Class 12th PCMB

Q51. The binary operation * is defined by a * b = a + b + ab + 1, then (2 * 3) * 2 is equal to:

1. 20

2. 40

3. 400

4. 445

Ans: 4. 445

Solution:

Given: a * b = a + b + ab + 1

2 * 3 = 2 + 3 + 2 × 3 + 1

= 4 + 9 + 6 + 1

= 20

(2 * 3) * 2 = 20 * 2

= 20 + 2 + 20 × 2 + 1

= 400 + 4 + 40 + 1

= 445

Q52. Let [x] denote the greatest integer less than or equal to x. If f(x) = sin x, g(x) = [x ] and

then

1.

2.

3.

4.

Ans: 3.

Solution:

hogof(x) = h(f(g(x)))

= h(f([x]))

= h(sin [x])

= 2sin [x]

= 2 × 0 = 0

f(x) = sin x

hogof(x) = hogo(x) = 0

f(x) = x + √x2 = x ± x = 0 or 2x

2 2

2 2

2 2

2 2

-1 2 h(x) = 2x, 1 ≤ x ≤ ,

2

1

√2

fogoh(x) = π

2

fogoh(x) = π

hofog = hogof

hofog ≠ hogof

hofog = hogof

-1

-1

-1

Q53. For real numbers x and y, define xRy if is an irrational number. Then the relation R is:

1. Reflexive.

2. Symmetric.

3. Transitive.

4. None of these.

Ans: 1. Reflexive.

Solution:

We have,

is an irrational number,

As, which is an irrational number

So, R is reflexive relation.

Since,

i.e. which is an irrational number

but which is a rational number

So, R is not symmetric relation.

Also, and

So, R is not transitive relation.

Q54. Let * be a binary operation on N defined by a * b = a + b + 10 for all a, b ∈ N. The identity element for * in N is:

1. −10

2. 0

3. 10

4. Non-existent.

Ans: 4. Non-existent.

Solution:

Given a * b = a + b + 10

Let the identity element be e, then

a * e = a

⇒ a + e + 10 = a

⇒ e = -10

But the operation is defined on the set of natural numbers.

So, the identity element doesn’t exist.

Q55. If A = {a, b, c}, then the relation R = {(b, c)} on A is:

1. Reflexive only.

2. Symmetric only.

3. Transitive only.

4. Reflexive and transitive only.

Ans: 3. Transitive only.

Solution:

The relation R = {(b, c)} is neither reflexive nor symmetric because every element of A is not related to itself. Also, the ordered pair

of R obtained by interchanging its elements is not contained in R.

We observe that R is transitive on A because there is only one pair.

Q56. Subtraction of integers is:

1. Commutative but no associative.

2. Commutative and associative.

3. Associative but not commutative.

4. Neither commutative nor associative.

Ans: 1. Neither commutative nor associative.

x − y + √2

R = {(x, y) : x − y + √2 x, y ∈ R}

x − x + √2 = √2,

⇒ (x, x) ∈ R

(√2, 2) ∈ R

√2 − 2 + √2 = 2√2 − 2,

2 − √2 + √2 = 2,

⇒ (2,√2) ∉ R

(√2, 2) ∈ R (2, 2√2) ∈ R

⇒ (√2, 2√2) ∉ R

Solution:

Let , then

a * b = a – b

b * a = b – a

⇒ a * b b * a

Substraction is not commutative.

(a * b) * c

= (a – b) * c

= a – b – c

a * (b * c)

= a * (b – c)

= a – b + c

⇒ (a * b) * c a * (b * c)

Substraction is not associative.

Q57. Let f : R → R be a function defined by Then, f is:

1. One-one but not onto.

2. One-one and onto.

3. Onto but not one-one.

4. Neither one-one nor onto.

Ans: 4. Neither one-one nor onto.

Solution:

Injectivity: Let x and y be two elements in the domain (R), such that

f(x) = f(y)

⇒ (x – 8)(y + 2) = (y – 8)(x + 2)

⇒ x y + 2x – 8y – 16 = x y + 2y – 8x – 16

⇒ 10x = 10y

⇒ x = y

So, f is not one-one.

Surjectivity:

and

⇒ f is not onto.

Q58. If is given by then f (x) equals:

1.

2.

3.

4.

Ans: 1.

