Maths MCQs Test P 10

1. A relation R in a set A is called _______, if (a1, a2) ∈ R implies (a2, a1) ∈ R, for all a1, a2 ∈ A.

(a) symmetric

(b) transitive

(c) equivalence

(d) non-symmetric

Correct option: (a) symmetric

Solution:

A relation R in a set A is called symmetric, if (a1, a2) ∈ R implies (a2, a1) ∈ R, for all a1, a2 ∈ A.

2. Let R be a relation on the set N of natural numbers defined by nRm if n divides m. Then R is

(a) Reflexive and symmetric

(b) Transitive and symmetric

(c) Equivalence

(d) Reflexive, transitive but not symmetric

Correct option: (d) Reflexive, transitive but not symmetric

Solution:

Given that n divides m, ∀ n ∈ N, R is reflexive.

Let n = 3 and m = 6

R is not symmetric since for 3, 6 ∈ N, 3 R 6 ≠ 6 R 3.

R is transitive since for n, m, r whenever n/m and m/r ⇒ n/r, i.e., n divides m and m divides r, then n will also divide r.

So, the given relation is reflexive, transitive, but not symmetric.

3. The maximum number of equivalence relations on the set A = {1, 2, 3} are

(a) 1

(b) 2

(c) 3

(d) 5

Correct option: (d) 5

Solution:

Given, set A = {1, 2, 3}

The equivalence relations for the given set are:

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), (1, 3), (3, 1)}

R4 = {(1, 1), (2, 2), (3, 3), (2, 3), (3, 2)}

R5 = {(1, 2, 3) ⇔ A x A = A2}

Therefore, the maximum number of an equivalence relation is ‘5’.

4. If set A contains 5 elements and the set B contains 6 elements, then the number of one-one and onto mappings from A to B is

(a) 720

(b) 120

(c) 0

(d) none of these

Correct option: (c) 0

Solution:

Given,

n(A) = 5

n(B) = 6

Each element in set B is assigned to only one element in set A for the one-one function.

Here, only ‘5’ elements of set B are assigned to ‘5’ elements of set ‘A’ and one element will be left in set B.

The range of the function must be equal to B. However, for the given sets, it is not possible.

Thus, the range of functions does not contain all ‘6’ elements of set ‘ B’. Therefore, if the function is one-one it cannot be onto.

Hence, the number of one-one and onto mappings from A to B is 0.

5. Let f : [2, ∞) → R be the function defined by f(x) = x2 – 4x + 5, then the range of f is

(a) R

(b) [1, ∞)

(c) [4, ∞)

(d) [5, ∞)

Correct option: (b) [1, ∞)

Solution:

Given,

f : [2, ∞) → R

f(x) = x2 – 4x + 5

Let f(x) = y

x2 – 4x + 5 = y

x2 – 4x + 4 + 1 = y

(x – 2)2 + 1 = y

(x – 2)2 = y – 1

x – 2 = √(y – 1)

x = 2 + √(y – 1)

If the function is a real-valued function, then y – 1 ≥ 0

So, y ≥ 1.

Therefore, the range is [1, ∞).

6. Let f : R → R be defined by f(x) = 1/x ∀ x ∈ R. Then f is

(a) one-one

(b) onto

(c) bijective

(d) f is not defined

Correct option: (d) f is not defined

Solution:

f(x) = 1/x ∀ x ∈ R

Suppose x = 0, then f is not defined.

i.e. f(0) = 1/0 = undefined

So, the function f is not defined.

7. Let A = {1, 2, 3} and consider the relation R = {1, 1), (2, 2), (3, 3), (1, 2), (2, 3), (1,3)}. Then R is

(a) reflexive but not symmetric

(b) reflexive but not transitive

(c) symmetric and transitive

(d) neither symmetric, nor transitive

Correct option: (a) reflexive but not symmetric

Solution:

Given,

A = {1, 2, 3}

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

Let us write the combination of elements to check whether the given relation is reflexive, symmetric, and transitive.

R is reflexive because (1, 1),(2, 2),(3, 3) ∈ R.

R is not symmetric because (1, 2), (2, 3), (1, 3) ∈ R but (2, 1), (3, 2), (3, 1) ∉ R.

