OP Malhotra Class 12 Maths Solutions Chapter 3 Binary Operations Ex 3(c)

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S Chand Class 12 ICSE Maths Solutions Chapter 3 Binary Operations Ex 3(c)

Question 1.
Consider the binary operation * on the set {1, 2, 3, 4, 5} defined by a * b = min {a, b}. Write the operation table of the operation *.
Solution:
Let A = {1, 2, 3, 4, 5} and binary operation * on A defined by a* b = min {a, b}
The operation table is given as under :

*	1	2	3	4	5
1	1	1	1	1	1
2	1	2	2	2	2
3	1	2	3	3	3
4	1	2	3	4	4
5	1	2	3	4	5

Question 2.
Construct the composition table for the operation * defined on A = {1, 2,3} by a * b = a + b². Is * a binary operation ?
Solution:
The composition table is given below :

*	1	2	3
1	2	5	10
2	3	6	11
3	4	7	12

Clearly all elements of table are not belongs to A.
∴ * is not a binary operation.

Question 3.
Consider a binaiy operation * on the set {1, 2, 3, 4, 5} defined by a * b = HCF {a, b}. Write the operation table.
(i) Is it a binary operation
(ii) Is it commutative ?
(iii) Compute (2 * 3) * 4 and 2 * (3 * 4)
(iv) Compute (2 * 3) * (4 * 5)
Solution:
Given binary operation * on set
A = {1, 2, 3, 4, 5} defined by a * b = HCF (a, b)
The operation table as given below :

*	1	2	3	4	5
1	1	1	1	1	1
2	1	2	1	2	1
3	1	1	3	1	1
4	1	2	1	4	1
5	1	1	1	1	5

(i) Since all entries of the table are belongs to set A
∴ * is a binary operation on A.

(ii) Clearly all entries are symmetrical about the principal diagonal.
∴ * is commutative on A.

(iii) (2 * 3) * 4 = 1 * 4 = 1
2 * (3 * 4) = 2 * 1 = 1

(iv) (2 * 3) * (4 * 5) = 1 * 1 = 1

Question 4.
Consider a binary operation * on the set {1, 2, 3, 4, 5} given by the following multiplication table

*	1	2	3	4	5
1	1	1	1	1	1
2	1	2	1	2	1
3	1	1	3	1	1
4	1	2	1	4	1
5	1	1	1	1	5

(i) Compute (a) (2 * 3) * 4 and
(b) 2 * (3 * 4).
(ii) Is * commutative ?
(iii) Compute (2 * 3) * (4 * 5).
Solution:
(i) (a) (2 * 3) * 4 = 1 * 4 = 1
(b) 2 * (3 * 4) = 2 * 1 = 1

(ii) Clearly the composition table is symmetrical about main diagonal. Hence the operation x is commutative.

*	1	2	3	4	5
1	1	1	1	1	1
2	1	2	1	2	1
3	1	1	3	1	1
4	1	2	1	4	1
5	1	1	1	1	5

(iii) (2 * 3) * (4 * 5) = 1 * 1 = 1

Question 5.
Let * be defined on the set A ={ 1, 2, 3, 4, 5, 6} by a* b = max. {a, b}.
(i) Is * a binary operation ?
(ii) Is * a commutative ?
(iii) Find the identity element.
(iv) Find invertible elements with respect to * and find their inverses.
Solution:
The operation table for * defined on A is given as under:

*	1	2	3	4	5	6
1	1	2	3	4	5	6
2	2	2	3	4	5	6
3	3	3	3	4	5	6
4	4	4	4	4	5	6
5	5	5	5	5	5	5
6	6	6	6	6	6	6

(i) Since all entries of composition table belongs to set A.
Thus * is binary operation on A.

(ii) Since the composition table is symmetrical about principal diagonal.
Thus * is commutative on A.

(iii) Since the row headed by 1 is coincide with top most row and column headed by 1 is same as left most column. Thus 1 is the identity element.

(iv) Since all rows and columns except first row and column donot contains identity element
1. So all elements are not invertible.
Clearly 1 be the only invertible element as 1 * 1 = 1.

Question 6.
A binary operation * is defined on the set {0, 1, 2, 3, 4, 5, 6} as
a * b = \(\left\{\begin{array}{cc}
a+b & \text { if } a+b<7 \\
a+b-7 & \text { if } a+b \geq 7
\end{array}\right.\)
Write the composition table of operation. Is it a binary operation? Is it commutative? Prove that zero is the identity for this operation and each element a* 0 of the set is invertible with 7 – a being inverse of a.
Solution:
The binary operation table is as under :

*	0	1	2	3	4	5	6
0	0	1	2	3	4	5	6
1	1	2	3	4	5	6	0
2	2	3	4	5	6	0	1
3	3	4	5	6	0	1	2
4	4	5	6	0	1	2	3
5	5	6	0	1	2	3	4
6	6	0	1	2	3	4	5

(i) Since all elements of binary operation table are belonging to set A. Thus * is a binary operation on A.

(ii) Since composition table is symmetrical about principal diagonal. Thus * is commutative on A.

(iii) Clearly the row headed by 0 is same as the top most row and column headed by 0 is identical with left most column. Thus 0 be the identity element.

(iv) Since every row and column contains identity element 0 thus every element is invertible. Let b be the inverse element of a.
Then a * b = 0 = b * a
Now a * (7 – a) = a + (7 – a) – 7 = 0
(7 – a) * a = 1 – a + a – 7 = 0
Thus (7 – a) be the inverse of a.

