** ** ** ** **NCERT Solutions for Maths Class 12 Exercise 1.1**

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**NCERT Solutions for Maths Class 12 Exercise 1.2**

**NCERT Solutions for Maths Class 12 Exercise 1.3**

**NCERT Solutions for Maths Class 12 Exercise 1.4**

**NCERT Solutions for Maths Class 12 Exercise 1.1**

**Maths Class 12 Ex 1.1 ****Question 1.**

**Question 1.**

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

**(i)
Relation R in the set A = {1, 2, 3, …. 13, 14} defined as R = {(x, y): 3x – y =
0}**

**(ii)
Relation R in the set N of natural numbers defined as R = {(x, y): y = x + 5
and x < 4}**

**(iii)
Relation R in the set A= {1, 2, 3, 4, 5, 6} as R = {(x, y): y is divisible by
x}**

**(iv)
Relation R in the set 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 at a particular time given by**

**(a)
R = {(x, y): x and y work at the same place}**

**(b)
R = {(x, y): x and y live in the same locality}**

**(c)
R = {(x, y): x is exactly 7 cm taller than y}**

**(d)
R = {(x, y): x is wife of y}**

**(e)
R = {(x, y): x is father of y}**

**Solution:**

**(i)** Relation R in the set A = **{1, 2, 3, …. 13, 14}** defined
as R = {(x, y): 3x – y = 0}

(a) Putting y = x, we get 3x – x ≠ 0 ⇒
R is not reflexive.

(b) If 3x – y = 0, then 3y – x ≠ 0 ⇒
R is not symmetric.

(c) If 3x – y = 0, 3y – z = 0,then 3x – z ≠ 0 ⇒ R
is not transitive.

**(ii)**
Relation R in the set N of natural numbers defined by R = {(x, y): y = x + 5
and x < 4}

(a) Putting y = x, we get x ≠ x + 5 ⇒
R is not reflexive.

(b) If y = x + 5, then x ≠ y + 5 ⇒ R
is not symmetric.

(c) If y = x + 5, z = y + 5, then z ≠ x + 5 ⇒ R
is not transitive.

**(iii)**
Relation R in the set A = {1, 2, 3, 4, 5, 6} as R = {(x, y): y is divisible by
x}

(a) Putting y = x, we get, x is divisible by x ⇒
R is reflexive.

(b) If y is divisible by x, then x is not divisible by y, when x ≠ y ⇒ R is not symmetric.

(c) If y is divisible by x and z is divisible by y, then z is divisible by x.

For example, if 8 is divisible by 4 and 16 is divisible by 8, then
16 is divisible by 4 ⇒ R is transitive.

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

(a) Putting y = x, we get, x – x = 0, which is an integer ⇒ R is reflexive.

(b) If x – y is an integer, then y – x is also an integer ⇒ R is symmetric.

(c) If x – y is an integer and y – z is an integer, then x – z is also an
integer ⇒ R is transitive.

**(v)** Relation
R in the set of human beings in a town at a particular time is given.

(a) R = {(x, y): x and y work at the same place}.

It is reflexive as x works at the same place.

It is symmetric since x and y or y and x work at the same place.

If x, y work at the same place and y, z work at the same place, then x, z also
work at the same place. Therefore, R is transitive.

(b) R = {(x, y): x and y live in the same locality}

It is reflexive as x lives in the same locality.

It is symmetric since x and y or y and x live in the same
locality.

If x, y live in the same locality and y, z live in the same locality, then x, z
also live in the same locality. Therefore, R is transitive.

(c) R = {(x, y): x is exactly 7 cm taller than y}.

It is not reflexive because x cannot be 7 cm taller than x.

It is not symmetric because x is exactly 7 cm taller than y but y
cannot be exactly 7 cm taller than x.

It is not transitive because if x is exactly 7 cm taller than y
and y is exactly 7 cm taller than z, then x is not exactly 7 cm taller than z.

(d) R = {(x, y): x is wife of y}

It is not reflexive because x cannot be wife of x.

