Q.Show that the Modulus Function $f: \mathbf{R} \rightarrow \mathbf{R}$, given by $f(x) = |x|$, is neither one-one nor onto, where $|x|$ is $x$, if $x$ is positive or 0 and $|x|$ is $-x$, if $x$ is negative.

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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