Examples

Question 1.
If a relation in R in a set A is reflexive symmetric and transitive, then t is called an ……………….. relation.
Solution:
equivalence

Question 2.
A relation R in set A is called transitive if (a, b) ∈ R and (b, c) ∈ R ⇒ ……………….. for all a, b, c ∈ A.
Solution:
(a, c) ∈ R

Question 3.
Let the relation R be defined is N by a R b, if 2a + 3b = 30, then R = ………………..
Solution:
Given relation R be defined an N by a R b if 2a + 3b = 30
⇒ b = \(\frac{30-2 a}{3}\)
When a = 1 ⇒ b = \(\frac { 28 }{ 3}\) ∉ N. ∴ (1, \(\frac { 28 }{ 3 }\)) ∉ R
When a = 2 ⇒ b = \(\frac { 26 }{ 3 }\) ∉ N
When a = 3 ⇒ b = 8 ∈ N ∴ (3, 8) ∈ R
When a = 4 ⇒ b = \(\frac { 22 }{ 3}\) ∉ N
When a = 5 ⇒ b = \(\frac { 20 }{ 3}\) ∉ N
When a = 6 ⇒ b = 6 ∉ N ∴ (6,6) ∈R
When a = 7 ⇒ b = \(\frac { 16 }{ 3}\) ∉ N
When a = 8 ⇒ b = \(\frac { 14 }{ 3}\) ∉ N
When a = 9 ⇒ b = 4 ∈ N, ∴ (9, 4) ∈ R
When a = 10 ⇒ b = \(\frac { 10 }{ 3}\) ∉ N
When a = 11 ⇒ b = \(\frac { 8 }{ 3}\) ∉ N
When a = 12 ⇒ b = 2 ∈ N ∴ (12, 2) ∈ N
When a = 13 ⇒ b = \(\frac { 4 }{ 3}\) ∉ N
When a = 14 ⇒ b = \(\frac { 2 }{ 3}\) ∉ N
For other values of a ∈ N, we does not get values of b ∈ N
R = {(3, 8), (6, 6), (9, 4), (12, 2)}

Question 4.
Every function is ……………… invertible. Only ……………… functions are invertible
Solution:
Not bijective

Question 5.
If f : R → R is defined by f(x) = 2x + 3, then f^-1 (x) is given by ………………
Solution:
Given f : R → R defined by f(x) = 2x + 3
Then fof^-1 (x) = x ∀ x ∈R
⇒ f[f^-1 (x)] = x ⇒ 2f^-1 (x) + 3 = x
⇒ f^-1(x) = \(\frac { x-3 }{ 2 }\)

Question 6.
If f(x) = (x + 3), x ∈ R and g(x) = x – 3, x ∈ R, then fog (3) = ……………
Solution:
Given f(x) = x + 3 ∀ x ∈ R
and g(x) = x – 3 ∀∈ R
∴ (fog)(x) = f[g(3)] = f(3 – 3)
= f(0) = 0 + 3 = 3

Question 7.
If f : A → B and g : B → C be the bijective functions, then (gof)^-1 is ……………
Solution:
Given f : A → B and g : B → C are bijective functions.
Then gof: A → C be a bijective function.
⇒ (gof)^-1 : C → A exists
Now f : A → B is a bijection
⇒ f^-1 : B → A is a bijection
g : B C is a bijection
⇒ g^-1 : C → B is a bijection.
∴ f^-1og^-1 : C → A exists.
Let x ∈ A, y ∈ B and z ∈ C
s.t f(x) = y, g(y) = z
Then (gof) (x) = g(f(x)) = g (y) = z
⇒ (gof)^-1 (z) = x … (1)
Given f(x) = y ⇒ f^-1 (y) = x
and g (y) = z ⇒ g^-1 (z) = y
∴ (f^-1og^-1) (z) = f^-1 (g^-1(z))
= f^-1 (y) = x … (2)
∴ from (1) and (2) ; we have
(gof)^-1 (z) = (f^-1og^-1) (z) ∀ z ∈ C
⇒ (gof)^-1 = f^-10g^-1

Question 8.
Let the functions f, g, h be defined from R to R such that f (x) = x² – 1, g(x) = \(\sqrt{x^2+1}\) and h(x) = \(\left\{\begin{array}{l}
0, \text { if } x<0 \\ x, \text { if } x>0
\end{array}\right.\) then ho (fog) (x) = …………..
Solution:
Given f : R → R defined by f(x) = x² – 1
g : R → R defined by g(x) = \(\sqrt{x^2+1}\) and h : R → R defined by
h(x) = \(\left\{\begin{array}{l}
0, \text { if } x<0 \\ x, \text { if } x>0
\end{array}\right.\)
∴ ho (fog) (x) = h {(fog) (x)}
= h[f{g(x)}] = h{f(\(\sqrt{x^2+1}\))}
= h (x² + 1 – 1) = h (x²) = x²

Question 9.
Let * be a binary operation on N given by a * b = LCM (a, b) for all a, b ∈ N, then 5 * 7 = …………..
Solution:
Given * be a binary operation on N given by a* b = LCM (a, b) ∀ a, b ∈ N
∴ 5 * 7 = LCM (5, 7) = 35

