Q.Are the following set of ordered pairs functions? If so, examine whether the mapping is injective or surjective.

Showing the 12 most recent of 14 on this concept.

1. AP EAPCET 2026Set eng-2026-05-14-AN1 markMCQ
   Q.The function $f: [0, \infty) \to [0, \infty)$ defined by $f(x) = \dfrac{x}{1+x}$ is (A) one - one and onto (B) one - one but not onto (C) onto but not one - one (D) neither one-one nor onto

   ›Reveal solutionSolution

   The function $f(x) = \frac{x}{1+x}$ is strictly increasing (hence one‑one) but never reaches the value 1, so it is not onto. The correct option is (B).

   We are asked to decide whether $f: [0,\infty) \to [0,\infty)$ given by $f(x) = \frac{x}{1+x}$ is one‑one (injective), onto (surjective), both, or neither.

   ---

   1. Intuition: What does the function look like?

   For $x \ge 0$, the denominator $1+x$ is always larger than the numerator $x$, so every output is less than 1. As $x$ grows very large, $f(x)$ gets closer and closer to 1 but never reaches it. At $x = 0$, $f(0) = 0$. So the range is $[0, 1)$, not the whole codomain $[0, \infty)$. That already suggests “not onto.”

   Also, the function is increasing: if you increase $x$, the fraction increases. That suggests “one‑one.”

   Let’s verify both properties rigorously.

   ---

   2. Checking one‑one (injectivity)

   A function is one‑one if different inputs give different outputs. Equivalently: if $f(a) = f(b)$ then $a = b$.

   Assume

   $\frac{a}{1+a} = \frac{b}{1+b}.$

   Cross‑multiply:

   $a(1+b) = b(1+a) \quad\Rightarrow\quad a + ab = b + ab.$

   Cancel $ab$ from both sides:

   $a = b.$

   So indeed $f(a) = f(b)$ forces $a = b$. Hence $f$ is one‑one.

   Tip

   A faster way: compute the derivative for $x > 0$:

   $f'(x) = \frac{1}{(1+x)^2} > 0$.

   A strictly positive derivative on an interval means the function is strictly increasing, which guarantees injectivity.

   ---

   3. Checking onto (surjectivity)

   A function is onto if every element of the codomain $[0, \infty)$ is actually attained as an output.

   We already suspect the range is only $[0,1)$. Let’s prove it.

   For any $x \ge 0$:

   $f(x) = \frac{x}{1+x} = 1 - \frac{1}{1+x}.$

   Since $1+x \ge 1$, we have $\frac{1}{1+x} \in (0, 1]$, so …

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2. AP EAPCET 2026Set eng-2026-05-15-FN1 markMCQ
   Q.If a real valued function $f: (1,2] \to B$ defined by $f(x) = \log_{10}(x-1)$ is a bijection, then $B =$ (A) $[0,\infty)$ (B) $\mathbb{R}$ (C) $(-\infty,0]$ (D) $(0,\infty)$

   ›Reveal solutionSolution

   A bijection requires the function to be both one‑to‑one and onto; here $f(x)=\log_{10}(x-1)$ on $(1,2]$ is strictly increasing and its range is $(-\infty,0]$, so $B$ must be $(-\infty,0]$.

   Concept and intuition (One‑One Onto)

   A bijection means every element of the domain maps to a distinct element of the codomain (one‑to‑one) and every element of the codomain is actually hit by some input (onto). For a strictly monotonic function on an interval, one‑to‑one is automatic. The real work is finding the range of $f$ on the given domain — that range must be exactly the set $B$ for the function to be onto. Here $f(x)=\log_{10}(x-1)$ is defined only for $x>1$, and the domain $(1,2]$ is a half‑open interval. We’ll compute the outputs at the endpoints and see how the function behaves between them.

   1. Understand the domain and function

      Domain: $(1,2]$, meaning $1 < x \le 2$.

