Rbse Solutions Class 12 Maths: Exercise 1.2 (Functions)

Are you struggling to prove if a function is Injective or Surjective? Exercise 1.2 is one of the most important sections of the RBSE Class 12 Maths syllabus. In the 2026 board exam, you can expect at least one 3-mark or 4-mark question from this specific exercise.

In this guide, we provide the NCERT-based solutions for Ex 1.2 with a focus on the logic required for the Rajasthan Board (BSER Ajmer) marking scheme.

💡 The “Cheat Sheet” for Exercise 1.2

Before solving the questions, memorize these two logical tests. If you can’t pass these, you can’t solve the exercise.

1. One-One (Injective): * The Test: Assume $f(x_1) = f(x_2)$.
   • The Result: If you solve this and get $x_1 = x_2$ only, it is One-one.
2. Onto (Surjective): * The Test: Pick an arbitrary element ‘$y$’ in the Co-domain.
   • The Result: If there exists an ‘$x$’ in the Domain such that $f(x) = y$, then it is Onto.

Topper’s Note: If a function is both One-one and Onto, it is called Bijective.

Question 1.

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

Solution:

Part 1: $f: \mathbb{R}_ \to \mathbb{R}_*$ defined by $f(x) = 1/x$.*

1. One-One (Injectivity):Let $f(x_1) = f(x_2)$ for $x_1, x_2 \in \mathbb{R}_*$.$$\frac{1}{x_1} = \frac{1}{x_2} \implies x_1 = x_2$$Thus, $\mathbf{f}$ is one-one.
2. Onto (Surjectivity):Let $y$ be any arbitrary element in the co-domain $\mathbb{R}_*$. We need to find $x \in \mathbb{R}_*$ such that $f(x) = y$.$$y = \frac{1}{x} \implies x = \frac{1}{y}$$Since $y$ is a non-zero real number, $x = 1/y$ is also a non-zero real number, so $x \in \mathbb{R}_*$.Thus, $f$ is onto.

Part 2: If the domain $\mathbb{R}_*$ is replaced by $N$, $f: N \to \mathbb{R}_*$.

1. One-One: The proof remains the same: $1/x_1 = 1/x_2 \implies x_1 = x_2$.The result is true for injectivity.
2. Onto: The co-domain is $\mathbb{R}_*$. Consider $2$ in the co-domain. If $f(x) = 2$, then $1/x = 2 \implies x = 1/2$.Since $x=1/2$ is not a natural number ($x \notin N$), $2$ has no pre-image in the domain $N$.The result is not true for surjectivity.

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Question 2.

Check the injectivity and surjectivity of the following functions:

(i) $f: N \to N$ given by $f(x) = x^2$

• Injective: $f(x_1) = f(x_2) \implies x_1^2 = x_2^2$. Since $x_1, x_2 \in N$, $x_1$ and $x_2$ must be positive, so $x_1 = x_2$. Injective.
• Surjective: Co-domain is $N$. Range is $\{1, 4, 9, 16, \dots\}$. E.g., $2 \in N$ has no pre-image. Not Surjective.

(ii) $f: Z \to Z$ given by $f(x) = x^2$

• Injective: $f(1) = 1$ and $f(-1) = 1$. Since $1 \neq -1$, $\mathbf{f}$ is not injective.
• Surjective: Co-domain is $Z$. Range contains only non-negative perfect squares. E.g., $-1 \in Z$ has no pre-image. Not Surjective.

(iii) $f: R \to R$ given by $f(x) = x^2$

• Injective: $f(1) = 1$ and $f(-1) = 1$. Not Injective.
• Surjective: Co-domain is $R$. Range is $[0, \infty)$. E.g., $-5 \in R$ has no pre-image. Not Surjective.

(iv) $f: N \to N$ given by $f(x) = x^3$

• Injective: $x_1^3 = x_2^3 \implies x_1 = x_2$ (since $x \in N$). Injective.
• Surjective: Co-domain is $N$. Range is $\{1, 8, 27, 64, \dots\}$. E.g., $2 \in N$ has no pre-image. Not Surjective.

(v) $f: Z \to Z$ given by $f(x) = x^3$

• Injective: $x_1^3 = x_2^3 \implies x_1 = x_2$ (since $x \in Z$). Injective.
• Surjective: Co-domain is $Z$. Range is $\{\dots, -8, -1, 0, 1, 8, \dots\}$. E.g., $2 \in Z$ has no pre-image (since $\sqrt[3]{2}$ is not an integer). Not Surjective.

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Question 3.

Show that the Signum Function $f: \mathbb{R} \to \mathbb{R}$, given by $f(x) = \begin{cases} 1, & \text{if } x > 0 \\ 0, & \text{if } x = 0 \\ -1, & \text{if } x < 0 \end{cases}$ is neither one-one nor onto.