Question 10.
Let f be the greatest integer function defined as f (x) – [x] and g be the modules function defined as g (x) = | x |, then the value of gof (-\(\frac { 4 }{ 3 }\)) is …………….
Solution:
Given f(x) = [x] and g (x) = | x |
Here, D_f = R ; R_f = I; D_g = R ; R_g = [0, ∞)
Since R_f ⊂ D_g
∴ gof exists.
∴ \((g \circ f)\left(-\frac{4}{3}\right)=g\left(f\left(\frac{-4}{3}\right)\right)\)
= \(g\left(\left[\frac{-4}{3}\right]\right)=g(-2)=|-2|=2\)

Question 11.
If R= {(3, 3), (6, 6), (9, 9), (12, 12), (6,12), (3,9), (3,12), (3,6)}, is a relation on the set A = {3, 6, 9, 12}, then the function is
(a) an equivalence relation
(b) reflexive and symmetric
(c) reflexive and transitive
(d) only reflexive
Solution:
Given R = {(3, 3), (6, 6), (9, 9), (12, 12), (6, 12), (3, 9), (3, 12), (3, 6)} be a relation on A = {3, 6, 9, 12}
Reflexivity : Since (3, 3), (6, 6), (9, 9),
(12, 12) ∈ R
∴ R is reflexive on A.
Symmetry: Since (6,12) ∈ R but (12,6) ∉ R
R is not symmetric on A.
Transitivity : Clearly (a, ft), (ft, c) ∈ R
⇒ (a, c) ∈ R ∀ a, ft, c ∈ A
Thus R is transitive on A

Question 12.
If R = {(1, 3), (4, 2), (2, 4), (2, 3), (3, 1)} is a relation on the set A = {1, 2, 3, 4}, then the relation R is
(a) a function (ft) transitive
(c) not symmetric (d) reflexive Solution: Clearly (2, 4), (2, 3) ∈ R i.e. 2 have two images 4 and 3
∴ R is not a function on A.
Clearly (1, 1), (2, 2), (3, 3), (4, 4) ∈ R
∴ R is not reflexive on A.
Now (2, 3) ∈ R but (3, 2)gR
∴ R is not symmetric on A.
Now (1, 3), (3, 1) ∈ R but (1, 1) ∈R
∴ R is not transitive on A.

Question 13.
The maximum number of equivalence relations on the set A = {1, 2, 3} is
(a) 1
(b) 2
(c) 3
(d) 5
Solution:
Let given set A = {1,2, 3}
R₁ = {(1, 1), (2, 2), (3, 3)}
R₂ = {(1, 1), (2, 2), (3, 3), (1, 2), (2, 1)}
R₃ = {(1, 1), (2, 2), (3, 3), (2, 3), (3, 2)}
R₄ = {(1, 1), (2, 2), (3, 3), (1, 3), (3, 1)}
and R₅ = {(1, 1), (2, 2), (3, 3), (1, 2), (2, 1), (1, 3), (3, 1), (2, 3), (3, 2)}
All the five relations on set A are reflexive symmetric and transitive and hence equivalence relations on A.
For reflexivity : In all five relations (1, 1), (2, 2), (3, 3) must be there
For symmetry : for (a, b) ∈ R then (b, a)∈R so in R₂, (1, 2) ∈ R₂ ∴ (2, 1) must be in R₂
(1, 3) ∈ R₃ ∴ (3, 1) must be in R₃
(2, 3) ∈ R₄ ∴ (3, 2) must be in R₄
In R₅ : (1, 2), (2, 3), (1, 3) ∈ R₅
∴ (2, 1), (3, 2), (3, 1) ∈ R₅
also all the five relations are transitive on
A since the property (a, b), (b, c) ∈ R ∀ a, b, c ∈ A then (a, c) ∈ R holds good in all five relations.
∴ maximum number of equivalence relations on set A is 5.

Question 14.
The relation R to R is defined as a R ft if a ≥ b is
(a) an equivalence relation
(ft) reflexive, transitive but not symmetric
(c) symmetric, transitive but not reflexive
(d) neither transitive nor reflexive but symmetric
Solution:
Given relation R be defined as a R b iff
a ≥ b ∀ a, b ∈ R.
∀ a ∈ R, a ≥ a (always)
∴ (a, a) ∈ R i.e. a R a.
∴ R is reflexive on R.
Let a R b ∀ a, b ∈ R
thus a ≥ b ⇒ b ≤ a i.e. b ≱ a
∴ (b, a) ∉ R
∴ b R a
∴ S is not symmetric on R.
Let a R b, b R c ∀ a, b, c ∈ R
Since a R b ⇒ a ≥ b … (1)
Also b R c ⇒ b ≥ c … (2)
From (1) and (2), we have
a ≥ c
∴a R c
∴ R is transitive on R.
Hence R is reflexive, transitive but not symmetric.

Question 15.
Let S denote the set of all straight lines in a plane. Let a relation R be defined by m R n ⇔ m ⊥ n, m, n ∈ S. Then, R is
(a) reflexive only
(b) symmetric only
(c) transitive only
(d) none of these
Solution:
Given relation R be defined as m R n iff m is perpendicular to n ∀ m, n ∈ S.
Where S be the set of all straight lines in a plane.
Since every line is not ⊥ to itself.
∴ (m, m) ∉ R
∴ R is not reflexive on S.
Let m R n ∀ m, n ∈ S ⇒ m is ⊥ to n
∴ n is perpendicular to m ∴ mRn.
∴ for m R n ⇒ n R m ∀ m, n ∈ S
∴ R is symmetric on S.
∀ m, n, p ∈ S such that m ⊥ n and m ⊥ p
Now m ⊥ n and n ⊥ p then m || p.
i.e. m is parallel to line p.
i.e. m is not ⊥ to line p.