It is not symmetric because x is wife of y but y is not wife of x.

It is transitive because if x is a wife of y then y is a male and cannot be the
wife of anybody else. Here, (x, y) ∈
R but (y, z) do not belong to R. Therefore, by the special case, it is
transitive.

(e) R = {(x, y): x is a father of y}

It is not reflexive because x cannot be the father of himself.

It is not symmetric because x is a father of y but y cannot be the
father of x.

It is not transitive because x is a father of y and y is a father of z but x
cannot be the father of z.

**Maths Class 12 Ex 1.1 ****Question 2.**

**Question 2.**

**Show
that the relation R in the set R of real numbers, defined as**

**R
= {(a, b): a ≤ b²} is neither reflexive nor symmetric nor transitive.**

**Solution:**

**(i)** R is not reflexive since a is
not less than or equal to a² for all a ∈
R, for example, 1/3 is not less
than or equal to 1/9.

**(ii)** R is not symmetric since a ≤ b²
but b is not less than or equal to a², for example, 1 < 2² but 2 is not less
than 1².

**(iii)** R is not transitive because
for a ≤ b² and b ≤ c², a is not less than or equal to c², for example, 3 ≤
(-2)², -2 ≤ (-1)², but 3 is not less than or equal to (-1)².

**Maths Class 12 Ex 1.1 ****Question 3.**

**Question 3.**

**Check
whether the relation R defined in the set {1, 2, 3, 4, 5, 6} as**

**R
= {(a, b): b = a + 1} is reflexive, symmetric or transitive.**

**Solution:**

**(i)** R is not reflexive because a ≠ a
+ 1.

**(ii)** R is not symmetric because if b
= a + 1, then a ≠ b + 1

**(iii)** R is not transitive because if
b = a + 1 and c = b + 1, then c ≠ a + 1.

**Maths Class 12 Ex 1.1 ****Question 4.**

**Question 4.**

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

**Solution:**

**(i)** R is reflexive because if b = a,
then a ≤ a ⇒ a = a is true.

**(ii)** R is not symmetric because a ≤
b but b is not less than or equal to a, for example, 3 ≤ 5, but 5 is not less
than or equal to 3.

**(iii)** R is transitive because if a ≤
b and b ≤ c, then a ≤ c, for example, 3 ≤ 5, 5 ≤ 6 ⇒ 3 ≤ 6.

**Maths Class 12 Ex 1.1 ****Question 5.**

**Question 5.**

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

**Solution:**

**(i)** R is not reflexive since a is
not less than or equal to a^{3} for all a ∈ R, for example, 1/3 is
not less than (1/3)^{3}.

**(ii)** R is not symmetric since a ≤ b^{3}
but b is not less than or equal to a^{3}, for example, 1 < 2^{3}
but 2 is not less than 1^{3}.

**(iii)** R is not transitive because
for a ≤ b^{3} and b ≤ c^{3}, a is not less than or equal to c^{3},
for example, 3 ≤ (-2)^{3}, -2 ≤ (-1)^{3}, but 3 is not less
than or equal to (-1)^{3}.

**Maths Class 12 Ex 1.1 ****Question 6.**

**Question 6.**

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

**Solution:**

**(i)** Since (1, 1), (2, 2), (3, 3) do
not belong to relation R.

∴ R is not reflexive.

**(ii)** Since (1, 2) and (2, 1) belong
to R. Therefore, R is symmetric.

**(iii)** Since (1, 2) and (2, 1) belong
to R but (1, 1) does not belong to R. Therefore, R is not transitive.

**Maths Class 12 Ex 1.1 ****Question 7.**

**Question 7.**

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

**Solution:**

**(i)** The number of pages in the same
book remains the same.

Therefore, the relation R is reflexive.

**(ii)** If the book x has the same
number of pages as the book y, then the book y has the same number of pages as
the book x.

Therefore, the relation R is symmetric.

**(iii)** If the books x and y have the
same number of pages and the books y and z have also the same number of pages,
then the books x and z have the same number of pages.

