Class 12th Maths Exercise 1.1 Solutions

Exercise 1.1 — Solutions

Q1.

Determine whether each of the following relations are reflexive, symmetric, and transitive:

(i) Relation R in A = {1, 2, 3, …, 14} defined as R = {(x, y): 3x – y = 0}.

Solution:
We have $y = 3 x$ .

Reflexive: For reflexivity, (x, x) ∈ R ⟺ 3x – x = 0 ⟺ 2x = 0 ⟺ x = 0.
But 0 ∉ A. Hence, R is not reflexive.
Symmetric: If (x, y) ∈ R, then y = 3x.
For symmetry, we need (x, y) ∈ R ⇒ (y, x) ∈ R.
Then (y, x) ∈ R ⟺ 3y – x = 0 ⇒ x = 3y.
But if y = 3x, x = 3y ⇒ y = 3x and x = 3(3x) = 9x ⇒ x = 0 (contradiction).
Thus, R is not symmetric.
Transitive: (x, y) ∈ R ⇒ y = 3x; (y, z) ∈ R ⇒ z = 3y = 9x.
Then (x, z) ∈ R ⟺ z = 3x ⇒ 9x = 3x ⇒ x = 0 (contradiction).
So, not transitive.

✅ Hence, R is neither reflexive, nor symmetric, nor transitive.

(ii) Relation R in N defined as R = {(x, y): y = x + 5 and x < 4}.

Solution:

Reflexive: (x, x) ∈ R ⟺ x = x + 5 ⟹ 5 = 0 (False).
Hence not reflexive.
Symmetric: (x, y) ∈ R ⇒ y = x + 5.
Then (y, x) ∈ R ⟺ x = y + 5 ⟺ x = x + 10 (False).
Hence not symmetric.
Transitive: (x, y) ∈ R ⇒ y = x + 5; (y, z) ∈ R ⇒ z = y + 5 = x + 10.
Then (x, z) ∈ R ⟺ z = x + 5 (False).
Hence not transitive.

✅ So, R is neither reflexive, symmetric, nor transitive.

(iii) Relation R in A = {1, 2, 3, 4, 5, 6} defined by R = {(x, y): y is divisible by x}.

Solution:

Reflexive: (x, x) ∈ R since x divides x. ✅ Reflexive.
Symmetric: (x, y) ∈ R ⇒ y divisible by x.
Then (y, x) ∈ R only if x divisible by y, which is not true (e.g., 2 divides 4 but 4 does not divide 2). ❌
So not symmetric.
Transitive: If y divisible by x and z divisible by y, then z divisible by x. ✅
Hence transitive.

✅ Therefore, R is reflexive and transitive but not symmetric.

(iv) Relation R in Z of all integers defined as R = {(x, y): x – y is an integer}.

Solution:
For any x, y ∈ Z, x – y is an integer since Z is closed under subtraction.
So R = Z × Z ⇒ every pair related.

✅ R is reflexive, symmetric, and transitive ⇒ an equivalence relation.

(v) Relation R in the set A of human beings in a town:

(a) R = {(x, y): x and y work at the same place}
→ Reflexive ✅ (a person works where they work),
Symmetric ✅ (if x works with y, then y works with x),
Transitive ✅ (if x, y, z work at same place → related).
✅ Equivalence relation.

(b) R = {(x, y): x and y live in same locality}
→ Reflexive ✅, Symmetric ✅, Transitive ✅.
✅ Equivalence relation.

(d) R = {(x, y): x is wife of y}
→ Reflexive ❌, Symmetric ❌, Transitive ❌.
❌ None.

(e) R = {(x, y): x is father of y}
→ Reflexive ❌, Symmetric ❌, Transitive ❌ (a father’s father ≠ father).
❌ None.

Q2.

Show that the relation R in R defined as R = {(a, b): a ≤ b²} is neither reflexive, nor symmetric, nor transitive.

Solution:

Reflexive: For reflexivity, (a, a) ∈ R ⇒ a ≤ a².
For a = ½, ½ ≤ ¼ ⇒ False. Hence not reflexive.
Symmetric: If (a, b) ∈ R ⇒ a ≤ b².
We need (b, a) ∈ R ⇒ b ≤ a².
Let a = 1, b = 2: (1, 2) ∈ R since 1 ≤ 4. But (2, 1) ∉ R since 2 ≤ 1 is False.
Hence not symmetric.
Transitive: (a, b) ∈ R ⇒ a ≤ b²; (b, c) ∈ R ⇒ b ≤ c².
Take a = 3, b = 2, c = 1. Then 3 ≤ 4 (True), 2 ≤ 1 (False).
Hence no guarantee for a ≤ c². Not transitive.

