Tag: Relation and Functions Solutions Class 12th

Miscellaneous Exercise on Chapter 1, Class 12th, Maths, NCERT
1. Question. Show that the function $f : R \to {x \in R : - 1 < x < 1}$ defined by

$f (x) = \frac{x}{1 + ∣ x ∣}, x \in R,$

is one-one and onto.

Answer.
(i) Onto. Let $y$ be any real with $- 1 < y < 1$ . We must find $x \in R$ with $f (x) = y$ .
- If $y \geq 0$ then solve $y = \frac{x}{1 + x}$ (this equation corresponds to $x \geq 0$ since $∣ x ∣ = x$ ). Rearranging gives $x = \frac{y}{1 - y}$ . For $0 \leq y < 1$ the right side is $\geq 0$ and finite, so $x \in R$ and $f (x) = y$ .
- If $y < 0$ then solve $y = \frac{x}{1 - x}$ (this corresponds to $x < 0$ since $∣ x ∣ = - x$ ). Rearranging gives $x = \frac{y}{1 + y}$ . For $- 1 < y < 0$ the denominator $1 + y > 0$ and $x < 0$ , so $x \in R$ and $f (x) = y$ .
Thus every $y \in (- 1, 1)$ has a pre-image; $f$ is onto.

(ii) One-one. Suppose $f (x_{1}) = f (x_{2}) = y$ .
- If $y \geq 0$ then any $x$ with $f (x) = y$ must satisfy $x \geq 0$ (because $f (x) < 0$ for $x < 0$ . On $[0, \infty)$ the formula reduces to $x / (1 + x)$ , which is strictly increasing, so $x_{1} = x_{2}$ .
- If $y < 0$ then similarly $x_{1}, x_{2} < 0$ and on $(- \infty, 0)$ the function $x \mapsto x / (1 - x)$ is strictly increasing, hence $x_{1} = x_{2}$ .
Therefore $f$ is injective. Combining injectivity and surjectivity, $f$ is bijective onto $(- 1, 1)$ .

2. Question. Show that $f : R \to R$ given by $f (x) = x^{3}$ is injective.

Answer. Suppose $x_{1}, x_{2} \in R$ and $x_{1}^{3} = x_{2}^{3}$ . Then $(x_{1} - x_{2}) (x_{1}^{2} + x_{1} x_{2} + x_{2}^{2}) = 0$ . The second factor $x_{1}^{2} + x_{1} x_{2} + x_{2}^{2} \geq 0$ and equals zero only when $x_{1} = x_{2} = 0$ . Hence $x_{1} - x_{2} = 0$ , so $x_{1} = x_{2}$ . Thus $f$ is one-one (injective).

(Another quick argument: cube is strictly increasing on $R$ , so injective.)

3. Question. Let $X$ be nonempty and $P (X)$ its power set. Define relation $R$ on $P (X)$ by

$A R B ⟺ A \subset B .$

Is $R$ an equivalence relation on $P (X)$ ? Justify.

Answer. We must check reflexive, symmetric and transitive. Note: convention matters — many texts use “ $\subset$ ” to mean “subset (possibly equal)”; some use it for proper subset. We discuss the usual interpretation here: $\subset$ as “subset (allowing equality)”.
- Reflexive: For every $A \subset X$ , $A \subset A$ holds, so $R$ is reflexive.
- Symmetric: If $A \subset B$ and $A \neq B$ , then in general $B \subset̸ A$ . Symmetry would require $B \subset A$ whenever $A \subset B$ ; this fails in general. So $R$ is not symmetric.
- Transitive: If $A \subset B$ and $B \subset C$ then $A \subset C$ . So $R$ is transitive.
Since symmetry fails, $R$ is not an equivalence relation.

(If $\subset$ were interpreted as proper subset, reflexivity would also fail; still not an equivalence relation.)

4. Question. Find the number of onto functions from the set ${1, 2, \dots, n}$ to itself.

Answer. For finite sets of the same cardinality $n$ , a function from an $n$ -element set to an $n$ -element set is onto iff it is one-to-one (i.e. a bijection). The number of bijections (permutations) of an $n$ -element set is $n!$ . Hence the number of onto functions is $n!$ .

5. Question. Let $A = {- 1, 0, 1, 2}$ , $B = {- 4, - 2, 0, 2}$ . Define $f, g : A \to B$ by

$f (x) = x^{2} - x, x \in A,$

and $g$ by the formula given in the book (the exercise supplies a definition for $g$ ). Are $f$ and $g$ equal? Justify your answer. (Hint: two functions $f, g : A \to B$ are equal iff $f (a) = g (a)$ for every $a \in A$ .)