      The argument of the log is $x-1$, so as $x$ approaches $1$ from the right, $x-1 \to 0^+$. As $x$ increases to $2$, $x-1$ increases to $1$.

   2. Find the range by evaluating endpoints

      • At $x=2$: $f(2) = \log_{10}(2-1) = \log_{10}(1) = 0$.
      • As $x \to 1^+$: $x-1 \to 0^+$, and $\log_{10}(t) \to -\infty$ when $t \to 0^+$. So $f(x) \to -\infty$.

   3. Monotonicity and continuity

      The function $g(t)=\log_{10}t$ is strictly increasing for $t>0$, and $t=x-1$ is also strictly increasing in $x$. Hence $f$ is strictly increasing on $(1,2]$.

      Since $f$ is continuous on $(1,2]$ (the log is continuous on its domain), it takes every value between its infimum ($-\infty$) and its maximum ($0$). Because the domain is open at $1$, the value $-\infty$ is never actually attained, but every finite real number less than or equal to $0$ is attained exactly once.

   4. Conclude the range …

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3. AP EAPCET 2026Set eng-2026-05-18-AN1 markMCQ
   Q.If $f: A \to B$, $g: B \to C$ are two functions such that $g \circ f: A \to C$ is an onto function, then it is necessary that (A) $f$ is onto function (B) $g$ is onto function (C) Both $f$ and $g$ are onto functions (D) $f$ is one-one and $g$ is onto

   ›Reveal solutionSolution

   If the composition $g \circ f$ is onto, then $g$ must be onto (surjective), but $f$ need not be onto or one‑one. The correct choice is (B).

   ---

   Why this approach works

   The key is to think about what “onto” (surjective) means for a composition.

   If $g \circ f$ hits every element of $C$, then every $c \in C$ has some $a \in A$ with $g(f(a)) = c$. That $c$ is reached by $g$ from the element $f(a) \in B$, so $g$ itself must be able to reach every $c$ — that is, $g$ is onto.

   But $f$ might not be onto: it could miss some parts of $B$, as long as the part it does hit is enough for $g$ to cover all of $C$.

   Also, $f$ need not be one‑one; multiple $a$’s could map to the same $b$, and that doesn’t break surjectivity of the composition.

   ---

   Step‑by‑step reasoning

   1. Recall the definition of onto (surjective)

      A function $h: X \to Y$ is onto if for every $y \in Y$, there exists some $x \in X$ such that $h(x) = y$.

   2. Apply this to $g \circ f$

      Since $g \circ f: A \to C$ is onto, for every $c \in C$ there exists some $a \in A$ with

   $(g \circ f)(a) = g(f(a)) = c.$

   3. What does this tell us about $g$?

      Let $b = f(a) \in B$. Then $g(b) = c$.

      So for every $c \in C$, we have found a $b \in B$ (namely $b = f(a)$) such that $g(b) = c$.

      That is exactly the definition of $g$ being onto.

      Hence $g$ must be onto.

   4. What about $f$?

      Could $f$ fail to be onto? Yes.

      Example: Let $A = \{1\}$, $B = \{x, y\}$, $C = \{z\}$.

      Define $f(1) = x$, and $g(x) = z$, $g(y) = z$.

      Then $g \circ f(1) = z$, so $g \circ f$ is onto $C$, but $f$ is not onto $B$ (it never hits $y$).

      So $f$ need not be onto.

   5. Could $f$ be one‑one?

      Not necessary.

      Example: Let $A = \{1,2\}$, $B = \{b\}$, $C = \{c\}$.

      Define $f(1)=b$, $f(2)=b$ (not one‑one), and $g(b)=c$. …

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4. AP EAPCET 2025Set eng-2025-05-21-AN1 markMCQ
   Q.If a function $f:\mathbb{Z}\to\mathbb{Z}$ is defined by $f(x)=x-(-1)^x$, then $f(x)$ 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

   Splitting $f(x)=x-(-1)^x$ by parity of $x$ shows even inputs map bijectively onto all odd integers and odd inputs map bijectively onto all even integers — together this is a bijection of $\mathbb{Z}$ onto itself.