Therefore, the relation R is transitive.

Thus, the relation R is an equivalence relation.

**Maths Class 12 Ex 1.1 ****Question 8.**

**Question 8.**

**Show that the relation R in the set A= {1, 2, 3, 4, 5} given by R = {(a, b): |a – b| is even}, is an equivalence relation. Show that all the elements of {1, 3, 5} are related to each other and all the elements of {2, 4} are related to each other. But no element of {1, 3, 5} is related to any element of {2, 4}.**

**Solution:**

We have the set A = {1, 2, 3, 4, 5} and the relation R = {(a, b): |a – b| is
even}

Therefore, R = {(1, 3), (1, 5), (3, 5), (2, 4)}

**(a)** (i) Suppose a is an element of the
set A. Then |a – a| = 0, which is even.

Therefore, R is reflexive.

(ii) If |a – b| is even, then |b – a| is also even. For example, if |1 – 3| is
even, then |3 – 1| is also even.

Therefore, R is symmetric.

(iii) If |a – b| and |b – c| are even, then |a – c| is also even. For example,
if |1 – 3| and |3 – 5| are even, then |1 – 5| is also even.

Therefore, R is transitive.

Hence, R is an equivalence relation.

**(b)** All
the elements of {1, 3, 5} are related to each other.

Because |1 – 3| = 2, |3 – 5| = 2, |1 – 5| = 4. All are even numbers.

Similarly, all the elements of {2, 4} are related to each other.

Because |2 – 4| = 2, which is an even number.

Now, no element of set {1, 3, 5} is related to any element of set {2, 4}.

Because |1 – 2| = 1, |1 – 4| = 3, |3 – 2| = 1, |3 – 4| = 1, |5 –
2| = 3, |5 – 4| = 1 are not even numbers.

**Maths Class 12 Ex 1.1 ****Question 9.**

**Question 9.**

**Show that each of the relation R in the set 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:**

The set A = {x ∈ Z: 0 ≤ x ≤ 12} = {0, 1, 2, ….., 12}

**(i)** R = {(a, b): |a – b| is a
multiple of 4}

∴ R = {(1, 5), (1, 9), (2, 6), (2,
10), (3, 7), (3, 11), (4, 8), (4, 12), (5, 9), (6, 10), (7, 11), (8, 12), (0, 0),
(1, 1), (2, 2), ….. (12, 12)}

(a) For any element a ∈ A, we have (a, a) ∈ R. For example, (0, 0), (1, 1), (2, 2), ….., (12, 12) ∈ R.

∴ R is reflexive.

(b) Here, (1, 5) ∈ R and (5, 1) ∈ R, i.e., (a, b) and (b, a) both belong to R.

Therefore, R is symmetric.

(c) Here, (1, 5) ∈ R and (5, 9) ∈ R implies (1, 9) ∈
R. It means for a, b, c ∈ A, (a, b) ∈ R and (b, c) ∈
R implies (a, c) ∈ R. This shows when |a – b| and |b
– c| are both multiples of 4, then |a – c| is also a multiple of 4. Therefore,
R is transitive.

∴ R is an equivalence relation.

The set of all elements related to 1 is {1, 5, 9}.

**(ii)**
R = {(a, b): a = b}

∴ R = {(0, 0), (1, 1), (2, 2), …..,
(12, 12)}

(a) For any element a ∈ A, we have a = a, it means (a, a)
∈ R

Therefore, R is reflexive.

(b) Again if (a, b) ∈ R, then (b, a) also ∈R. It means if a = b, then b = a.

Therefore, R is symmetric.

(c) If (a, b) ∈ R and (b, c) ∈ R, then (a, c) ∈ R.
It means if a = b and b = c, then a = c.

Therefore, R is transitive.

The set of all elements related to 1 is {1}.

**Maths Class 12 Ex 1.1 ****Question 10.**

**Question 10.**

**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)** Let A = {straight lines in a
plane}

R = {(a, b): a is perpendicular to b}

(a) A line a is not perpendicular to itself. Therefore, R is not
reflexive.