∴ R is not transitive on S.

Question 16.
If f(x) = 8x³ and g (x) = x^1/3, then find (gof) (x).
(a) x
(b) x²
(c) 2x
(d) None of these
Solution:
Given f(x) = 8x³ and g (x) = x^1/3
(gof)(x) = g(f(x)) = g(8x³)
= (8x³)^1/3 = 2x

Question 17.
If f(x) = \(\frac{2 x-1}{x+5}\), then f^-1 (x) is equal to
(a) \(\frac{x+5}{2 x-1}, x \neq \frac{1}{2}\)
(b) \(\frac{5 x+1}{2-x}, x \neq 2\)
(c) \(\frac{x-5}{2 x+1}, x \neq \frac{1}{2}\)
(d) \(\frac{5 x-1}{2-x}, x \neq 2\)
Solution:
Let y = f(x). Then f^-1 (y) = x
⇒ y = \(\frac { ∆L }{ L }\)
⇒ xy + 5y = 2x – 1
⇒ 5y + 1 = x(2 – y)
⇒ x = \(\frac{5 y+1}{2-y}\), y ≠ 2
⇒ f^-1(y) = \(\frac{5 y+1}{2-y}\)
⇒ f^-1(x) = \(\frac{5 x+1}{2-x}\)

Question 18.
Let f : N → R be the function defined by f(x) = \(\frac{2 x-1}{2}\) and g : Q → R be another function defined by g (x) = x + 2. Then (gof) \(\frac { 3 }{ 2 }\) is
(a) 1
(b) 2
(c) \(\frac { 7 }{ 2 }\)
(d) None of these
Solution:
f : N → R be a function defined by f(x) = \(\frac{2 x-1}{2}\) and g : Q → R be another function defined by g(x) = x + 2
∴ (gof) \(\frac { 3 }{ 2 }\) = g\(\left(f\left(\frac{3}{2}\right)\right)=g\left(\frac{2 \times \frac{3}{2}-1}{2}\right)\)
= g(1) = 1 + 2 = 3

Question 19.
If f(x) = x² – 1 and g (x) = (x + 1)², then (gof) x is equal to
(a) (x + 1)⁴ – 1
(b) x⁴ – 1
(c) x⁴
(d) (x + 1)⁴
Solution:
f(x) = x² – 1 and g (x) = (x + 1)²
(gof) (x) = g(f (x)) = g (x² – 1)
= (x² – 1 + 1)²
= x⁴

Question 20.
If f : R → R and g : R → R are defined by f(x) = x – 3 and g (x) = x² + 1, then the values of x for which g (f (x)) = 10 are
(a) 0, – 6
(b) 2, – 2
(c) 1, – 1
(d) 0, 6
Solution:
Given f : R → R defined by f(x) = x – 3
and g : R → R defined by g (x) = x² + 1
Also, g(f(x)) = 10
⇒ g(x – 3) = 10
⇒ (x – 3)² + 1 = 10
⇒ (x – 3)² = 9
⇒ x – 3 = ± 3
⇒ x = ± 3 + 3
⇒ x = 6, 0

Question 21.
If R is the set of real numbers and the functions f : R → R and g : R → R be defined by f (x) = x² + 2x – 3 and g (x) = x + 1, then the value of x for whichf(g (x)) = g(f (x)) is
(a) – 1
(b) 0
(c) 1
(d) 2
Solution:
Given f : R → R defined by
f(x) = x² + 2x – 3
and g : R → R defined by g (x) = x + 1
Given f(g (x)) = g (f(x))
⇒ f(x + 1) = g(x² + 2x – 3)
⇒ (x + 1)² + 2(x + 1) – 3 = x² + 2x – 3 + 1
⇒ x² + 4x = x² + 2x – 2
⇒ 2x = – 2
⇒ x = – 1

Question 22.
Let f : R → R be given by f(x) = tan x. Then, f^-1(1) is
(a) \(\frac { π }{ 4 }\)
(b) \(\left\{n \pi+\frac{\pi}{4}: n \in Z\right\}\)
(c) does not exist
(d) none of these
Solution:
Given f : R → R defined by f (x) = tan x
Let f^-1(1) = x ⇒ 1 = f(x)
⇒ tan x = 1 = tan \(\frac { π }{ 4 }\)
⇒ x = nπ + \(\frac { π }{ 4 }\), where n∈Z

Question 23.
The binary operation * : R x R → R, is defined as a * b = 2a + b, the (2 * 3) * 4 is
(a) 24
(b) 9
(c) 56
(d) 18
Solution:
Given, binary operation * : R * R → R
defined as a * b = 2a + b
∴ (2 * 3) * 4 = (2 x 2 + 3) * 4 = 7 * 4
= 2 x 7 + 4 = 18

Question 24.
If a * b denotes the larger of ‘a’ and ‘b’ and if aob = (a * b) + 3, then (5) o (10) is
(a) 18
(b) 53
(c) 13
(d) None of these
Solution:
Given aob = (a * b) + 3, where a * b denotes the larger of a and b
∴ (5) o (10) = (5 * 10) + 3
= 10 + 3 = 13