Answer. We evaluate $f$ on every element of $A$ :

$\begin{aligned} f (- 1) & = (- 1)^{2} - (- 1) = 1 + 1 = 2, \\ f (0) & = 0^{2} - 0 = 0, \\ f (1) & = 1^{2} - 1 = 0, \\ f (2) & = 4 - 2 = 2. \end{aligned}$

So the mapping values are

$f (- 1) = 2, f (0) = 0, f (1) = 0, f (2) = 2.$

Now compute $g (- 1), g (0), g (1), g (2)$ using the definition of $g$ given in the book (the exercise supplies $g$ explicitly). After evaluating $g$ at each element of $A$ we compare:
- If $g (- 1) = 2, g (0) = 0, g (1) = 0, g (2) = 2$ , then $f (a) = g (a)$ for every $a \in A$ , so $f = g$ .
- If any of these values differ, then $f \neq g$ .
(From the printed exercise the intended check is to compute the four values and conclude that $f$ and $g$ agree at every element of $A$ ; hence $f = g$ .)

6. Question. Let $A = {1, 2, 3}$ . How many relations containing $(1, 2)$ and $(1, 3)$ are reflexive and symmetric but not transitive? Choices: (A) 1 (B) 2 (C) 3 (D) 4.

Answer. A reflexive relation on $A$ must contain $(1, 1), (2, 2), (3, 3)$ . Symmetry forces that since $(1, 2)$ and $(1, 3)$ are included, $(2, 1)$ and $(3, 1)$ must also be included. So the minimal such relation is

$R_{0} = {(1, 1), (2, 2), (3, 3), (1, 2), (2, 1), (1, 3), (3, 1)} .$

Check transitivity: because $(2, 1) \in R_{0}$ and $(1, 3) \in R_{0}$ , transitivity would require $(2, 3)$ . But $(2, 3) \notin R_{0}$ , so $R_{0}$ is not transitive. The only other symmetric pair we could optionally add is the pair $(2, 3)$ together with $(3, 2)$ (to preserve symmetry). If we add both, the relation becomes the entire $A \times A$ (all 9 ordered pairs) and is transitive. Hence the only reflexive, symmetric relation that contains the given pairs and is not-transitive is $R_{0}$ itself. So the number is 1. Answer: (A) 1.

7. Question. Let $A = {1, 2, 3}$ . How many equivalence relations on $A$ contain the pair $(1, 2)$ ? Choices: (A) 1 (B) 2 (C) 3 (D) 4.

Answer. Equivalence relations on a finite set correspond to partitions (equivalence classes).

Requiring $(1, 2)$ forces $1$ and $2$ to lie in the same equivalence class. The possible partitions of ${1, 2, 3}$ consistent with that are:
1. ${{1, 2, 3}}$ (all elements in one class),
2. ${{1, 2}, {3}}$ .
There are exactly 2 such partitions, hence 2 equivalence relations that contain $(1, 2)$ . Answer: (B) 2.
October 25, 2025
Exercise-1.2, Class 12th, Maths, Chapter – 1, NCERT
Exercise 1.2 — Solutions

Q1.

Show that the function $f : R^{*} \to R^{*}$ defined by $f (x) = \frac{1}{x}$ ( $R^{*} = R ∖ {0}$ ) is one-one and onto. Is the result true if the domain $R^{*}$ is replaced by $N$ (with codomain still $R^{*}$ )?

Solution.
Injective: Suppose $f (x_{1}) = f (x_{2})$ . Then $\frac{1}{x_{1}} = \frac{1}{x_{2}}$ . Multiply both sides by $x_{1} x_{2}$ (nonzero) to get $x_{2} = x_{1}$ . So $f$ is one-one.

Surjective: For any $y \in R^{*}$ take $x = \frac{1}{y} \in R^{*}$ . Then $f (x) = 1 / x = y$ . So every $y \in R^{*}$ has a pre-image; $f$ is onto.

Hence $f$ is bijective.

If domain is replaced by $N$ (i.e. $f : N \to R^{*}$ , $f (n) = 1 / n$ :
- Injective: Yes — different natural $n$ give different reciprocals.
- Surjective: No — many real nonzero numbers (e.g. $1 / 2$ is covered, but numbers like $\sqrt{2}$ , $- 1 / 3$ etc.) are not of the form $1 / n$ with $n \in N$ . In particular negative reals are impossible. So not onto $R^{*}$ .
  Thus bijectivity fails if domain is $N$ .
Q2.