   Concept and Intuition

   $(-1)^x$ depends only on the parity of $x$: it's $+1$ for even $x$ and $-1$ for odd $x$. So $f$ behaves like two separate linear functions glued together by parity, and checking one-one/onto means checking both the "same output can't come from two different parities" question and the "does every integer get hit" question.

   Step-by-Step Solution

   1. For $x=2k$ (even): $f(2k)=2k-(-1)^{2k}=2k-1$.
   2. For $x=2k+1$ (odd): $f(2k+1)=2k+1-(-1)^{2k+1}=2k+1-(-1)=2k+2$.
   3. As $k$ ranges over all integers, $2k-1$ takes every odd integer exactly once (injective in $k$), and $2k+2$ takes every even integer exactly once (injective in $k$).
   4. Odd-image and even-image sets are disjoint, and their union is all of $\mathbb{Z}$ — so every integer in the codomain is hit exactly once. …

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5. AP EAPCET 2024Set eng-2024-05-18-FN1 markMCQ
   Q.If a function $f: R \to R$ is defined by $f(x) = x^3 - x$, then $f$ is (A) one – one and onto (B) one – one but not onto (C) onto but not one – one (D) neither one – one nor onto

   ›Reveal solutionSolution

   This tests checking one-one (injectivity) and onto (surjectivity) for the cubic function $f(x)=x^3-x$ on $\mathbb{R}$; it fails to be one-one (multiple roots share the same output) but, being a continuous unbounded cubic, it is onto.

   Concept and Intuition

   A function $f:\mathbb{R}\to\mathbb{R}$ is one-one (injective) if distinct inputs always give distinct outputs, and onto (surjective) if every real number in the codomain is actually achieved by some input. For a cubic polynomial like $x^3-x$, the derivative test tells us whether the function is monotonic (which would guarantee one-one), while the end behaviour (as $x\to\pm\infty$) combined with continuity guarantees onto for any odd-degree polynomial.

   Step-by-Step Solution

   1. Check one-one directly. Factor: $f(x) = x^3 - x = x(x-1)(x+1)$. So $f(0) = 0$, $f(1) = 0$, and $f(-1) = 0$ — three distinct values of $x$ all map to the same output $0$. This single counterexample is enough to conclude $f$ is not one-one.
   2. Alternatively, check via calculus: $f'(x) = 3x^2 - 1$, which is zero at $x = \pm\frac{1}{\sqrt3}$ and changes sign there — so $f$ increases, then decreases, then increases again, confirming it is not monotonic and therefore not injective. …

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6. AP EAPCET 2024Set eng-2024-05-21-AN1 markMCQ
   Q.The real valued function $f: R \to \left[\dfrac{5}{2}, \infty\right)$ defined by $f(x) = |2x+1| + |x-2|$ is (A) One-one function but not onto (B) Onto function but not one-one (C) Bijection (D) Neither one-one function nor onto

   ›Reveal solutionSolution

   This piecewise-linear V-shaped-with-a-kink function has minimum value $5/2$ (matching the stated codomain) and covers all values from $5/2$ to $\infty$, so it's onto — but the decreasing branch and the increasing branch overlap in output values, so it's not one-one.

   Concept and Intuition

   A sum of absolute values $|2x+1|+|x-2|$ is piecewise linear, changing slope at the "kink" points where each absolute-value expression changes sign ($x=-1/2$ and $x=2$). Between consecutive kinks the function is linear; outside the kinks it becomes steeper (sum of both slopes). Onto-ness is checked by comparing the range of $f$ to the stated codomain; one-one-ness is checked for any repeated output value.