Thus R is neither reflexive, symmetric, nor transitive.

Q3.

Check whether relation R defined in {1, 2, 3, 4, 5, 6} as R = {(a, b): b = a + 1} is reflexive, symmetric or transitive.

Solution:

Reflexive: (a, a) ∈ R ⟺ a = a + 1 (False). ❌
Symmetric: If b = a + 1, then a = b + 1 (False). ❌
Transitive: If (b = a + 1) and (c = b + 1) ⇒ c = a + 2 ≠ a + 1 ⇒ False. ❌

✅ Hence R is neither reflexive, symmetric, nor transitive.

Q4.

Show that the relation R in R defined as R = {(a, b): a ≤ b} is reflexive and transitive but not symmetric.

Solution:

Reflexive: a ≤ a ✅ True.
Symmetric: a ≤ b does not imply b ≤ a (e.g., 2 ≤ 3 but 3 ≤ 2 False). ❌
Transitive: If a ≤ b and b ≤ c ⇒ a ≤ c. ✅

✅ Hence R is reflexive and transitive but not symmetric.

Q5.

Check whether the relation R in R defined by R = {(a, b): a ≤ b³} is reflexive, symmetric or transitive.

Solution:

Reflexive: (a, a) ∈ R ⟺ a ≤ a³.
True for a ≥ 1 and a ≤ –1, but False for 0 < a < 1 (e.g., ½ ≤ ⅛ False). ❌
So not reflexive.
Symmetric: a ≤ b³ does not imply b ≤ a³ (e.g., a = 1, b = 2). ❌
Transitive: If a ≤ b³ and b ≤ c³ → no guarantee that a ≤ c³. ❌

✅ Hence R is neither reflexive, symmetric, nor transitive.

Q6.

Show that the relation R in {1, 2, 3} given by R = {(1, 2), (2, 1)} is symmetric but neither reflexive nor transitive.

Solution:

Reflexive: (1, 1), (2, 2), (3, 3) ∉ R ❌
Symmetric: (1, 2) ∈ R ⇒ (2, 1) ∈ R. ✅
Transitive: (1, 2), (2, 1) ∈ R ⇒ (1, 1) ∉ R ❌

✅ Hence R is symmetric but neither reflexive nor transitive.

Q7.

Show that the relation R in the set A of all books in a college library, defined as R = {(x, y): x and y have same number of pages}, is an equivalence relation.

Solution:

Reflexive: Every book has same number of pages as itself.
Symmetric: If x and y have same pages, then y and x also do.
Transitive: If x and y same, and y and z same ⇒ x and z same.

Hence R is an equivalence relation.

Q8.

Show that the relation R in A = {1, 2, 3, 4, 5} given by R = {(a, b): |a – b| is even} is an equivalence relation.
Also show that all elements of {1, 3, 5} are related and of {2, 4} are related, but none between them.

Solution:

Reflexive: |a – a| = 0 (even).
Symmetric: |a – b| = |b – a|.
Transitive: If |a – b| and |b – c| even ⇒ (a – b) and (b – c) even ⇒ (a – c) even.

Equivalence relation.
Classes: {1, 3, 5} (odd) and {2, 4} (even). No cross-relation.

Q9.

Show that each relation R in A = {x ∈ Z | 0 ≤ x ≤ 12}, given by:
(i) R = {(a, b): |a – b| is a multiple of 4},
(ii) R = {(a, b): a = b},
is an equivalence relation. Find the set of all elements related to 1 in each case.

Solution:

(i)

Reflexive: |a – a| = 0 ⇒ multiple of 4.
Symmetric: |a – b| = |b – a|.
Transitive: If |a – b| and |b – c| multiples of 4 ⇒ |a – c| multiple of 4.
✅ Equivalence.

Elements related to 1: b such that |1 – b| multiple of 4 → b = 1, 5, 9 (within 0–12).

(ii)

a = b ⇒ reflexive, symmetric, transitive
Elements related to 1 ⇒ {1}.

Q10.

