) \to -\infty$.
      • As $x \to +\infty$, $x^3 \to +\infty$, so $f(x) \to +\infty$. Since $f$ is continuous (every polynomial is continuous), by the Intermediate Value Theorem it takes every value between $-\infty$ and $+\infty$. That is, the range is $\mathbb{R}$.
      Tip

      A quick check: for any real $y$, the equation $x^3 + 5x + 1 = y$ is a cubic in $x$. Every cubic with real coefficients has at least one real root, so there is always some $x$ solving it. That alone proves surjectivity onto $\mathbb{R}$.

   3. Putting it together

      • One-one: yes, because $f'(x) > 0$ everywhere.
      • Onto: yes, because it's a continuous odd-degree polynomial with positive leading coefficient. Hence $f$ is both one-one and onto on $\mathbb{R}$.

   Watch out

   A common mistake is to think that a cubic is always one-one. That is false — for example, $f(x)=x^3 - x$ has derivative $3x^2 - 1$, which changes sign, so it is not one-one. Always check the derivative (or monotonicity) before concluding injectivity.

   ✓Final answer

   The correct option is (C) — $f$ is one-one and onto on $\mathbb{R}$.

8. CBSE 2025Set 65/4/11 markMCQ
   Q.If $f : N \to W$ is defined as $f(n) = \begin{cases} \frac{n}{2}, & \text{if } n \text{ is even} \\ 0, & \text{if } n \text{ is odd} \end{cases}$, then $f$ is : (A) injective only (B) surjective only (C) a bijection (D) neither surjective nor injective

   ›Reveal solutionSolution

   The function maps all odd naturals to 0 and each even natural to half its value, so it is surjective onto $W$ (every whole number is hit) but not injective (many inputs give the same output). The correct option is (B).

   The core idea here is to understand what the function does to its domain, $N$ (the set of natural numbers, typically $\{1,2,3,\dots\}$), and where it lands, $W$ (the set of whole numbers, $\{0,1,2,3,\dots\}$). The definition splits the domain into two clear cases: odd numbers and even numbers.

   For every odd natural number — 1, 3, 5, 7, … — the output is 0. That means infinitely many inputs all map to the single output 0. That alone kills injectivity: a function is injective (one-to-one) only if different inputs always give different outputs. Here, $f(1)=0$, $f(3)=0$, $f(5)=0$, and so on, so it is clearly not injective.

   For every even natural number — 2, 4, 6, 8, … — the output is half of that number. So $f(2)=1$, $f(4)=2$, $f(6)=3$, $f(8)=4$, and so on. This gives us every positive whole number exactly once. And the odd numbers already cover 0. So every whole number — 0, 1, 2, 3, … — appears as an output at least once. That makes the function surjective (onto).

   Let’s walk through it step by step.

   1. Check injectivity (one-to-one)

      Take two different inputs, say $n=1$ and $n=3$. Both are odd, so $f(1)=0$ and $f(3)=0$. Since $1 \neq 3$ but $f(1)=f(3)$, the function is not injective.

      Watch out

      A common mistake is to only check the even case and think the function looks one-to-one. But the odd case collapses everything to 0 — that’s the trap.

   2. Check surjectivity (onto)

      We need to see if every element of $W$ (the codomain) is actually hit by some $n$ in $N$.

      • For $0 \in W$: pick any odd $n$, say $n=1$, then $f(1)=0$.
      • For any positive whole number $k \in W$ (i.e., $k=1,2,3,\dots$): pick $n=2k$, which is an even natural number. Then $f(2k) = \frac{2k}{2} = k$. So every whole number has a preimage in $N$. Hence $f$ is surjective.

   3. Conclusion

      Since $f$ is surjective but not injective, it is not a bijection. The correct classification is “surjective only”.

   Tip

   A quick mental check: if a function from an infinite set to another infinite set sends infinitely many inputs to a single output, it cannot be injective. Here, the odd numbers are an infinite set all mapping to 0 — injectivity is impossible.