Solution:

1. One-One (Injectivity):We have $f(1) = 1$ and $f(2) = 1$.Since $f(1) = f(2)$ but $1 \neq 2$, $\mathbf{f}$ is not one-one.
2. Onto (Surjectivity):The co-domain is $\mathbb{R}$. The range of $f$ is $\{-1, 0, 1\}$.Since the range $\neq$ co-domain (e.g., $2$ is in the co-domain but not in the range), $\mathbf{f}$ is not onto.

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Question 4.

Show that the Modulus Function $f: \mathbb{R} \to \mathbb{R}$, given by $f(x) = |x|$, is neither one-one nor onto.

Solution:

1. One-One (Injectivity):We have $f(1) = |1| = 1$ and $f(-1) = |-1| = 1$.Since $f(1) = f(-1)$ but $1 \neq -1$, $\mathbf{f}$ is not one-one.
2. Onto (Surjectivity):The co-domain is $\mathbb{R}$. The range of $f$ is $[0, \infty)$ (all non-negative real numbers).The co-domain contains negative numbers (e.g., $-2$). Since there is no real number $x$ such that $|x| = -2$, the range $\neq$ co-domain.$\mathbf{f}$ is not onto.

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Question 5.

Show that the Greatest Integer Function $f: \mathbb{R} \to \mathbb{R}$, given by $f(x) = [x]$, is neither one-one nor onto.

Solution:

1. One-One (Injectivity):We have $f(1.2) = [1.2] = 1$ and $f(1.7) = [1.7] = 1$.Since $f(1.2) = f(1.7)$ but $1.2 \neq 1.7$, $\mathbf{f}$ is not one-one.
2. Onto (Surjectivity):The co-domain is $\mathbb{R}$. The range of $f$ is $Z$ (the set of integers).The co-domain contains non-integer numbers (e.g., $1.5$). Since there is no $x \in \mathbb{R}$ such that $f(x) = [x] = 1.5$, the range $\neq$ co-domain.$\mathbf{f}$ is not onto.

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Question 6.

Let $A = \{1, 2, 3\}$ and $B = \{4, 5, 6, 7\}$. Let $f = \{(1, 4), (2, 5), (3, 6)\}$ be a function from $A$ to $B$. Show that $f$ is one-one.

Solution:

The function is given by the mappings: $f(1)=4$, $f(2)=5$, $f(3)=6$.

A function is one-one if every distinct element in the domain has a distinct image in the co-domain.

• $f(1) = 4$
• $f(2) = 5$
• $f(3) = 6$

Since $1 \neq 2 \implies f(1) \neq f(2)$, $1 \neq 3 \implies f(1) \neq f(3)$, and $2 \neq 3 \implies f(2) \neq f(3)$, all elements in $A$ map to distinct elements in $B$.

Thus, $\mathbf{f}$ is one-one.

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Question 7.

In each of the following cases, state whether the function is one-one, onto or bijective. Justify your answer.

(i) $f: \mathbb{R} \to \mathbb{R}$ defined by $f(x) = 3 – 4x$

1. One-One (Injectivity):Let $f(x_1) = f(x_2)$.$$3 – 4x_1 = 3 – 4x_2 \implies -4x_1 = -4x_2 \implies x_1 = x_2$$$f$ is one-one.
2. Onto (Surjectivity):Let $y$ be any element in the co-domain $\mathbb{R}$. We find $x$ such that $f(x) = y$.$$y = 3 – 4x \implies 4x = 3 – y \implies x = \frac{3 – y}{4}$$Since $y$ is a real number, $x$ is also a real number ($x \in \mathbb{R}$).$f$ is onto.

Conclusion: $f$ is bijective (one-one and onto).

(ii) $f: \mathbb{R} \to \mathbb{R}$ defined by $f(x) = 1 + x^2$

1. One-One (Injectivity):$f(1) = 1 + 1^2 = 2$ and $f(-1) = 1 + (-1)^2 = 2$.Since $f(1) = f(-1)$ but $1 \neq -1$, $f$ is not one-one.
2. Onto (Surjectivity):Since $x^2 \geq 0$, $f(x) = 1 + x^2 \geq 1$.The Range of $f$ is $[1, \infty)$. The co-domain is $\mathbb{R}$.E.g., $0$ is in the co-domain, but $f(x) = 0 \implies x^2 = -1$, which has no real solution.$f$ is not onto.

Conclusion: $f$ is neither one-one nor onto.

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Question 8.

Let $A$ and $B$ be sets. Show that $f: A \times B \to B \times A$ such that $f(a, b) = (b, a)$ is a bijective function.

Solution:

1. One-One (Injectivity):Let $f(a_1, b_1) = f(a_2, b_2)$ for $(a_1, b_1), (a_2, b_2) \in A \times B$.$$(b_1, a_1) = (b_2, a_2)$$By equality of ordered pairs, $b_1 = b_2$ and $a_1 = a_2$.Thus, $(a_1, b_1) = (a_2, b_2)$. $f$ is one-one.
2. Onto (Surjectivity):Let $(b, a)$ be an arbitrary element in the co-domain $B \times A$.We need to find a pre-image $(x, y) \in A \times B$ such that $f(x, y) = (b, a)$.Since $b \in B$ and $a \in A$, the pair $(a, b)$ exists in the domain $A \times B$.$f(a, b) = (b, a)$.Thus, for every element in the co-domain, there exists a pre-image in the domain. $f$ is onto.