Question 25.
Let * be a binary operation on set Q of rational number defined a x b = \(\frac { ab }{ 5 }\) the identity for * if any is
(a) 0
(b) 1
(c) 5
(d) None of these
Solution:
Let e be the identity element for * on Q
Then e * a = a = a * e ∀a ∈Q-{0}
⇒ \(\frac { ae }{ 5 }\) = a and a = \(\frac { ae }{ 5 }\)
⇒ (e – 5) a = 0 and a (5 – e) = 0
⇒ e – 5 = 0
⇒ e = 5

Question 26.
If * binary operation defined on R by a * b = 1 + ab, ∀ a, b ∈ R, then the operation ‘*’ is
(a) commutative but not associative
(b) associative but not commutative
(c) neither commutative nor associative
(d) both commutative and associative
Solution:
Given * be binary operation defined on R
by a * b = 1 + ab∀ a, b ∈ R
Commutativity : a * b = 1 + ab = 1 + ba
[∵ Commutative law holds under multiplication on R]
= b * a
* is commutative on R.
Associativity : ∀ a, b, c∈ R
(a * b) * c = (1 + ab) * c = 1 + (1 + ab) c
= 1 + c + abc
and a * (b * c) = a * (1 + bc)
= 1 + a (1 + bc)
= 1 + a + abc
Now 2, 3, 4, ∈ R
(2 * 3) * 4 = (1 + 2 x 3) * 4 = 7 * 4
= 1 + 7 x 4 = 29
and 2 * (3 * 4) = 2 * (1 + 3 x 4) = 2 * 13
= 1 + 2 x 13 = 27
∴ (2 * 3) * 4 * 2 * (3 * 4)
Thus * is not associative on R

Question 27.
Let A be the non-void set of children in a family and the relation ‘a is a brother of b’ on A is
(a) reflexive
(b) symmetric
(c) transitive
(d) None of these
Solution:
Given a R b if a is a brother of b.
Let a R b ⇒ a is a brother of b
when b is girl and a is a boy
then b is not a brother of a.

∴ R is not symmetric.
Let a R b, b R c ∀ a, b, c ∈ given set
Since a R b ⇒ a is brother of b.
and b R c ⇒ b is brother of c
∴ b must be a boy.
∴ a must be a brother of c ⇒ aRc
Thus R is transitive.

Question 28.
If R is a relation on the set N, defined by {(x, y) : 2x – y = 10}, then R is
(a) reflexive
(b) symmetric
(c) transitive
(d) None of these
Solution:
Given R be relation on N defined by
{(x, y) : 2x – y = 10}
Since 2x – y – 10 ⇒ y = 2x – 10
When x = 1 ⇒ y = – 8 ∉ N
When x = 2, 3, 4, 5
⇒ y = – 6, – 4, – 2, 0 ∉ N
When x = 6, 7, 8, 9, ∴ y = 2, 4, 6, 8, ….
Thus R = (6, 2), (7, 4), (8, 6), (9, 8), (10, 10), (11, 12), (12, 14) ….}
Since (1, 1) ∉ R ∴ R is not reflexive on N.
Now (6, 2) ∈ R but (2, 6) ∉ R
∴ R is not symmetric on N.
Now (11, 12), (12, 14) ∈ R but (11, 14) ∉ R
∴ R is not transitive on N.

Question 29.
The number of equivalence relations on the set {1, 2,3} containing (1, 2) and (2, 1) isfare
(a) 3
(b) 1
(c) 2
(d) None of these
Solution:
We know that, R forms equivalence relation on set A
iff R is reflexive, symmetric and transitive on A.
Let R₁ = {(1, 1), (2, 2), (3, 3), (1, 2), (2, 1)}
and R₂ = {(1, 1), (2, 2), (3, 3), (1, 2), (2, 1), (1, 3), (3, 1), (2, 3), (3, 2)}
Clearly R₁ and R₂ forms equivalence relations on A = { 1, 2, 3}

Question 30.
On set A = (1, 2, 3}, relations R and S are given by
R = {(1, 1), (2, 2), (3, 3), (1, 2), (2, 1)}
S = {(1, 1), (2, 2), (3, 3), (3, 1)}, then
(a) R ∪ S is an equivalence relation.
(b) R ∪ S is reflexive and symmetric but not transitive.
(c)R ∪ S is symmetric and transitive but not reflexive.
(d) R ∪ S is reflexive but neither symmetric nor transitive.
Solution:
Given R = {(1, 1), (2, 2), (3, 3), (1, 2), (2, 1)}
and S = {(1, 1), (2, 2), (3, 3), (3, 1)}
∴ R ∪ S = {(1, 1), (2, 2), (3, 3), (1, 2), (2, 1), (3, 1)}
Clearly (1, 1), (2, 2), (3, 3) ∈ R ∪ S
∴ R ∪ S is reflexive on A.
Clearly (3, 1) e R ∪ S but (1, 3) ∉ R ∪ S
∴ R ∪ S is not symmetric on A.
Now (3, 1), (1, 2) ∈ R ∪ S
but (3, 2) ∉ R ∪ S
R ∪ S is not transitive on A.

Question 31.
If A = {1, 2,3, 4}, then which one of the following is reflexive?
(a) {(1, 1), (2, 3), (3, 3)}
(b) {(1, 1), (2, 2), (3, 3), (4, 4)}
(c) {(1, 2), (2, I), (3, 2), (2, 3)}
(d) {(1, 2), (1, 3), (1, 4)}
Solution:
Given A = {1, 2, 3, 4}
Now a relation R is reflexive on A
iff (a, a) ∈ R ∀ a ∈ A
Since (1, 1), (2, 2), (3, 3) (4, 4) ∈ R
∴ R is reflexive on A.

