Q.Give an example of a map

Showing the 12 most recent of 45 on this concept.

1. CBSE 2024Set 65/3/11 markMCQ
   Q.Let $\mathbb{R}_+$ denote the set of all non-negative real numbers. Then the function $f : \mathbb{R}_+ \to \mathbb{R}_+$ defined as $f(x) = x^2 + 1$ is: (A) one-one but not onto (B) onto but not one-one (C) both one-one and onto (D) neither one-one nor onto

   ›Reveal solutionSolution

   The function $f(x) = x^2 + 1$ maps non-negative real numbers to non-negative real numbers. It is one-one because distinct non-negative inputs always produce distinct outputs, but it is not onto because values in the codomain between $0$ and $1$ (exclusive) do not have a pre-image. The function is one-one but not onto.

   To determine if a function is one-one (injective) and/or onto (surjective), we need to understand what these terms mean in the context of the given domain and codomain.

   The function is $f : \mathbb{R}_+ \to \mathbb{R}_+$ defined as $f(x) = x^2 + 1$.

   Here, $\mathbb{R}_+$ denotes the set of all non-negative real numbers, which is the interval $[0, \infty)$.

   So, the domain is $[0, \infty)$ and the codomain is also $[0, \infty)$.

   Understanding One-one (Injectivity):

   A function $f: A \to B$ is one-one if every distinct element in the domain $A$ maps to a distinct element in the codomain $B$. In other words, no two different inputs produce the same output.

   Mathematically, this means: If $f(x_1) = f(x_2)$ for any $x_1, x_2 \in A$, then it must imply $x_1 = x_2$.

   Understanding Onto (Surjectivity):

   A function $f: A \to B$ is onto if every element in the codomain $B$ has at least one corresponding element in the domain $A$ that maps to it. This means the range of the function must be equal to its codomain.

   Mathematically, this means: For every $y \in B$, there exists at least one $x \in A$ such that $f(x) = y$.

   Let's analyze the given function step-by-step.

   1. Check for One-one (Injectivity): We assume $f(x_1) = f(x_2)$ for any $x_1, x_2$ in the domain $\mathbb{R}_+$.

   $x_1^2 + 1 = x_2^2 + 1$

   Subtracting $1$ from both sides:
   

   $x_1^2 = x_2^2$

   Taking the square root of both sides:
   

   $x_1 = \pm x_2$

   Now, we must consider the domain. Since $x_1, x_2 \in \mathbb{R}_+$, both $x_1$ and $x_2$ must be non-negative.
   If $x_2 > 0$, then $x_1 = -x_2$ would mean $x_1$ is negative, which is not allowed in $\mathbb{R}_+$.
   Therefore, the only possibility for $x_1, x_2 \in \mathbb{R}_+$ is $x_1 = x_2$.
   This confirms that if the outputs are the same, the inputs must also be the same.
   
   > [!WARNING]
   > If the domain were $\mathbb{R}$ (all real numbers) instead of $\mathbb{R}_+$, then $x_1 = \pm x_2$ would mean the function is *not* one-one. For example, $f(2) = 2^2+1 = 5$ and $f(-2) = (-2)^2+1 = 5$, but $2 \neq -2$. The restriction of the domain to $\mathbb{R}_+$ is crucial here.
   
   Thus, the function $f(x) = x^2 + 1$ is one-one on $\mathbb{R}_+$.
   

   2. Check for Onto (Surjectivity):

   For the function to be onto, every element $y$ in the codomain $\mathbb{R}_+$ must have a pre-image $x$ in the domain $\mathbb{R}_+$ such that $f(x) = y$.

   Let $y \in \mathbb{R}_+$ be an arbitrary element in the codomain. We set $f(x) = y$:

   $x^2 + 1 = y$ …

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2. CBSE 2024Set 65/1/11 markMCQ
   Q.A function $f: \mathrm{R}_{+} \rightarrow \mathrm{R}$ (where $\mathrm{R}_{+}$is the set of all non-negative real numbers) defined by $f(x)=4x+3$, then this function is : (A) One-one but not onto (B) Onto but not one-one (C) Both one-one and onto (D) Neither one-one nor onto

   ›Reveal solutionSolution

   A linear function with positive slope is always one-one (strictly increasing). But here the codomain is all real numbers, while the range is only numbers $\ge 3$, so it is not onto. The function is one-one but not onto.