Conclusion: $f$ is bijective (one-one and onto).

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Question 9.

Let $f: N \to N$ be defined by $f(n) = \begin{cases} \frac{n+1}{2}, & \text{if } n \text{ is odd} \\ \frac{n}{2}, & \text{if } n \text{ is even} \end{cases}$ for all $n \in N$. State whether the function $f$ is bijective. Justify your answer.

Solution:

1. One-One (Injectivity):Check for two distinct inputs that give the same output.Let $n_1 = 1$ (odd) and $n_2 = 2$ (even).$$f(1) = \frac{1+1}{2} = 1$$$$f(2) = \frac{2}{2} = 1$$Since $f(1) = f(2)$ but $1 \neq 2$, $\mathbf{f}$ is not one-one.
2. Onto (Surjectivity):Let $m$ be any arbitrary element in the co-domain $N$.
   • If $m$ is obtained from an odd $n$: $m = \frac{n+1}{2} \implies n = 2m – 1$. Since $m \geq 1$, $2m-1 \geq 1$, and $2m-1$ is always odd. $n \in N$ exists.
   • If $m$ is obtained from an even $n$: $m = \frac{n}{2} \implies n = 2m$. Since $m \geq 1$, $2m \geq 2$, and $2m$ is always even. $n \in N$ exists.Thus, for every $m \in N$, there is at least one pre-image $n \in N$. $f$ is onto.

Conclusion: $f$ is onto but not one-one, hence it is not bijective.

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Question 10.

Let $A = \mathbb{R} – \{3\}$ and $B = \mathbb{R} – \{1\}$. Consider the function $f: A \to B$ defined by $f(x) = \left(\frac{x-2}{x-3}\right)$. Is $f$ one-one and onto? Justify your answer.

Solution:

1. One-One (Injectivity):Let $f(x_1) = f(x_2)$.$$\frac{x_1 – 2}{x_1 – 3} = \frac{x_2 – 2}{x_2 – 3}$$$$(x_1 – 2)(x_2 – 3) = (x_2 – 2)(x_1 – 3)$$$$x_1 x_2 – 3x_1 – 2x_2 + 6 = x_1 x_2 – 3x_2 – 2x_1 + 6$$$$-3x_1 – 2x_2 = -3x_2 – 2x_1 \implies x_2 = x_1$$$f$ is one-one.
2. Onto (Surjectivity):Let $y$ be an arbitrary element in the co-domain $B = \mathbb{R} – \{1\}$.$$y = \frac{x – 2}{x – 3} \implies y(x – 3) = x – 2$$$$xy – 3y = x – 2 \implies xy – x = 3y – 2$$$$x(y – 1) = 3y – 2 \implies x = \frac{3y – 2}{y – 1}$$Since $y \in B$, $y \neq 1$, so $x$ is well-defined. Also, $x \in A$ since $x \neq 3$.$f$ is onto.

Conclusion: $f$ is one-one and onto (bijective).

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Question 11.

Let $f: \mathbb{R} \to \mathbb{R}$ be defined as $f(x) = x^4$. Choose the correct answer.

(A) $f$ is one-one, onto

(B) $f$ is many-one, onto

(C) $f$ is one-one, not onto

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

Solution:

1. One-One: $f(1) = 1^4 = 1$ and $f(-1) = (-1)^4 = 1$. Not one-one (many-one).
2. Onto: The range is $[0, \infty)$ (non-negative numbers), but the co-domain is $\mathbb{R}$. E.g., $-2$ has no pre-image. Not onto.

Conclusion: $f$ is neither one-one nor onto.

Correct Answer: (D)

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Question 12.

Let $f: \mathbb{R} \to \mathbb{R}$ be defined as $f(x) = 3x$. Choose the correct answer.

(A) $f$ is one-one, onto

(B) $f$ is many-one, onto

(C) $f$ is one-one, not onto

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

Solution:

1. One-One: $f(x_1) = f(x_2) \implies 3x_1 = 3x_2 \implies x_1 = x_2$. One-one.
2. Onto: Let $y \in \mathbb{R}$. $y = 3x \implies x = y/3$. Since $y$ is a real number, $x$ is a real number. Onto.

Conclusion: $f$ is one-one and onto.

Correct Answer: (A)

FAQs

❓ FAQ: Frequently Asked Questions

Q: Is Exercise 1.2 harder than 1.1? A: Not necessarily. While 1.1 is about checking properties of relations, 1.2 is about “Mapping.” If you understand the difference between Domain and Range, it’s very easy.

Q: What is a Bijective function? A: A function that is both One-one (Injective) and Onto (Surjective) is Bijective. Only Bijective functions have an “Inverse.”