(b) If a is perpendicular to b, then b is also perpendicular to a.
Therefore, R is symmetric.

(c) If a is perpendicular to b and b is perpendicular to c, then a
is not perpendicular to c. Therefore, R is not transitive.

Hence, R is symmetric but neither reflexive nor transitive.

**(ii)**
Let A = {real numbers}

R = {(a, b): a > b}

(a) An element is not greater than itself.

Therefore, R is not reflexive.

(b) If a > b, then b is not greater than a. Therefore, R is not symmetric.

(c) If a > b and b > c, then a > c. Therefore, R is transitive.

Hence, R is transitive but neither reflexive nor symmetric.

**(iii)** Let
A = {1, 2, 3}

R = {(a, b): |a – b| is either 0 or 1}

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

(a) (1, 1), (2, 2), (3, 3) ∈
R, therefore, R is reflexive.

(b) (1, 2), (2, 1), (1, 3), (3, 1) ∈
R, therefore, R is symmetric.

(c) Since (1, 2) ∈ R and (2, 3) ∈ R but (1, 3) ∉ R.
Therefore, R is not transitive.

Hence, R is reflexive and symmetric but not transitive.

**(iv)** Let
A = {1, 2, 3}

R = {(a, b): a ≤ b}

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

(a) (1, 1), (2, 2), (3, 3) ∈
R, therefore, R is reflexive.

(b) (1, 2) ∈ R, but (2, 1) ∉ R, therefore, R is not symmetric.

(c) (1, 2) ∈ R and (2, 3) ∈ R implies (1, 3) ∈ R,
therefore, R is transitive.

Hence, R is reflexive and transitive but not symmetric.

**(v)** Let
A = {1, 2, 3}

R = {(a, b): 0 < |a – b| ≤ 2}

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

(a) Since (1, 1), (2, 2), (3, 3) ∉
R, therefore, R is not reflexive.

(b) Since (1, 2), (2, 1), (1, 3), (3, 1), (2, 3), (3, 2) ∈R, therefore, R is symmetric.

(c) Since (1, 2) ∈ R and (2, 3) ∈ R implies (1, 3) ∈ R,
therefore, R is transitive.

Hence, R is symmetric and transitive but not reflexive.

**Maths Class 12 Ex 1.1 ****Question 11.**

**Question 11.**

**Show that the relation R in the set A of points in a plane given by R = {(P, Q): distance of the point P from the origin is same as the distance of the point Q from the origin}, is an equivalence relation. Further, show that the set of all points related to a point P ≠ (0, 0) is the circle passing through P with origin as centre.**

**Solution:**

Let O be the origin. Then the relation R is defined as

R = {(P, Q): OP = OQ}

(i) R is reflexive. For any distance OP, OP = OP. Therefore, R is reflexive.

(ii) R is symmetric. If OP = OQ, then OQ = OP. Therefore, R is symmetric.

(iii) R is transitive. Let OP = OQ and OQ = OR, then OP = OR. Therefore, R is
transitive.

Hence, the relation R is an equivalence relation.

For all points **P ≠ (0, 0)**,
OP = constant.

Therefore, P lies on a circle with origin as centre.

**Maths Class 12 Ex 1.1 ****Question 12.**

**Question 12.**

**Show that the relation R defined in the set A of all triangles as R= {(T**

_{1}, T_{2}): T_{1}is similar to T_{2}}, is equivalence relation. Consider three right angle 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}and T_{3}are related?

**Solution:**

(i) In a set A of all triangles, the relation R is defined as

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

(a) R is reflexive. A triangle T is similar to itself. Therefore, (T, T) ∈ R for all T ∈
A.

Therefore, R is reflexive.

(b) R is symmetric. If triangle T_{1} is similar to triangle T_{2},
then triangle T_{2} is similar to triangle T_{1}. Therefore, R
is symmetric.