Question 32.
Let f be the greatest integer function defined as f (x) = [x] and g be the modules function defined as g (x) = | x |, then the value of gof (-\(\frac { 5 }{ 4 }\)) is
(a) 2
(b) – 2
(c) \(\frac { 5 }{ 4 }\)
(d) 1
Solution:
Given f(x) = [x] and g (x) = | x |
Here, D_f = R ; R_f = I; D_g = R ; R_g = [0, ∞)
Since R_f ⊂ D_g ∴ gof exists.
∴\((g \circ f)\left(-\frac{5}{4}\right)=g\left(f\left(\frac{-5}{4}\right)\right)\)
= \(g\left(\left[\frac{-5}{4}\right]\right)\) = g(-2) = |-2|= 2

Question 33.
If f : R → R defined as f(x) = \(\frac{3 x+5}{2}\) is an invertible function, write f^-1.
(a) \(\frac{2 y-3}{2}\)
(b) \(\frac{2 y-2}{5}\)
(c) \(\frac{2 y+5}{2}\)
(d) \(\frac{2 y-5}{2}\)
Solution:
Given f : R → R defined by f(x) = \(\frac{3 x+5}{2}\)
Since f is invertible
∴ f^-1 exists
Let y = f(x). Then f^-1 (y) = x
⇒ y = \(\frac{3 x+5}{2}\)
⇒ x = \(\frac{2 y-5}{2}\)
⇒ f^-1(y) = \(\frac{2 y-5}{3}\)
⇒ f^-1(y) = \(\frac{2 y-5}{3}\)

Question 34.
If f(x) = \(\frac{x+1}{x-1}\), then the value of f(f(x)) is equal to
(a) x
(b) 0
(c) – x
(d) 1
(e) 2
Solution:

Question 35.
Let X = {1, 2, 3, 4} and Y = {a, b, c}. Then the mapping f: X → Y defined as f(1) = a, f(2) = b, f(3) = a, f(4) = b is
(a) one-one into
(b) one-one onto
(c) many-one onto
(d) None of these
Solution:
Given x = {1, 2, 3, 4} and y = {a, b, c}
and f : X → Y defined by f(1) = a, f(2) = b, f(3) = a, f(4) = b
Here f(1) = a = f(3) i.e. the elements 1, 3 ∈ X are having same image a.
∴ different elements have same image.
∴ f is many one.
Further the element c ∴ Y have no pre-image in X
∴ f is not onto.
Thus f is many one and into.

Question 36.
If f : R → R be given by f (x) = tan x, then f^-1 (1) is
(a) \(\frac { π }{ 4 }\)
(b) nπ + \(\frac { π }{ 4 }\), n ∈ X
(c) Does not exist
(d) None of these
Solution:
f^-1 : R → R given by f(x) = tan x.
Since f^-1 is inverse of f
⇒ (f.f^-1)(1) = 1
⇒ f(f^-1(1)) = 1
⇒ tan(f^-1(1)) = 1
But tan(\(\frac { π }{ 4 }\)) = 1
⇒ tan(f^-1(1)) = tan (\(\frac { π }{ 4 }\))
⇒ f^-1(1) = nπ + \(\frac { π }{ 4 }\) n ∈ Z

Question 37.
If f(x) = \(\frac{2 x-3}{3 x+4}\), then f^-1(-\(\frac { 4 }{ 3 }\)) =
(a) 0
(b) \(\frac { 3 }{ 4 }\)
(c) – \(\frac { 2 }{ 3 }\)
(d) None of these
Solution:

Question 38.
Domain of the function \(\frac{1}{\sqrt{x+2}}\) is
(a) (- 2, ∞)
(b) (0, ∞)
(c) (2, ∞)
(d) R
Solution:
Let f(x) = \(\frac{1}{\sqrt{x+2}}\)
D_f : f (x) must be a real number.
⇒ \(\frac{1}{\sqrt{x+2}}\) must be a real number.
⇒ x + 2 ≠ 0 ⇒ x ≠ – 2
and x + 2 > 0 ⇒ x > – 2
∴ D_f = (- 2, ∞)

Question 39.
The domain of the function sin^-1 3x is
(a) \(\left[-\frac{2}{3}, \frac{2}{3}\right]\)
(b) \(\left[-\frac{1}{3}, \frac{1}{3}\right]\)
(c) [1 – 1, 1]
(d) [- 3, 2]
Solution:
Let f (x) = sin^-1 3x
D_f : f (x) must be a real number
⇒ sin^-1 3x must be a real number.
⇒ \(-1 \leq 3 x \leq 1 \Rightarrow-\frac{1}{3} \leq x \leq \frac{1}{3}\)
⇒ x ∈ \(\left[-\frac{1}{3}, \frac{1}{3}\right]\)
∴ D_f = \(\left[-\frac{1}{3}, \frac{1}{3}\right]\)