Give an example of a relation which is:

(i) Symmetric but neither reflexive nor transitive.
(ii) Transitive but neither reflexive nor symmetric.
(iii) Reflexive and symmetric but not transitive.
(iv) Reflexive and transitive but not symmetric.
(v) Symmetric and transitive but not reflexive.

Solution:

(i) R = {(1, 2), (2, 1)} on A = {1, 2, 3}.
(ii) R = {(1, 2), (2, 3), (1, 3)}
(iii) R = {(1, 1), (2, 2), (1, 2), (2, 1)}
(iv) R = {(1, 1), (2, 2), (1, 2), (1, 3)}
(v) R = {(1, 2), (2, 1)}

Q11.

Show that the relation $R$ in the set $A$ of points in the plane given by

$R = {(P, Q) : distance of P from the origin = distance of Q from the origin}$

is an equivalence relation. Further, show that the set of all points related to a point $P \neq (0, 0)$ is the circle with centre at the origin passing through $P$ .

Solution.
Let the origin be $O$ . For points $P, Q, R$ :

Reflexive: Distance of $P$ from $O$ equals itself, so $(P, P) \in R$ for every $P$ . Thus $R$ is reflexive.
Symmetric: If $(P, Q) \in R$ then $∣ O P ∣ = ∣ O Q ∣$ . That equality is symmetric, so $∣ O Q ∣ = ∣ O P ∣$ and $(Q, P) \in R$ . Thus $R$ is symmetric.
Transitive: If $(P, Q) \in R$ and $(Q, R) \in R$ , then $∣ O P ∣ = ∣ O Q ∣$ and $∣ O Q ∣ = ∣ O R ∣$ . Hence $∣ O P ∣ = ∣ O R ∣$ and $(P, R) \in R$ . Thus $R$ is transitive.

Therefore $R$ is reflexive, symmetric and transitive ⇒ an equivalence relation.

Now fix $P \neq (0, 0)$ and let $r = ∣ O P ∣ > 0$ . The equivalence class $[P] = {Q \in R^{2} : ∣ O Q ∣ = r}$ is precisely the circle centered at the origin with radius $r$ , which passes through $P$ .

Q12.

Show that the relation $R$ on the set $A$ of all triangles defined by

$R = {(T_{1}, T_{2}) : T_{1} is similar to T_{2}}$

is an equivalence relation. Consider the right-angled triangles $T_{1}$ with sides $3, 4, 5;$ $T_{2}$ with sides $5, 12, 13$ and $T_{3}$ with sides $6, 8, 10$ . Which triangles among $T_{1}, T_{2}, T_{3}$ are related?

Solution.

Reflexive: Any triangle is similar to itself, so $(T, T) \in R$
Symmetric: If $T_{1}$ is similar to $T_{2}$ , then $T_{2}$ is similar to $T_{1}$
Transitive: If $T_{1} \sim T_{2}$ and $T_{2} \sim T_{3}$

Hence $R$ is an equivalence relation.

Now check similarity of the given triangles: Two triangles are similar if their corresponding side lengths are proportional.

$T_{1} : (3, 4, 5)$ and $T_{3} : (6, 8, 10)$ Each side of $T_{3}$ is exactly $2$ times the corresponding side of $T_{1}$ . So $T_{1} \sim T_{3}$ .
$T_{2} : (5, 12, 13)$ is not proportional to $T_{1}$ (ratios $5 / 3, 12 / 4, 13 / 5$ are not equal), so $T_{2}$ is not similar to $T_{1}$ nor to $T_{3}$ .

Therefore $T_{1}$ and $T_{3}$ are related; $T_{2}$ is not related to them.

Q13.

Show that the relation $R$ on the set $A$ of all polygons defined by

$R = {(P_{1}, P_{2}) : P_{1} and P_{2} have the same number of sides}$ is an equivalence relation. What is the set of all elements in $A$ related to the right-angle triangle $T$ with sides $3, 4, 5$ ?

Solution.

Reflexive: Any polygon has same number of sides as itself ⇒ $(P, P) \in R$ .
Symmetric: If $P_{1}$ has same number of sides as $P_{2}$ , then $P_{2}$ has same number as $P_{1}$ .
Transitive: If $P_{1}$ and $P_{2}$ have the same number of sides and $P_{2}$ and $P_{3}$ have the same number, then $P_{1}$ and $P_{3}$ do too.

Thus $R$ is an equivalence relation. A right-angled triangle is a polygon with 3 sides, so the equivalence class of $T$ is the set of all triangles (all polygons with 3 sides) in $A$ .