R is transitive because (1, 2) ∈ R and (2, 3) ∈ R ⇒ (1, 3) ∈ R

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

8. If f : R → R be defined by f(x) = 3x2 – 5 and g : R → R by g(x) = x/(x2 + 1), then g o f is

(a) (3x2 – 5)/(9x4 – 30x2 + 26)

(b) (3x2 – 5)/(9x4 – 6x2 + 26)

(c) 3x2/(x4 + 2x2 – 4)

(d) 3x2/(9x4 + 30x2 – 2)

Correct option: (a) (3x2 – 5)/(9x4 – 30x2 + 26)

Solution:

Given,

f(x) = 3x2 – 5

g(x) = x/(x2 + 1)

g o f = g[f(x)]

= g(3x2 – 5)

= (3x2 – 5)/ [(3x2 – 5)2 + 1]

= (3x2 – 5)/(9x4 – 30x2 + 25 + 1)

= (3x2 – 5)/(9x4 – 30x2 + 26)

9. Let f : R → R be given by f (x) = tan x. Then f–1(1) is

(a) π/4

(b) {n π + π/4 : n ∈ Z}

(c) does not exist

(d) none of these

Correct option: (a) π/4

Solution:

Given,

f(x) = tan x

Let f(x) = y

⇒ y = tan x

⇒ x = tan-1y

⇒ f-1(y) = tan-1y [since f(x) = y ⇒ x = f-1(y) ]

⇒ f-1(x) = tan-1x

Similarly,

f-1(1) = tan-1(1)

= tan-1(tan π/4)

= π/4

10. If f: R → R be given by f(x) = (3 – x3)1/3, then fof(x) is

(a) x1/3

(b) x3

(c) x

(d) (3 – x3)

Correct option: (c) x

Solution:

Given,

f(x) = (3 – x3)1/3, then

fof(x) = f[(3 – x3)1/3]

= [3 – {(3 – x3)1/3)}3]1/3

= [3 – (3 – x3)]1/3

= [3 – 3 + x3]1/3

= (x3)1/3

= x

11. Let f : A → B and g : B → C be the bijective functions. Then (g ∘ f)–1 is

(a) f–1 ∘ g–1

(b) f ∘ g

(c) g–1 ∘ f–1

(d) g ∘ f

Correct option: (a) f–1 ∘ g–1

Solution:

Given that, f : A → B and g : B → C be the bijective functions.

Let us consider three sets such as A = {1,3,4}, B ={2,5,1} and C = {3,4,2}.

f : A → B is a bijective function.

∴ f = {(1, 2), (3, 5), (4, 1)}

f-1 = {(2,1),(5,3),(1,4)}

g : B → C is a bijective function.

∴ g = {(2, 3), (5, 4), (1, 2)}

g-1 ={(3,2),(4, 5),(2, 1)}

Now,

g∘f (1) = g[f(1)] = g(2) = 3

g∘f (3) = g[f(3)] = g(5) = 4

g∘f (4) = g[f(4)] = g(1) = 2

∴ g∘f = {(1,3),(3,4),(4,2)} ….(1)

Then (g∘f)-1 = {(3,1),(4, 3),(2, 4)} ….(2)

Now, f-1∘g-1(3) = f-1[g-1(3)] = f-1(2) = 1
f-1∘g-1(4) = f-1[g-1(4)] = f-1(5) = 3
f-1∘g-1(2) = f-1[g-1(2)] = f-1(1) = 4

∴ f-1∘g-1 = {(3, 1), (4, 3), (2, 4)} ….(3)

From the (2) and (3), we can conclude that (g ∘ f)–1 = f-1∘g-1.

13. Consider a binary operation ∗ on N defined as a ∗ b = a3 + b3. Choose the correct answer.

(a) ∗ is both associative and commutative

(b) ∗ is commutative but not associative

(c) ∗ is associative but not commutative

(d) ∗ is neither commutative nor associative

Correct option: (b) ∗ is commutative but not associative

Solution:

Given that the binary operation ∗ on N is defined as a∗b= a3 + b3.

Apply the given binary operation on b∗a.

b∗a= b3 + a3 = a3 + b3

It shows that the value of a∗b is equal to that of b∗a.

So, the operation is commutative.

Consider different values of the variable as a = 1, b = 2 and c = 3.

Apply the given binary operation on (a∗b)∗c.

(a∗b)∗c = ( 1∗2 )∗3 = ( 13 + 23 )∗3 = 93 + 33 = 729 + 27 = 756

Apply the given binary operation on a∗( b∗c ).

( a∗b )∗c = 1∗( 2∗3 ) =1∗( 23 + 33 ) = 13 + 353 = 42876

( a∗b )∗c ≠ a∗( b∗c )

So the operation is not associative.

Therefore, the given operation is commutative but not associative.