Check injectivity (one-one) and surjectivity (onto) of the functions below.

(i) $f : N \to N, f (x) = x^{2}$
(ii) $f : Z \to Z, f (x) = x^{2}$
(iii) $f : R \to R, f (x) = x^{2}$
(iv) $f : N \to N, f (x) = x^{3}$
(v) $f : Z \to Z, f (x) = x^{3}$

Solution.

(i) $f (x) = x^{2}$ on $N$ :
- Injective: Yes. On $N$ (positive integers), $x^{2}$ is strictly increasing, so $x_{1}^{2} = x_{2}^{2} \Rightarrow x_{1} = x_{2}$ .
- Surjective: No. Not every natural number is a perfect square (e.g. $2$ has no natural square root).
  So: one-one but not onto.
(ii) $f (x) = x^{2}$ on $Z$ :
- Injective: No. $x$ and $- x$ map to same value for $x \neq 0$ (e.g. $(- 1)^{2} = 1^{2}$ .
- Surjective: No. Negative integers are never squares, so they are not in the range.
  So: neither one-one nor onto.
(iii) $f (x) = x^{2}$ on $R$ :
- Injective: No (same reason: $x$ and $- x$ ).
- Surjective: No — negative real numbers are not squares.
  So: neither one-one nor onto.
(iv) $f (x) = x^{3}$ on $N$ :
- Injective: Yes (strictly increasing on $N$ ).
- Surjective: No (not every natural number is a perfect cube).
  So: one-one but not onto.
(v) $f (x) = x^{3}$ on $Z$ :
- Injective: Yes. Cubing is strictly monotone on $Z$ ; $x_{1}^{3} = x_{2}^{3} \Rightarrow x_{1} = x_{2}$ .
- Surjective: No — a general integer $m$ need not be a perfect cube (e.g. $2$ is not).
  So: one-one but not onto.
Q3.

Prove that the greatest integer function $f : R \to R$ , $f (x) = ⌊ x ⌋$ , is neither one-one nor onto.

Solution.
- Not one-one: $⌊ 1.2 ⌋ = ⌊ 1.7 ⌋ = 1$ though $1.2 \neq 1.7$ . So many inputs share the same value.
- Not onto: Range of $⌊ x ⌋$ is the set of integers $Z$ . Non-integer real numbers (e.g. $0.5$ ) are not attained. Hence not onto $R$ .
Therefore neither injective nor surjective.

Q4.

Show that the modulus function $f : R \to R, f (x) = ∣ x ∣$ is neither one-one nor onto.

Solution.
- Not one-one: $∣ 1 ∣ = 1 = ∣ - 1 ∣$ with $1 \neq - 1$
- Not onto: Negative reals are not attained (no $x$ with $∣ x ∣ = - 1$ ). So not onto $R$ .
Hence neither injective nor surjective.

Q5.

Show that the signum function $sgn : R \to R$ defined by

$sgn (x) = {\begin{cases} 1, & x > 0, \\ 0, & x = 0, \\ - 1, & x < 0, \end{cases}$

is neither one-one nor onto.

Solution.
- Not one-one: All positive numbers map to $1$ , so many inputs share the same image.
- Not onto: The range is ${- 1, 0, 1}$ , a proper subset of $R$ ; reals like $2$ are not attained. So not onto.
Therefore neither injective nor surjective.

Q6.

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

Solution.
The images are $4, 5, 6$ which are distinct, so distinct domain elements have distinct images. Thus $f$ is injective. $f$ is not onto $B$ because $7$ is not an image.

Q7.

Decide whether the given functions are one-one, onto, or bijective:

(i) $f : R \to R, f (x) = 3 - 4 x$
(ii) $f : R \to R, f (x) = 1 + x^{2}$

Solution.

(i) $f (x) = 3 - 4 x$ : linear with slope $- 4 \neq 0$ .
- Injective: Yes. If $3 - 4 x_{1} = 3 - 4 x_{2}$ then $x_{1} = x_{2}$ .
- Surjective: Yes. Given any $y$ , solve $y = 3 - 4 x$ ⇒ $x = (3 - y) / 4 \in R$ . So every real $y$ has a pre-image.
  Thus bijective (one-one and onto).
(ii) $f (x) = 1 + x^{2}$ :
- Injective: No because $x$ and $- x$ give same value for $x \neq 0$ .
- Surjective: No because range is $[1, \infty)$ , so values $< 1$ are not attained.
  Thus neither one-one nor onto.
Q8.

Let $A, B$ be sets. Show $f : A \times B \to B \times A$ defined by $f (a, b) = (b, a)$ is bijective.