(c) R is transitive. If triangle T_{1} is similar to triangle T_{2}
and triangle T_{2} is similar to triangle T_{3}, then triangle
T_{1} is similar to triangle T_{3}. Therefore, R is transitive.

Hence, R is an equivalence relation.

(ii) According to the proportionality theorem, two triangles are similar, if
their sides are proportional. Now, sides 3, 4, 5 of triangle T_{1} are
proportional to the sides 6, 8, 10 of triangle T_{3}.

Therefore, triangle T_{1} is related to triangle T_{3}.

**Maths Class 12 Ex 1.1 ****Question 13.**

**Question 13.**

**Show that the relation R defined in the set A of all polygons as R = {(P**

_{1}, P_{2}): P_{1}and P_{2}have 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 and 5?

**Solution:**

Let *n* be the number of sides of
polygon P_{1}. Then, the relation R is defined as

R = {(P_{1}, P_{2}): P_{1} and P_{2} are *n*-sided polygons}

(i) (a) R is reflexive. Any polygon P_{1} has *n* sides. Therefore, R is reflexive.

(b) R is symmetric. If P_{1} has *n*
sides, P_{2} also has *n* sides
and if P_{2} has *n* sides, P_{1}
also has *n* sides. Therefore, R is
symmetric.

(c) R is transitive. If P_{1}, P_{2} are *n*-sided and P_{2}, P_{3} are *n*-sided polygons, then P_{1} and P_{3} are also *n*-sided polygons. Therefore, R is
transitive.

Hence, R is an equivalence relation.

(ii) The set of all elements in A related to the right angle triangle T with
sides 3, 4, and 5 = {all the triangles in a plane}

**Maths Class 12 Ex 1.1 ****Question 14.**

**Question 14.**

**Let L be the set of all lines in XY plane and R be the relation in L defined as 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 = 2x + 4.

**Solution:**

It is given that the set L = {all the lines in XY plane}

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

(i) (a) R is reflexive. The line L_{1} is parallel to itself. Therefore,
R is reflexive.

(b) R is symmetric. If L_{1} is parallel to L_{2}, then L_{2}
is parallel to L_{1}. Therefore, R is symmetric.

(c) R is reflexive. If L_{1} is parallel to L_{2} and L_{2}
is parallel to L_{3}, then L_{1} is parallel to L_{3}.
Therefore, R is transitive.

Hence, R is an equivalence relation.

(ii) The set of all lines related to the line y = 2x + 4 is y = 2x + c, where c
∈ R, because y = 2x + c is always
parallel to y = 2x + 4.

**Maths Class 12 Ex 1.1 ****Question 15.**

**Question 15.**

**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:**

**(b)**** R is reflexive and transitive but not
symmetric.**

(i) (1, 1), (2, 2), (3, 3), (4, 4) ∈ R.
Therefore, R is reflexive.

(ii) (1, 2) ∈ R but (2, 1) does not belong to
R. Therefore, R is not symmetric.

(iii) (1, 2), (2, 2) ∈ R
implies (1, 2) ∈ R. (1, 3), (3, 3) ∈ R implies (1, 3) ∈ R.
Again, (1, 3), (3, 2) ∈ R implies (1, 2) ∈ R. Therefore, R is transitive.

**Maths Class 12 Ex 1.1 ****Question 16.**

**Question 16.**

**Let R be the relation in the set N given by R = {(a, b): a = b – 2, b > 6}. Choose the correct answer.**

**(a) (2, 4)**

**∈**

**R (b) (3, 8)**

**∈**

**R (c) (6, 8)**

**∈**

**R (d) (8, 7)**

**∈**

**R**

**Solution:**

**(c)** **(6, 8) ****∈**** R**
satisfies the condition a = b – 2, because if b = 8, a = 8 – 2 = 6 and b >
6.

If we take b = 7, a = 7 – 2 = 5 and (5, 7) is not given in the
option.

Therefore, option (c) is correct.

**NCERT Solutions for Maths Class 12 Exercise 1.2**