14. Let f : R → R be defined by f(x) = x2 + 1. Then, pre-images of 17 and – 3, respectively, are

(a) φ, {4, – 4}

(b) {3, – 3}, φ

(c) {4, –4}, φ

(d) {4, – 4}, {2, – 2}

Correct option: (c) {4, –4}, φ

Solution:

Given,

f(x) = x2 + 1

Let f –1(17) = x

⇒ f(x) = 17 or

x2 + 1 = 17

x2 = 16

⇒ x = ± 4

or f –1(17) = {4, – 4}

Again consider f –1(–3) = x

⇒ f(x) = – 3

⇒ x2 + 1 = – 3

⇒ x2 = – 4

So, f –1(– 3) = φ.

15. Set A has 3 elements, and set B has 4 elements. Then the number of injective mappings that can be defined from A to B is

(a) 144

(b) 12

(c) 24

(d) 64

Correct option: (c) 24

Solution:

The total number of injective mappings from the set containing 3 elements into the set containing 4 elements is 4P3 = 4! = 4 × 3 × 2 × 1 = 24.

16. Let f : R → R be defined by f(x) = 3x – 4. Then f–1 (x) is given by

(a) (x + 4)/3

(b) (x/3) – 4

(c) 3x + 4

(d) None of these

Correct option: (a) (x + 4)/3

Solution:

Given,

f(x) = 3x – 4

Let f(x) = y

3x – 4 = y

3x = y + 4

x = (y + 4)/3

From f(x) = y, we can write as x = f-1(y).

So, f-1(y) = (y + 4)/3

Therefore, f-1(x) = (x + 4)/3

17. The identity element for the binary operation * defined on Q ~ {0} as a * b = ab/2 ∀ a, b ∈ Q ~ {0} is

(a) 1

(b) 0

(c) 2

(d) none of these

Correct option: (c) 2

Solution:

Given,

Given,

Binary operation * defined on Q ~ {0} as a * b = ab/2 ∀ a, b ∈ Q ~ {0}

Let e be the identity element for * such that

a*e = e*a = a….(1)

So, a*e = ae/2

⇒ a = ae/2 [From (1)]

⇒ 2a = ae

⇒ e = 2

18. Number of binary operations on the set {a, b} are

(a) 10

(b) 16

(c) 20

(d ) 8

Correct option: (b) 16

Solution:

Let the given set be A = {a, b}

n(A) = 2

Total number of binary operations = 2(2 × Number of elements in the set)

= 2(2 × 2)

= 24

= 16

Therefore, the number of binary operations on the set {a, b} is 16.

19. Let A = {1, 2, 3}. Then the number of relations containing (1, 2) and (1, 3), which are reflexive and symmetric but not transitive is

(a) 1

(b) 2

(c) 3

(d) 4

Correct option: (a) 1

Solution:

Relation R is reflexive as (1, 1), (2, 2), (3, 3) ∈ R.

Relation R is symmetric since (1, 2), (2, 1) ∈ R and (1, 3), (3, 1) ∈ R.

But relation R is not transitive as (3, 1), (1, 2) ∈ R but (3, 2)∉ R.

When we add any one of the two pairs, i.e. (3,2) and (2,3) or both, to relation R, it will become transitive.

Hence, the total number of desired relations is 1.

20. Let f , g : R → R be defined by f(x) = 3x + 1 and g(x) = x2 – 2, ∀ x ∈ R, respectively. Then, f o g is

(a) 9x2 + 6x – 1

(b) 3x2 – 5

(c) 9x2 – 6x – 3

(d) 3x2

Correct option: (b) 3x2 – 5

Solution:

Given,

f(x) = 3x + 1

g(x) = x2 – 2

f o g = f[g(x)]

= f(x2 – 2)

= 3(x2 – 2) + 1

= 3x2 – 6 + 1

= 3x2 – 5

21. The principal value of tan-1(tan 3π/5) is

(a) 2π/5

(b) -2π/5

(c) 3π/5

(d) -3π/5

Correct option: (b) -2π/5

Solution:

tan-1 (tan 3π/5)

This can be written as:

tan-1 (tan 3π/5) = tan-1 (tan[π – 2π/5])

= tan-1 (- tan 2π/5) {since tan(π – x) = -tan x}

= –tan-1 (tan 2π/2)

= –2π/5

22. sin[π/3 – sin-1(-½)] is equal to:

(a) ½

(b) ⅓

(c) -1

(d) 1

Correct option: (d) 1

Solution:

sin[π/3 – sin-1(-½)]

= sin[π/3 – sin-1[sin (-π/6))]

sin[π/3 – (-π/6)]

= sin(π/3 + π/6)

= sin π/2

= 1

23. The domain of sin–1(2x) is

(a) [0, 1]

(b) [– 1, 1]

(c) [-1/2, 1/2]

(d) [–2, 2]

Correct option: (c) [-1/2, 1/2]

Solution:

Let sin–1(2x) = θ.