   Step-by-Step Solution

   1. Break into three pieces based on the sign changes at $x=-1/2$ (where $2x+1=0$) and $x=2$ (where $x-2=0$):
      • $x < -1/2$: both expressions negative inside: $f(x) = -(2x+1) - (x-2) = -3x+1$.
      • $-1/2 \le x \le 2$: $2x+1\ge0$, $x-2\le0$: $f(x) = (2x+1) - (x-2) = x+3$.
      • $x > 2$: both non-negative: $f(x) = (2x+1)+(x-2) = 3x-1$.
   2. Evaluate at kinks: $f(-1/2) = -1/2+3 = 5/2$; $f(2) = 2+3 = 5$.
   3. Behaviour: on $(-\infty,-1/2)$, $f$ decreases from $+\infty$ down to $5/2$ (slope $-3$). On $[-1/2,2]$, $f$ increases from $5/2$ to $5$ (slope $+1$). On $(2,\infty)$, $f$ increases from $5$ to $+\infty$ (slope $+3$). …

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7. AP EAPCET 2023Set eng-2023-05-15-AN1 markMCQ
   Q.If a real valued function $f$ is defined by $f(x) = \dfrac{ax + \sqrt{a^2 - x^2}}{bx}$, then $f$ is (A) only one-one (B) only onto (C) both one-one and onto (D) neither one-one nor onto

   ›Reveal solutionSolution

   $f(x)=\frac{ax+\sqrt{a^2-x^2}}{bx}$ has a bounded domain $[-a,a]-\{0\}$. Checking

   the endpoints shows $f(a)=f(-a)$ (not one-one), while the function's range across

   the two branches covers all of $\mathbb{R}$ (onto).

   Concept and Intuition

   Whenever $\sqrt{a^2-x^2}$ appears, the domain is forced to be the bounded interval

   $[-a,a]$ (need $a^2-x^2\ge0$). Here we additionally need $x\ne0$ (division by $bx$).

   So $f$ lives only on $[-a,a]-\{0\}$ — a small, symmetric, bounded set. On such a

   restricted domain, checking one-one/onto is best done by direct evaluation at a few

   strategic points (especially the endpoints, where $\sqrt{a^2-x^2}=0$ and the formula

   collapses) rather than trying to argue in the abstract.

   Step-by-Step Solution

   1. Take representative values $a=1,\ b=1$ (the qualitative behaviour — bounded domain, symmetric endpoints — is the same for any nonzero $a,b$): domain is $[-1,1]-\{0\}$, $f(x) = \dfrac{x+\sqrt{1-x^2}}{x} = 1 + \dfrac{\sqrt{1-x^2}}{x}$.
   2. Evaluate at the endpoints, where $\sqrt{1-x^2}=0$:

   $f(1) = 1 + 0 = 1, \qquad f(-1) = 1 + 0 = 1.$

   Two distinct domain points, $x=1$ and $x=-1$, both map to the same value $1$

   $\Rightarrow$ $f$ is not one-one.

   3. Check the range on $(0,1)$: as $x\to0^+$, $\sqrt{1-x^2}/x \to +\infty$, so

   $f\to+\infty$; at $x=1$, $f=1$. Since $f$ is continuous and monotonic on this

   branch, it sweeps out $(1,\infty)$.

   4. Check the range on $(-1,0)$: as $x\to0^-$, $\sqrt{1-x^2}/x\to-\infty$, so

   $f\to-\infty$; at $x=-1$, $f=1$. This branch sweeps out $(-\infty,1)$.

   5. Combining both branches and the shared endpoint value $1$: the total range is …

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8. AP EAPCET 2022Set eng-2022-07-06-FN1 markMCQ
   Q.Let f: R→R defined by f(x) = 5x^4+2. Then (A) f is one-one but not onto (B) f is onto but not one-one (C) f is both one-one and onto (D) f is neither one-one nor onto

   ›Reveal solutionSolution

   $f(x)=5x^4+2$ fails to be one-one because even powers make $f(x)=f(-x)$, and it fails to be onto because $x^4\ge0$ restricts the range to $[2,\infty)$, missing most of $\mathbb{R}$.

