types-of-functions MCQs for UPSC Prelims

37 practice questions on types-of-functions from the Relations and Functions section of the UPSC Prelims syllabus. 37 come with a written explanation. Try the sample set below - the answer stays hidden until you ask for it.

11 Easy 18 Medium 8 Hard

Sample questions

Q1
medium

A set $A$ has $3$ elements and a set $B$ has $5$ elements. How many one-one functions $f : A \to B$ are there?

  1. A $125$
  2. B $243$
  3. C $10$
  4. D $60$
Show answer and explanation

Correct answer: D - $60$

Give the elements of $A$ images one at a time: the first has $5$ choices, the second must avoid that image and has $4$, and the third has $3$, so the count is $5 \times 4 \times 3 = 60$. The value $125 = 5^3$ counts all functions, repeats allowed, and $243 = 3^5$ also exchanges the roles of the two sets. The value $10$ chooses which three elements of $B$ get used but ignores the order in which the inputs take them.

Q2
medium

Let $f : \mathbb{R} \to [0,\ \infty)$ be given by $f(x) = x^2$. Which statement is correct?

  1. A $f$ is neither one-one nor onto, because $f(0) = 0$
  2. B $f$ is onto but not one-one, because $[0,\ \infty)$ is exactly the set of squares
  3. C $f$ is one-one but not onto, because $f$ never takes a negative value
  4. D $f$ is both one-one and onto, because every non-negative number has a square root
Show answer and explanation

Correct answer: B - $f$ is onto but not one-one, because $[0,\ \infty)$ is exactly the set of squares

Every $y \ge 0$ is the square of $\sqrt{y}$, and the declared codomain holds exactly the non-negative reals, so each element of the codomain is an image and $f$ is onto. Injectivity fails because $f(-3) = 9 = f(3)$, so having a square root for every codomain element is not enough for a bijection. Judging onto against $\mathbb{R}$ instead of the declared codomain is what makes the negative values look like a gap. That $f(0) = 0$ breaks neither property.

Q3
easy

A function $f : A \to B$ is called one-one (injective) when which condition holds for all $a_1, a_2 \in A$?

  1. A $f(a_1) \ne f(a_2)$ for every choice of $a_1$ and $a_2$
  2. B Every element of $B$ equals $f(a)$ for some $a \in A$
  3. C $f(a_1) = f(a_2)$ forces $a_1 = a_2$
  4. D $a_1 = a_2$ forces $f(a_1) = f(a_2)$
Show answer and explanation

Correct answer: C - $f(a_1) = f(a_2)$ forces $a_1 = a_2$

A one-one function sends different inputs to different outputs, and the usual test states that the other way round: if two inputs share an image, they were the same input. The condition that equal inputs give equal outputs holds for every function, so it tests nothing. Asking that every element of $B$ be an image is the definition of onto. Demanding unequal outputs for every pair fails immediately when $a_1$ and $a_2$ are the same element.

Q4
medium

Let $f : \mathbb{R} \to \mathbb{R}$ be defined by $f(x) = 7 - 2x$. Which statement is correct?

  1. A $f$ is onto but not one-one, since the negative coefficient makes values repeat
  2. B $f$ is neither one-one nor onto
  3. C $f$ is one-one but not onto, since $f(x)$ is never $7$
  4. D $f$ is a bijection
Show answer and explanation

Correct answer: D - $f$ is a bijection

If $7 - 2a = 7 - 2b$ then $-2a = -2b$, so $a = b$ and $f$ is one-one. For any real $y$ the number $x = \frac{7 - y}{2}$ satisfies $f(x) = y$, so $f$ is onto as well and therefore a bijection. The value $7$ is reached, at $x = 0$. A negative slope reverses the order of the values but never repeats one, since a straight line of non-zero slope crosses each horizontal level exactly once.