Thus, 2x = sin θ.

As we know, – 1 ≤ sin θ ≤ 1

We can write this as– 1 ≤ 2x ≤ 1, which gives -1/2 ≤ x ≤ 1/2.

Therefore, the domain of sin-1(2x) is [-½, ½].

24. If sin–1x + sin–1y = π/2, then value of cos–1x + cos–1y is

(a) π/2

(b) π

(c) 0

(d) 2π/3

Correct option: (a) π/2

Solution:

Given,

sin–1 x + sin–1 y = π/2

[(π/2) – cos-1x] + [(π/2) – cos-1y] = π/2

(π/2) + (π/2) – (π/2) = cos-1x + cos-1y

Therefore, cos–1x + cos–1y = π/2.

25. Which of the following is the principal value branch of cos–1x?

(a) [–π/2, π/2]

(b) (0, π)

(c) [0, π]

(d) (0, π) – {π/2}

Correct option: (c) [0, π]

Solution:

The principal value branch of cos–1x is [0, π].

26. The value of the expression sin [cot–1 (cos (tan–1 1))] is

(a) 0

(b) 1

(c) 1/√3

(d) √(2/3)

Correct option: (d) √(2/3)

Solution:

sin [cot–1 (cos (tan–1 1))]

= sin[cot-1 {cos (tan-1 (tan π/4))}] {since tan π/4 = 1}

= sin[cot-1 (cos π/4)]

= sin[cot-1(1/√2)]

= sin [sin-1(√(⅔))] {by Pythagoras theorem}

= √(⅔)

27. The domain of y = cos–1 (x2 – 4) is

(a) [3, 5]

(b) [0, π]

(c) [-√5, -√3] ∩ [-√5, √3]

(d) [-√5, -√3] ∪ [√3, √5]

Correct option: (d)

Solution:

Given,

y = cos–1 (x2 – 4 )

⇒ cos y = x2 – 4

As we know, –1 ≤ cos y ≤ 1

So, – 1 ≤ x2 – 4 ≤ 1

Adding 4 on both sides, we get;

⇒ 3 ≤ x2 ≤ 5

Taking square root on both sides, we get;

⇒ √3 ≤ x ≤ √5

⇒ x∈ [-√5, -√3] ∪ [√3, √5]

28. If α ≤ 2 sin–1x + cos–1x ≤ β, then

(a) α = -π/2, β = π/2

(b) α = 0, β = π

(c) α = -π/2, β = 3π/2

(d) α = 0, β = 2π

Correct option: (b) α = 0, β = π

Solution:

Given,

α ≤ 2 sin–1x + cos–1x ≤ β

We know that,

-π/2 ≤ sin–1 x ≤ π/2

⇒ (-π/2) + (π/2) ≤ sin–1x + (π/2) ≤ (π/2) + (π/2)

⇒ 0 ≤ sin–1x + (sin–1x + cos–1x) ≤ π

⇒ 0 ≤ 2 sin–1x + cos–1x ≤ π

By comparing with α ≤ 2 sin–1x + cos–1x ≤ β, we get α = 0, β = π.

29. The value of sin (2 tan–1 (.75)) is equal to

(a) .75

(b) 1.5

(c) .96

(d) sin 1.5

Correct option: (c) .96

Solution:

sin (2tan–1 (.75))

Let, tan–1 (.75) = θ

tan θ = 0.75

tan θ = 3/4

Thus by Pythagoras theorem, we get;

sin θ = 3/5 and cos θ = 4/5.

Now,

sin (2tan–1 (.75)) = sin 2θ {as tan-1(.75) = θ}

= 2 sin θ cos θ

= 2 × (3/5) × (4/5)

= 24/25

= 0.96

Therefore, sin (2tan–1 (.75)) = .96.

30. sin(tan-1 x), where |x| < 1, is equal to:

(a) x/√(1 – x2)

(b) 1/√(1 – x2)

(c) 1/√(1 + x2)

(d) x/√(1 + x2)

Correct option: (d) x/√(1 + x2)

Solution:

Let tan-1x = θ.

So, tan θ = x = x/1

From this, we can write the sin θ and cos θ values as:

sin θ = x/√(1 + x2)

cos θ = 1/√(1 + x2)

Now,

sin(tan-1 x) = sin θ = x/√(1 + x2).