   Concept and Intuition

   A function is one-one (injective) if distinct inputs always give distinct outputs, and onto (surjective) if every element of the codomain is actually achieved by some input. Even-degree power functions like $x^4$ are inherently "two-to-one" away from zero, since $(-x)^4=x^4$ — this immediately breaks injectivity for any function built purely from an even power (plus a constant/linear combination that doesn't fix this symmetry). Also, because $x^4$ can never be negative, adding a positive constant just shifts the range upward, but the range is still bounded below — it can never cover the whole real line, so it cannot be onto $\mathbb{R}$.

   Step-by-Step Solution

   1. Check one-one: pick $x=1$ and $x=-1$. $f(1)=5(1)^4+2=7$ and $f(-1)=5(-1)^4+2=5(1)+2=7$. Since $f(1)=f(-1)$ but $1\neq-1$, $f$ is not one-one.
   2. Check onto: for any real $x$, $x^4\ge0$, so $f(x)=5x^4+2\ge2$. This means $f$'s range is $[2,\infty)$, a proper subset of the codomain $\mathbb{R}$ (e.g. $f(x)=0$ has no real solution). So $f$ is not onto. …

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9. AP EAPCET 2022Set eng-2022-07-07-AN1 markMCQ
   Q.If f is a relation from set of positive real numbers to the set of positive real numbers defined by $f(x) = 3x^2 - 2$ then f is (A) one-one but not onto (B) onto but not one-one (C) a bijection (D) not a function

   ›Reveal solutionSolution

   For $f(x)=3x^2-2$ to be a function $\mathbb{R}^+ \to \mathbb{R}^+$, every output for a positive-real input must itself be a positive real — but this fails for small $x$, so the relation is not a valid function on the stated codomain.

   Concept and Intuition

   A relation qualifies as a function from a set $A$ to a set $B$ only if every element of $A$ maps to an element that actually lies in $B$ (as well as each input having a unique output). Here $A=B=\mathbb{R}^+$ (positive reals). The rule is $f(x)=3x^2-2$.

   Check whether $f(x) \in \mathbb{R}^+$ for every $x \in \mathbb{R}^+$:

   $f(x) > 0 \iff 3x^2 > 2 \iff x > \sqrt{2/3} \approx 0.816$

   So for any $x$ in $(0, \sqrt{2/3})$ — which is a perfectly valid part of the domain $\mathbb{R}^+$ — the output $f(x)$ is negative, i.e. it does not belong to the codomain $\mathbb{R}^+$. Since the rule fails to land every domain element inside the stated codomain, it does not define a valid function from $\mathbb{R}^+$ to $\mathbb{R}^+$ at all — the question about one-one/onto doesn't even arise until the codomain condition is satisfied.

   Step-by-Step Solution

   1. Domain and codomain are both stated as $\mathbb{R}^+$ (strictly positive reals).
   2. Test a small positive $x$, e.g. $x=0.5$: $f(0.5) = 3(0.25)-2 = 0.75-2=-1.25$, which is negative. …

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10. AP EAPCET 2022Set eng-2022-07-08-AN1 markMCQ
    Q.If the function f : R → R is defined by f(x) = x|x|, then (A) f is one-one but not onto (B) f is onto but not one-one (C) f is both one-one and onto (D) f is neither one-one nor onto

    ›Reveal solutionSolution

    This tests recognizing $f(x)=x|x|$ as a strictly monotonic, unbounded, continuous function — hence a bijection on $\mathbb{R}$.

    Concept and Intuition

    The absolute value splits the domain, but here it does so in a way that keeps the function moving in the same direction throughout — that's the key insight, not the piecewise formula itself.