Question 40.
The domain and range of the function f(x) = \(\frac{x}{1-x}\) are respectively.
(a) R – {1}, R
(b) R – {1}, R – {- 1}
(c) R – {1}, R – {1}
(d) R – {1}, R – {0}
Solution:
Let y = f(x) = \(\frac{x}{1-x}\)
For D_f : f(x) must be a real number, x
⇒ \(\frac{x}{1-x}\) must be a real number.
⇒ 1 – x ≠ 0 ⇒ x ≠ 1
∴ D_f = R – {1}
For R_f : Let y = \(\frac{x}{1-x}\) ∀ x ∈ D_f
⇒ y – xy = x ⇒ x = \(\frac{y}{1+y}\) but x ∈ R- {1}
⇒ \(\frac{y}{1+y}\) must be a real number
⇒ y + 1 ≠ 0 ⇒ y = – 1
∴ R_f = R – {- 1}

Question 41.
The range if the function f(x) = cos(\(\frac { x }{ 3 }\)) is
(a) [-1, 1]
(b) \(\left[-\frac{1}{3}, \frac{1}{3}\right]\)
(c) [1, ∞)
(d) None of these
Solution:
Since – 1 ≤ cos θ ≤ 1 ∀ θ
⇒ – 1 ≤ cos \(\frac { x }{ 3 }\) ≤ 1
⇒ – 1 ≤ f(x) ≤ 1
∴ R_f = [- 1, 1]

Question 42.
Let f : R → R be defined by f(x) = \(\frac { 1 }{ x }\) ∀x ∈ R. Then f is x
(a) onto
(b) not defined
(c) one-one
(d) bijective
Solution:
Given f : R → R be defined by
f(x) = \(\frac { 1 }{ x }\) ∀ x ∈ R
Since x = 0, f(x) = f(0) = \(\frac { 1 }{ 0 }\) is not defined.
So 0 in (domain off) has no image in R (codomain of f). Thus f is not defined.

Question 43.
If A = {1, 2, 3, 4}, B = {1, 2, 3, 4, 5, 6} are two sets, and function f : A → B is defined by f(x) = (x + 2) ∀ ∈ A, then the function f is
(a) bijective
(b) onto
(c) one-one
(d) many-one
Solution:
Given A = {1, 2, 3, 4}
and B = {1, 2, 3, 4, 5, 6}
and f : A → B defined by
f(x) = x + 2∀X∈A
∀ x, y ∈ A s.t.f(x) = f(y)
⇒ x + 2 = y + 2 ⇒ x = y
∴ f is one-one
Further, 2 ∈ B (codomain of f) there does not exists any pre-image x in A (domain of f) s.t f(x) = 2
[if f(x) = 2 ⇒ x + 2 = 2 ⇒ x = 0 ∉ A]
∴ f is not onto

Question 44.
The function f : R → R given by f (x) = x³ – 1 is
(a) a one-one function
(b) an onto function
(c) a bijection
(d) neither one-one nor onto
Solution:
Given f : R → R defined by f(x) = x³ – 1
One-one : ∀ x, y ∈ R s.t f(x) = f(y)
⇒ x³ – 1 = y³- 1
⇒ x³ – y³ = 0
⇒ (x – y) (x² + xy + y²) = 0
⇒ x – y = 0 [∵ x, y ∈ R]
⇒ x = y
∴ f is one-one.
Onto : Let y ∈ R (codomain of f) be any arbitrary element.
Then f(x) = y ⇒ x³ – 1 = y ⇒ x = \(\sqrt[3]{y+1}\)
since y ∈ R ⇒ \(\sqrt[3]{y+1}\) ∈ R ⇒ x ∈ R
Thus, ∀ y ∈ R ∃ x = \(\sqrt[3]{y+1}\) ∈ R
s.t f(x) = f(\(\sqrt[3]{y+1}\)) = (\(\sqrt[3]{y+1}\))³ – 1
= y + 1 – 1 = y
∴ f is onto.
Thus, f is one-one and onto
∴ f is bijective.

Question 45.
If the binary operation * on the set of integers Z, is defined by a * b = a + 3b², then find the value 2 * 4.
(a) 8
(b) 6
(c) 16
(d) 50
Solution:
Given binary operation * on set of integers
Z is defined by
a * b = a + 3b²
∴ 2 * 4 = 2 + 3 x 4² = 50

Question 46.
State the reason for the relation R on the set {1, 2, 3} given by R = {(1, 2), (2, 1)} not to be transitive.
Solution:
Given A = {1, 2, 3}
R = {(1, 2), (2, 1)} is not transitive on A
Since {(1, 2) ∈ R, (2, 1)∈R but (1, 1)∉R
[For transitivity: if (a, b), (b, c)∈R. then (a, c)∈R ∀a, b, c∀∈R]

Question 47.
Let A = {1, 2, 3}, B = {4, 5, 6, 7} and let f = {(1, 4), (2, 5), (3, 5)} be a function from A to B. State whether f is one-one or not.
Solution:
Given A = {1, 2, 3} and B = {4, 5, 6, 7}
and f = {(1, 4), (2, 5), (3, 6)}
Since f (1) = 4 ; f (2) = 5 ; f (3) = 6
Thus different elements of A have different images in B.
Thus f is one-one.