31. If A is a square matrix such that A2 = A, then (I – A)3 + A is equal to

(a) I

(b) 0

(c) I – A

(d) I + A

Correct option: (a) I

Solution:

Given that, A is a square matrix and A2 = A.

Consider (I + A)3, where I is the identity matrix.

Using the identity of (a + b)3 = a3 + b3 + 3ab (a + b), we get;

(I + A)3 = I3 + A3 + 3A2I + 3AI2

= I + A2(A) + 3AI + 3A

= I + A2 + 3A + 3A

= 7A + I {since it is given that A2 = A}

So, (I + A)3 = 7A + I….(1)

Now,

(I + A)3 – 7A = 7A + I – 7A [From (1)]

= I

32. If A = [aij] is a square matrix of order 2 such that aij = 1, when i ≠ j and aij = 0, when i = j, then A2 is

Solution:

Given,

A = [aij] is a square matrix of order 2 such that aij = 1, when i ≠ j and aij = 0, when i = j.

So, a11 = 0, a12 = 1, a21 = 1 and a22 = 0.

Thus,

33. Total number of possible matrices of order 3 × 3 with each entry 2 or 0 is

(a) 9

(b) 27

(c) 81

(d) 512

Correct option: (d) 512

Solution:

We know that a matrix 3× 3 contains 9 elements.

Given that each entry of this 3× 3 matrix is either 0 or 2.

Thus, by simple counting principle, we can calculate the total number of possible matrices as:

Total number of possible matrices = Total number of ways in which 9 elements can take possible values

= 29

= 512

34. If A and B are two matrices of the order 3 × m and 3 × n, respectively, and m = n, then the order of matrix (5A – 2B) is

(a) m × 3

(b) 3 × 3

(c) m × n

(d) 3 × n

Correct option: (d) 3 × n

Solution:

Given that, the order of matrix A is 3 × m, and the order of B is 3 × n.

Also, m = n.

So, the order of matrix A and B is the same, i.e. 3 × m.

Thus, subtraction of matrices is possible and (5A – 3B) also has the same order, i.e. 3 × n.

35. If

(a) 8

(b) 10

(c) 4

(d) -8

Correct option: (a) 8

Solution:

Given,

Now, by equating the corresponding elements of these two matrices, we get;

2p + q = 4….(1)

p – 2q = -3….(2)

5r – s = 11….(3)

4r + 3s = 24….(4)

By equ (1) × 2 + equ (2), we get;

4p + 2q + p – 2q = 8 – 3

5p = 5

p = 1

Substituting p = 1 in (1),

2 + q = 4

q = 4 – 2 = 2

By equ (3) × 3 + equ (4), we get;

15r – 3s + 4r + 3s = 33 + 24

19r = 57

r = 3

Substituting r = 3 in (3),

15 – s = 11

s = 15 – 11 = 4

Now,

p + q – r + 2s = 1 + 2 – 3 + 2(4) = 8

36. 

37. For any two matrices A and B, we have

(a) AB = BA

(b) AB ≠ BA

(c) AB = O

(d) None of the above

Correct option: (d) None of the above

Solution:

For any two matrices A and B,

AB = BA and AB ≠ BA are not valid unless they follow the condition of matrix multiplication.

Also, AB = O is not true in all cases.

38. If A and B are symmetric matrices of the same order, then (AB′ –BA′) is a

(a) Skew symmetric matrix

(b) Null matrix

(c) Symmetric matrix

(d) None of these

Correct option: (a) Skew symmetric matrix

Solution:

Given that A and B are symmetric matrices of the same order.

Let’s find the transpose of (AB′ –BA′).

(AB′ –BA′)′ = (AB′)′ – (BA′)′

= (BA′ – AB′)

= – (AB′ –BA′)

As (AB′ –BA′)′ = – (AB′ –BA′), the matrix (AB′ –BA′) is skew symmetric.

39. If A is a skew-symmetric matrix, then A2 is a

(a) Skew symmetric matrix

(b) Symmetric matrix

(c) Null matrix

(d) Cannot be determined

Correct option: (b) Symmetric matrix

Solution:

Given that A is a skew-symmetric matrix, so A′ =-A.

Consider the transpose of A2.

(A2)′ = (AA)′

= A′A′

=(-A)(-A)

= A2

⇒ (A2)′ = A2

Therefore, A2 is a symmetric matrix.

40.

k = -6

Also, 2k = 3a

2(-6) = 3a

3a = -12

a = -4

And

3k = 2b

3(-6) = 2b

2b = -18

b = -9

Therefore, k = -6, a = -4, and b = -9.