    Step-by-Step Solution

    1. Write $f(x)=x^2$ for $x\ge 0$ and $f(x)=-x^2$ for $x<0$.
    2. On $x\ge0$, $f$ is increasing (it's $x^2$ restricted to non-negative $x$). On $x<0$, $f(x)=-x^2$ is also increasing as $x$ increases (e.g. $f(-2)=-4$, $f(-1)=-1$, increasing toward $0$).
    3. At the junction $x=0$, both pieces give $f(0)=0$, so the function is continuous and strictly increasing across all of $\mathbb{R}$.
    4. A strictly monotonic function is automatically one-one (injective). …

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11. AP EAPCET 2022Set eng-2022-07-08-FN1 markMCQ
    Q.If f: R→R is defined as f(x) = x^2-2x-3 then f is (A) one-one but not onto (B) onto but not one-one (C) neither one-one nor onto (D) a bijection

    ›Reveal solutionSolution

    $f(x)=x^2-2x-3$ is a parabola with vertex at $(1,-4)$; it fails injectivity (symmetric pairs give equal outputs) and fails surjectivity onto $\mathbb{R}$ (its range is bounded below at $-4$), so it is neither one-one nor onto.

    Concept and Intuition

    A quadratic function $f:\mathbb{R}\to\mathbb{R}$ can never be one-one over all of $\mathbb{R}$, because a parabola is symmetric about its vertex — for any value above the minimum, there are always two distinct $x$-values (symmetric about the vertex) giving the same $y$-value. Also, since the parabola opens upward, its range is bounded below by the vertex's $y$-value and never reaches values below that, so it cannot be onto $\mathbb{R}$ either.

    Step-by-Step Solution

    1. Complete the square: $f(x) = x^2-2x-3 = (x^2-2x+1)-4 = (x-1)^2-4$.
    2. This is an upward parabola with vertex at $(1,-4)$; its minimum value is $-4$, so the range of $f$ is $[-4,\infty)$.
    3. Since the range $[-4,\infty) \ne \mathbb{R}$ (codomain), $f$ is NOT onto. …

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12. AP EAPCET 2021Set eng-2021-08-23-AN1 markMCQ
    Q.Define $f:\mathbb{R}\to\mathbb{R}$ by $f(x)=\max\{x+1,\,1-x,\,2\}$. Then $f$ is ____ (A) One – one but not onto (B) Onto but not one – one (C) Neither one – one nor onto (D) Both one – one and onto

    ›Reveal solutionSolution

    Piecewise analysis shows $f$ is constant ($=2$) on $[-1,1]$ (so not injective) and its range is only $[2,\infty)$, never covering all of $\mathbb{R}$ (so not surjective onto $\mathbb{R}$). Hence neither one-one nor onto.

    Concept and Intuition

    $f(x)=\max\{x+1,\ 1-x,\ 2\}$ picks, at each $x$, the largest of three lines/constant. To understand its shape, find where each pair of expressions crosses, since the max switches from one expression to another exactly at those crossing points.

    Step-by-Step Solution

    1. $x+1$ vs $1-x$: equal when $x+1=1-x \Rightarrow x=0$; for $x>0$, $x+1>1-x$.
    2. $x+1$ vs $2$: equal when $x=1$; for $x>1$, $x+1>2$.
    3. $1-x$ vs $2$: equal when $x=-1$; for $x<-1$, $1-x>2$.
    4. Combine: for $x\le-1$: $1-x\ge2$ and $1-x\ge x+1$ (since $x\le0$), so $f(x)=1-x$. For $-1\le x\le 1$: check e.g. $x=0$: values are $1,1,2$, so $f=2$; in fact throughout this interval both $x+1\le2$ and $1-x\le2$, so $f(x)=2$ (constant). For $x\ge1$: $x+1\ge2$ and $x+1\ge1-x$ (since $x\ge0$), so $f(x)=x+1$.
    5. So $f(x) = \begin{cases}1-x, & x\le-1\\ 2, & -1\le x\le1 \\ x+1, & x\ge1\end{cases}$, and $f(x)\ge2$ everywhere.
    6. Not one-one: infinitely many $x$ in $[-1,1]$ all map to the same value $2$. …

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