Q5
hard

Consider two statements about functions between finite sets. Statement 1: If $A$ and $B$ are finite sets with the same number of elements, then every one-one function $f : A \to B$ is also onto. Statement 2: If $A$ and $B$ are finite sets and $A$ has fewer elements than $B$, then no function $f : A \to B$ can be one-one. Which is correct?

  1. A Statement 1 is true and Statement 2 is false
  2. B Statement 1 is false and Statement 2 is true
  3. C Both statements are true
  4. D Both statements are false
Show answer and explanation

Correct answer: A - Statement 1 is true and Statement 2 is false

Let $A$ and $B$ both have $n$ elements. A one-one $f$ then produces $n$ distinct images inside a set of size $n$, so every element of $B$ is used and $f$ is onto, which settles the first statement as true. The second swaps the two ideas: when $A$ is smaller than $B$ it is being onto that becomes impossible, while one-one functions are plentiful, for example $1 \mapsto 1$, $2 \mapsto 2$ from $\{1,2\}$ into $\{1,2,3\}$.

Q6
medium

Consider two statements. Statement 1: The function $g : \mathbb{R} \to \mathbb{R}$ given by $g(x) = x^4$ is onto. Statement 2: The function $f : \mathbb{R} \to \mathbb{R}$ given by $f(x) = x^3$ is onto. Which is correct?

  1. A Both statements are true
  2. B Both statements are false
  3. C Statement 1 is false and Statement 2 is true
  4. D Statement 1 is true and Statement 2 is false
Show answer and explanation

Correct answer: C - Statement 1 is false and Statement 2 is true

A fourth power is never negative, so $-1$ lies in the codomain of $g$ with no preimage, and the statement about $x^4$ fails. Every real number has exactly one real cube root, so for any $y$ the input $x = \sqrt[3]{y}$ gives $f(x) = y$, and the statement about $x^3$ holds. The difference is that an even power folds the negative half of the domain onto values already produced, while an odd power keeps the sign.

Q7
medium

Let $f : \mathbb{R} - \{0\} \to \mathbb{R} - \{0\}$ be given by $f(x) = \dfrac{1}{x}$. Which statement is correct?

  1. A $f$ is neither one-one nor onto
  2. B $f$ is one-one but not onto, since the value $0$ is never taken
  3. C $f$ is both one-one and onto
  4. D $f$ is onto but not one-one, since $f(2) = f(-2)$
Show answer and explanation

Correct answer: C - $f$ is both one-one and onto

If $\frac{1}{a} = \frac{1}{b}$ with $a$ and $b$ non-zero, then $a = b$, so $f$ is one-one. For any non-zero $y$ the number $x = \frac{1}{y}$ is non-zero and gives $f(x) = y$, so $f$ is onto. The value $0$ has been removed from the codomain, so never producing it costs nothing. The pairing of $2$ with $-2$ is wrong, since $f(2) = \frac{1}{2}$ while $f(-2) = -\frac{1}{2}$.

Q8
medium

Consider $f : \mathbb{R} \to \mathbb{R}$ defined by $f(x) = x^2 - 4x + 5$. Which statement is correct?

  1. A $f$ is one-one, because $f(x) > 0$ for every real $x$
  2. B $f$ is one-one, because its graph is a parabola with a single lowest point
  3. C $f$ is not one-one, because $f(1) = f(3) = 2$
  4. D $f$ is not one-one, because $f(0) = f(5)$
Show answer and explanation

Correct answer: C - $f$ is not one-one, because $f(1) = f(3) = 2$

Injectivity fails as soon as two different inputs share an image, and here $f(1) = 1 - 4 + 5 = 2$ while $f(3) = 9 - 12 + 5 = 2$. The parabola shape and the fact that the values stay positive say nothing about whether values repeat. The pairing of $0$ with $5$ is wrong arithmetic: $f(0) = 5$ but $f(5) = 25 - 20 + 5 = 10$. The input sharing the image of $0$ is $4$, since the graph is symmetric about $x = 2$.

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