Q14.

Let $L$ be the set of all lines in the $x y$ -plane and define $R$ on $L$ by

$R = {(L_{1}, L_{2}) : L_{1} is parallel to L_{2}}$

Show that $R$ is an equivalence relation. Find the set of all lines related to the line $y = 2 x + 4$ .

Solution.

Reflexive: Any line is parallel to itself ⇒ $(L, L) \in R$ .
Symmetric: If $L_{1}$ is parallel to $L_{2}$ , then $L_{2}$ is parallel to $L_{1}$ .
Transitive: If $L_{1} ∥ L_{2}$ and $L_{2} ∥ L_{3}$ , then $L_{1} ∥ L_{3}$

Hence $R$ is an equivalence relation.

The line $y = 2 x + 4$ has slope $2$ . All lines parallel to it are exactly the lines with slope $2$ ; i.e. all lines of the form $y = 2 x + c$ (for any real constant $c$ ). That collection is the equivalence class of $y = 2 x + 4$

Q15.

Let $R$ be the relation in the set ${1, 2, 3, 4}$ given by

$R = {(1, 2), (2, 2), (1, 1), (4, 4), (1, 3), (3, 3), (3, 2)}$

Choose the correct answer:

(A) $R$ is reflexive and symmetric but not transitive.
(B) $R$ is reflexive and transitive but not symmetric.
(C) $R$ is symmetric and transitive but not reflexive.
(D) $R$ is an equivalence relation.

Solution.
First check reflexivity: do we have $(1, 1), (2, 2), (3, 3), (4, 4)$ all in $R$ ? Yes — they are listed. So $R$ is reflexive.

Check symmetry: $(1, 2) \in R$ but $(2, 1)$ is not listed. So $R$ is not symmetric.

Check transitivity: we must check whenever $(a, b) \in R$ and $(b, c) \in R$ , then $(a, c) \in R$ .

Key checks:

$(1, 2) \in R$ and $(2, 2) \in R$ ⇒ $(1, 2)$ is required — it is present.
$(1, 3) \in R$ and $(3, 2) \in R$ ⇒ $(1, 2)$ is required — it is present.
$(3, 2) \in R$ and $(2, 2) \in R$ ⇒ $(3, 2)$ required — present.
Other combinations similarly hold; no transitivity violation appears.

Thus $R$ is transitive.

So $R$ is reflexive and transitive but not symmetric, which is option (B).

Q16.

Let $R$ be the relation in $N$ given by

$R = {(a, b) : a = b - 2, b > 6}$ Which of the following are true?

(A) $(2, 4) \in R$
(B) $(3, 8) \in R$
(C) $(6, 8) \in R$
(D) $(8, 7) \in R$

Solution.
We check each ordered pair:

(A) $(2, 4)$ : Here $b = 4$ but requirement is $b > 6$ . So $(2, 4) \notin R$
(B) $(3, 8)$ : We need $a = b - 2$ . For $b = 8$ , $b - 2 = 6$ So $(3, 8) \notin R$
(C) $(6, 8)$ : For $b = 8$ , $b - 2 = 6$ so $a = 6$ matches, and $b > 6$ holds. Hence $(6, 8) \in R$ .
(D) $(8, 7)$ : For $b = 7$ , $b - 2 = 5$ ; but $a = 8$ , so $a \neq b - 2$ . Also $b > 6$ is true but the equality fails. Thus $(8, 7) \notin R$ .

Only (C) is correct.

Tag: Class 12th Maths Exercise 1.1 Solutions

Exercise-1, Class 12th, Mathematics, NCERT, Solutions

Exercise 1.1 — Solutions

Q1.

(i) Relation R in A = {1, 2, 3, …, 14} defined as R = {(x, y): 3x – y = 0}.

(ii) Relation R in N defined as R = {(x, y): y = x + 5 and x < 4}.

(iii) Relation R in A = {1, 2, 3, 4, 5, 6} defined by R = {(x, y): y is divisible by x}.

(iv) Relation R in Z of all integers defined as R = {(x, y): x – y is an integer}.

(v) Relation R in the set A of human beings in a town:

Q2.

Q3.

Q4.

Q5.

Q6.

Q7.

Q8.

Q9.

Q10.

Q11.

Q12.

Q13.

Q14.

Q15.

Q16.