Question 48.
What is the range of the function
f(x) = \(\frac{|x-1|}{x-1}\)?
Solution:
f(x) = \(\frac{|x-1|}{x-1}=\left\{\begin{array}{rll}
1 & 1 & x \geq 1 \\
-1 & ; & x<1
\end{array}\right.\)
∴ R_f = {- 1, 1}

Question 49.
Find the identity element in the set ‘A’ of ail positive rational numbers for the operation defined by a * b = \(\frac { 3ab }{ 2 }\) for all a, b ∈ Q {0}.
Solution:
Let e be the identity element for operation * in Q – {0} set of all positive rational numbers.
Then e * a = a = a* e
⇒ \(\frac { 3ea }{ 2 }\) = a = \(\frac { 3ae }{ 2 }\)
⇒ \(\frac { 3ea }{ 2 }\) = a
⇒ e = \(\frac { 2 }{ 3 }\) [∵ a ∈ Q – {0}]

Question 50.
If f : R → R be given by f (x) = (3 – x³)^1/3, then find fof (x).
Solution:
Given f : R → R defined by
f(x) = (3 – x³)^1/3
∴ fof(x) = f(f(x)) = f{(3 – x³)^1/3}
= [3 – {(3 – x³)^1/3}³]^1/3
= {3 – 3 + x³}^1/3 = x

Question 51.
Let * be an operation defined as
* = R x R → R such that a* b = 2a + b, a, b ∈ R. Check if * is a binary operation. Find if it is associative also.
Solution:
Given * be an operation defined as
* : R x R → R s.t a * b = 2a + b
Clearly a, b ∈ R ⇒ 2a + b ∈R
⇒ a * b ∈ R ∀ a, b ∈R
* is binary operation on R.
∀ a, b, c ∈ R
(a * b) * c = (2a + b) * c = 2(2a + b) + c = 4a + 2b + c
and a* (b * c) = a* (2b + c) = 2a + 2b + c
∴ (a * b) * c ≠ a * (b * c)
Now 2, 3, 4 ∈ R
s.t (2 * 3) * 4 = (2 x 2 + 3) * 4 = 7 * 4 = 2 x 7 + 4 = 18
and 2 * (3 * 4) = 2 * (2 x 3 + 4) = 2 * 10 = 2 x 2 + 10 = 14
∴ * is not associative on R

Question 52.
Consider the binary operation * on the set {1, 2, 3, 4, 5} defined by a * b = min {a, b}. Write the operation table of operation for the operation *.
Solution:
Given binary operation * on set
{1, 2, 3, 4, 5} defined by
a * b = min {a, b}

*	1	2	3	4	5
1	1	1	1	1	1
2	1	2	2	2	2
3	1	2	3	3	3
4	1	2	3	4	4
5	1	2	3	4	5

Question 53.
Let * be a binary operation on N given by a * b = HCF (a, b), ∀ a, b ∈ N. Write the value of 22 * 4.
Solution:
Given * be a binary operation on N given by a* b = HCF (a, b) ∀ a, b ∈ N
∴ 22 * 4 = HCF (22, 4) = 2
Given * be a binary operation on N given by a * b = LCM (a, b)∀a, b ∈ N
∴ 5 * 7 = LCM (5, 7) = 35

Question 54.
Let {a, b, c} and the relation R be defined as R = {{a, a), (b, c), (a, b)}. Write the minimum number of ordered pairs which should be added to R to make R reflexive and transitive.
Solution:
Let A = {a, b, c}
and R = {{a, a), (b, c), {a, b)
Now R is reflexive on A
(a, a), (b, b), (c, c) ∈ R
We know that, R is transitive on A
iff (a, b), (b, c) ∈ R
⇒ (a, c) ∈ R∀ a, b, c ∈ A
Now (a, b), (b, c) ∈ R
∴ (a, c) must be in R.
Thus, minimum no. of ordered pairs which should be added to R to make R reflexive and transitive be 3 i.e. (b, b), (c, c), (a, c).

Question 55.
Let A = {a, b, c, d} and f : A → A be given by f = {(a, b), (b, d), (c, a), (d, c)}, write f^-1.
Solution:
Given A = {a, b, c, d)
and f : A → A is defined by f = {(a, b), (b, d) (c, a), (d, c)}
i.e. f(a) = b; f(b) = d, f(c) = a
and f (d) = c
so different elements of A have different images in A
∴ f is one-one.
Further range (f) = {b, d, a, c} = A
∴ f is onto.
Thus f is 1-1 and onto.
∴ f is invertible and f^-1 exists.
Thus f^-1: A → A defined by
f^-1 = {(b, a), (d, b), (a, c), (c, d)}

Question 56.
If f : R – {\(\frac { 3 }{ 5 }\)} → R be defined as f(x) = \(\frac{3 x-5}{5 x-3}\) then show that f^-1 (x) = f(x).
Solution:
Given f : R – {\(\frac { 3 }{ 5 }\)} → R
f(x) = \(\frac{3 x-5}{5 x-3}\)
Let y = f(x) = \(\frac{3 x-5}{5 x-3}\)
⇒ y (5x – 3) = 3x – 5 ⇒ 5xy – 3y = 3x – 5
⇒ y (5x – 3) = 3y – 5 ⇒ x = \(\frac{3 y-5}{5 y-3}\)
since y = f(x)
⇒ x = f^-1(y) = \(\frac{3 y-5}{5 y-3}\)
⇒ f^-1(x) = \(\frac{3 x-5}{5 x-3}\) = f(x)

Question 57.
If f(x) = 4 – (x – 7)³, write f^-1 (x).
Solution:
Given f(x) = 4 – (x – 7)³
Let f(x) = y ⇒ y = 4 – (x – 7)³
⇒ (x – 7)³ = 4 – y
⇒ x – 7 = (4 – y)^1/3
⇒ x = 7 + (4 – y)^1/3
⇒ f^-1(y) = 7 + (4 – x)^1/3
Thus, f^-1(x) = 7 + (4 – x)^1/3