   We need to check two properties: one-one (injective) and onto (surjective). The function is $f(x) = 4x + 3$, with domain $\mathbb{R}_+$ (all non-negative reals, including zero) and codomain $\mathbb{R}$ (all real numbers).

   The key idea: a linear function with non-zero slope is always one-one. But whether it is onto depends on whether its range equals the entire codomain. Here the domain is restricted to $x \ge 0$, so the range is $[3, \infty)$, not all of $\mathbb{R}$.

   Let’s go step by step.

   1. Checking one-one (injective) A function is one-one if different inputs give different outputs. For $f(x) = 4x + 3$, suppose $f(a) = f(b)$. Then:

   $4a + 3 = 4b + 3 \implies 4a = 4b \implies a = b$

   So $f$ is injective.

   Alternatively, the derivative $f'(x) = 4 > 0$ means $f$ is strictly increasing, which also guarantees one-one.

   Tip

   For linear functions $f(x) = mx + c$, if $m \neq 0$, it is always one-one on any domain. The sign of $m$ tells you if it's increasing or decreasing.

   2. Checking onto (surjective) A function is onto if every element in the codomain $\mathbb{R}$ has a preimage in the domain $\mathbb{R}_+$. That is, for any $y \in \mathbb{R}$, we need some $x \ge 0$ such that $f(x) = y$. Solve $4x + 3 = y$ for $x$: $x = \frac{y - 3}{4}$ …

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3. 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 $\mathbb{R}$. …

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4. CBSE 2024Set 65/2/11 markMCQ
   Q.Let $f : R_+ \to [-5, \infty)$ be defined as $f(x) = 9x^2 + 6x - 5$, where $R_+$ is the set of all non-negative real numbers. Then, $f$ is: (A) one-one (B) onto (C) bijective (D) neither one-one nor onto

   ›Reveal solutionSolution

   A quadratic function on $[0,\infty)$ is strictly increasing, hence one-one; checking the range shows it maps onto $[-5,\infty)$ exactly. The function is bijective.

   The question asks us to determine whether $f$ is one-one (injective), onto (surjective), both (bijective), or neither. Understanding these properties for a quadratic restricted to non-negative reals requires examining both monotonicity and range.

   A function is one-one if distinct inputs always produce distinct outputs: $f(x_1) = f(x_2) \implies x_1 = x_2$. For continuous functions, strict monotonicity (always increasing or always decreasing) guarantees this property.

   A function is onto if every element in the codomain $[-5,\infty)$ is actually achieved by some input from the domain $R_+$.

   Checking if $f$ is one-one

   The derivative tells us about monotonicity:

   $f'(x) = 18x + 6$

   For all $x \in R_+ = [0,\infty)$, we have $f'(x) = 18x + 6 \geq 6 > 0$. The function is strictly increasing on its entire domain.

   Tip

   A strictly monotonic function on any interval is automatically one-one — if $x_1 < x_2$, then $f(x_1) < f(x_2)$, so they can never be equal.

   Therefore $f$ is one-one.

   Checking if $f$ is onto

   We need to verify that the range of $f$ (the set of all actual outputs) equals the codomain $[-5,\infty)$.

   1. Find the minimum value. Since $f$ is strictly increasing on $[0,\infty)$, the minimum occurs at the left endpoint $x=0$:

   $f(0) = 9(0)^2 + 6(0) - 5 = -5$

   2. Find the behavior as $x \to \infty$. The leading term $9x^2$ dominates: …

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5. 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.

   …

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6. 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.

   …

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7. 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.

   …

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8. 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): …

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9. 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 $f(x)=y$.

   …

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10. 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).

    …

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11. 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 $N$. …

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12. 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.

    …

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