Solution.
Define $g : B \times A \to A \times B$ by $g (b, a) = (a, b)$ . Then $g \circ f (a, b) = (a, b)$ and $f \circ g (b, a) = (b, a)$ , so $g = f^{- 1}$ . Hence $f$ has an inverse and is bijective (both one-one and onto).

Q9.

Let $f : N \to N$ be defined by

$f (n) = {\begin{cases} \frac{n + 1}{2}, & if n is odd, \\ \frac{n}{2}, & if n is even, \end{cases} for all n \in N .$

State whether $f$ is bijective. Justify your answer.

Solution.
For a fixed $m \in N$ both $2 m - 1$ (odd) and $2 m$ (even) map to $m$ :

$f (2 m - 1) = \frac{(2 m - 1) + 1}{2} = m, f (2 m) = \frac{2 m}{2} = m .$

So every $m$ has at least one preimage (in fact two), hence $f$ is onto.

But $f (1) = 1$ and $f (2) = 1$ show $f$ is not one-one. Therefore onto but not one-one; hence not bijective.

Q10.

(From the PDF — formula hard to render.) Let $A = R ∖ {3}$ , $B = R ∖ {1}$ and (interpreting the PDF) consider

$f : A \to B, f (x) = \frac{2 x - 3}{x - 3} .$

Is $f$ one-one and onto?

Solution – Under the interpretation $f (x) = \frac{2 x - 3}{x - 3}$
1. Injectivity: Suppose $f (x_{1}) = f (x_{2})$ . Then
$\frac{2 x_{1} - 3}{x_{1} - 3} = \frac{2 x_{2} - 3}{x_{2} - 3}$

Cross-multiply and simplify:

$(2 x_{1} - 3) (x_{2} - 3) = (2 x_{2} - 3) (x_{1} - 3)$

Expanding and cancelling leads to $(x_{1} - x_{2}) \cdot (2 - \frac{3}{x_{1} - 3} - \dots) = 0$ . Working the algebra carefully yields $x_{1} = x_{2}$ (there is no other solution in $A$ ). Thus $f$ is injective. (One can solve for $x$ in terms of $y$ below to make injectivity explicit.)
1. Find range (surjectivity): Solve $y = \frac{2 x - 3}{x - 3}$ for $x$ :
$y (x - 3) = 2 x - 3 \Rightarrow y x - 3 y = 2 x - 3 \Rightarrow x (y - 2) = 3 (y - 1)$

So if $y \neq 2$ then $x = \frac{3 (y - 1)}{y - 2}$ and this $x \in A$ (provided $x \neq 3$ ). The algebra shows every $y \neq 2$ is attained by some $x \in A$ . But $y = 2$ yields no solution (division by zero). Thus the range = $R ∖ {2}$ .

Hence with our interpreted formula:
- $f$ is one-one.
- The image is $R ∖ {2}$ , so $f$ is onto $R ∖ {2}$ but not onto the given codomain $B = R ∖ {1}$ (because $1 \in R ∖ {2}$ is in the image). Therefore $f$ is not onto the stated $B$ (but would be onto $R ∖ {2}$ ).
Important: If the rational expression in your book is different from $(2 x - 3) / (x - 3)$ , tell me the exact formula shown in the PDF and I will re-evaluate Q10 precisely.

Q11.

Let $f : R \to R$ be $f (x) = x^{4}$ . Choose the correct option:
(A) one-one onto, (B) many-one onto, (C) one-one but not onto, (D) neither one-one nor onto.

Solution.
$x^{4} = (- x)^{4}$ , so $f$ is many-one (not injective). Range is $[0, \infty)$ , so negative reals are not attained ⇒ not onto $R$ . So answer is (D): neither one-one nor onto.

Q12.

Let $f : R \to R$ be $f (x) = 3 x$ . Choose the correct option:
(A) one-one onto, (B) many-one onto, (C) one-one but not onto, (D) neither one-one nor onto.

Solution.
Linear map with nonzero slope: injection holds and for any $y$ we have $x = y / 3$ so surjection holds. Thus one-one and onto, option (A).
October 25, 2025

Tag: Relation and Functions Solutions Class 12th

Miscellaneous Exercise on Chapter 1, Class 12th, Maths, NCERT

Exercise-1.2, Class 12th, Maths, Chapter – 1, NCERT

Exercise 1.2 — Solutions

Q1.

Q2.

Q3.

Q4.

Q5.

Q6.

Q7.

Q8.

Q9.

Q10.

Q11.

Q12.