   ✓Final answer

   The function is surjective only, so the correct option is (B).

9. CBSE 2025Set ANNUAL1 markMCQ
   Q.Let $f : \mathbb{R} \to \mathbb{R}$ be defined as $f(x) = x^2$, where $\mathbb{R}$ is the set of real numbers. Choose the correct answer:

   (a) f is one-one onto

   (b) f is many-one onto

   (c) f is one-one but not onto

   (d) f is neither one-one nor onto

   ›Reveal solutionSolution

   $f(x)=x^2$ on $\mathbb{R}\to\mathbb{R}$ is neither one-one nor onto.

   Not one-one: $f(-1)=1=f(1)$ but $-1\neq 1$, so two different inputs give the same output.

   Not onto: $x^2\ge 0$ for every real $x$, so the range is $[0,\infty)$, which is not all of $\mathbb{R}$ (e.g. $-1$ has no pre-image).

   ✓Final answer

   (iv) f is neither one-one nor onto.

10. CBSE 2025Set IX1 markMCQ
    Q.The modulus function $f: R \to R^{+}$ given by $f(x) = |x|$ is

    (a) one-one and onto

    (b) many-one and onto

    (c) one-one but not onto

    (d) neither one-one nor onto

    ›Reveal solutionSolution

    $f(x)=|x|$ is many-one (as $f(2)=f(-2)$) and onto $R^{+}$; option (b).

    Concept. A function is one-one if different inputs give different outputs, and onto if every element of the codomain is actually attained.

    One-one? Take $x=2$ and $x=-2$: $f(2)=|2|=2$ and $f(-2)=|-2|=2$. Two different inputs share the same image, so $f$ is many-one.

    Onto? The codomain here is $R^{+}=[0,\infty)$. For any $y\ge 0$ we can pick $x=y$, giving $f(x)=|y|=y$. So every element of $R^{+}$ is hit, and $f$ is onto.

    ✓Final answer

    (b) many-one and onto.

11. CBSE 2025Set A1 markMCQ
    Q.If $f: R \to R$ be defined as $f(x) = 3x$, then

    (a) $f$ is one-one onto.

    (b) $f$ is many one onto.

    (c) $f$ is one-one but not onto.

    (d) $f$ is neither one-one nor onto.

    ›Reveal solutionSolution

    A linear function $f(x) = 3x$ with non-zero slope is always a bijection on $\mathbb{R}$.

    Checking one-one (injective): Suppose $f(x_1) = f(x_2)$ for $x_1, x_2 \in R$.

    $3x_1 = 3x_2 \implies x_1 = x_2$

    So distinct inputs always give distinct outputs — $f$ is one-one.

    Checking onto (surjective): Let $y \in R$ be any real number. We need $x \in R$ such that $f(x) = y$, i.e. $3x = y$, so $x = y/3$. Since $y/3$ is always a real number, every $y \in R$ has a pre-image. Hence $f$ is onto.

    Since $f$ is both one-one and onto, it is a bijection.

    ✓Final answer

    (a) $f$ is one-one onto.

12. CBSE 2025Set A1 mark
    Q.Write True or False: If a function $f$ is invertible, then $f$ must be one-one and onto.

    ›Reveal solutionSolution

    Invertibility of a function is EQUIVALENT to it being a bijection; this is a standard NCERT theorem.

    A function $f: A \to B$ is invertible if there exists $g: B \to A$ such that $g\circ f = I_A$ and $f\circ g = I_B$. It is a proven theorem that $f$ is invertible if and only if $f$ is a bijection (both one-one and onto):

    • One-one ensures each output has a unique pre-image, so $g$ is well-defined.
    • Onto ensures every element of $B$ has a pre-image, so $g$ can be defined on the whole of $B$.

    Hence the statement is correct.

    ✓Final answer

    True.