Solution:

Let f (x) = y

a, b ∈ Z

≠

≠

f(x) = . x2−8

x2+2

= x2−8

x2+2

y2−8

y2+2

2 2 2 2

2 2 2 2 2 2 2 2

2 2

2 2

⇒ x = ±y

f(−1) = = =

(−1)2−8

(−1)2+2

1−8

1+2

−7

3

f(1) = = =

(1)2−8

(1)2+2

1−8

1+2

−7

3

⇒ f(−1) = f(1) = −7

3

F : [1,∞) → [2,∞) f(x) = x + 1 ,

x

-1

x+√x2−4

2

x

1+x2

x−√x2−4

2

1 + √x2 − 4

x+√x2−4

2

-1

⇒ f(y) = x

⇒ y + = x 1

y

⇒ y2 + 1 = xy

⇒ y2 − xy + 1 = 0

⇒ y2 − 2 × y × + ( )2

− ( )2

+ 1 = 0 x

2

x

2

x

2

⇒ y2 − 2 × y × + ( )2 x =

2

x

2

x2−1

4

⇒ (y − )2

x =

2

x2−1

4

⇒ y − x =

2

√x2−4

2

⇒ y = x +

2

√x2−4

2

Q59. The number of binary operation that can be defined on a set of 2 elements is:

1. 8

2. 4

3. 16

4. 64

Ans: 3. 16

Solution:

Total number of binary operations on a set containing n elements is

so for n = 2 we have

Q60. R is a relation on the set Z of integers and it is given by (x, y) ∈ R ⇔ | x – y | ≤ 1. Then, R is:

1. Reflexive and transitive.

2. Reflexive and symmetric.

3. Symmetric and transitive.

4. An equivalence relation.

Ans: 2. Reflexive and symmetric.

Solution:

Reflexivity: Let Then,

for all

So, R is reflexive on Z.

Symmetry: Let Then,

[Since |x – y| = |y – x|]

for all

So, R is symmetric on Z.

Transitivity: Let and Then,

and

⇒ It is not always true that

So, R is not transitive on Z.

Q61. Let A = {1, 2, 3} and B = {(1, 2), (2, 3), (1, 3)} be a relation on A. Then, R is:

1. Neither reflexive nor transitive.

2. Neither symmetric nor transitive.

3. Transitive.

4. None of these.

Ans: 3. Transitive.

Solution:

Reflexivity: Since B is not reflexive on A.

Symmetry: Since but B is not symmetric on A.

Transitivity: Since and B is transitive on A.

Q62. If the binary operation is defined on the set Q of all positive rational numbers by . Then,

is equal to:

1.

2.

3.

4.

⇒ y = x+√x2−4

2

⇒ f−1(x) = x+√x2−4

2

(n)n2

(2)22

= 24 = 16

x ∈ R.

x − x = 0 < 1

⇒ |x − x| ≤ 1

⇒ (x, x) ∈ R x ∈ Z

x, y ∈ R.

|x − y| ≤ 0

⇒ | − (y − x)| ≤ 1

⇒ |(y − x)| ≤ 1

⇒ (y, x) ∈ R x, y ∈ Z

(x, y) ∈ R (y, z) ∈ R.

|x − y| ≤ 1 |y − z| ≤ 1

|x − y| ≤ 1.

⇒ (x, z) ∉ R

(1, 1) ∉ B,

1, 2 ∈ B 2, 1 ∉ B,

1, 2 ∈ B, 2, 3 ∈ B 1, 3 ∈ B,

⊙ + a ⊙ b = ab

4

3 ⊙ ( 1 ⊙ )

5

1

2

3

160

5

160

3

10

3

40

Ans: 1.

Solution:

Given

Q63. Choose the correct answer from the given four options.

If the 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:

1. 720

2. 120

3. 0

4. none of these.

Ans: 3. 0

Solution:

Since, the number of elements in B is more than A.

Hence, there cannot be any one-one and onto mapping from A to B.

Q64. Choose the correct answer from the given four options.

Let us define a relation R in R as aRb if a ≥ b. Then R is:

1. An equivalence relation.

2. Reflexive, transitive but not symmetric.

3. Symmetric, transitive but not reflexive.

4. Neither transitive nor reflexive but symmetric.

Ans: 2. Reflexive, transitive but not symmetric.

Solution:

We are given that, aRb if a ≥ b

⇒ aRa ⇒ a ≥ a which is true.

For relation aRb to be symmetric, we must have a ≥ b and b ≥ a which can’t be possible.

Hence, R is not symmetric.

For relation aRb to be transitive, we must have aRb and bRc.

⇒ a ≥ b and b ≥ c

⇒ a ≥ c

Hence, R is transitive.

Q65. The function f : R → R defined by f(x) = 2 + 2 is:

1. One-one and onto.

2. Many-one and onto.

3. One-one and into.

4. Many-one and into.

Ans: 3. One-one and into.

Solution:

The function f : R → R defined by f(x) = 2 + 2

Here, for each value of x we will get different values of f(x).

Hence, it is one-one function.

Also, each element of codomain is mapped to at most one element of the domain.

Function is one-one and into.

Q66. Which of the following functions from to itself are bijections?

3

160

a ⊙ b = ab

4

⇒ ( 1 ⊙ )

5

1

2

=

. 1

5

1

2

4

= 1

40

3 ⊙ ( 1 ⊙ )

5

1

2

= 3 ⊙ 1

40

=

.3 1

40

4

= 3

160

x |x|

x |x|

A = {x : −1 ≤ x ≤ 1}

1.

2.

3.

4.

Ans: 2.

Solution:

1. Range of

So, f is not a bijection.

2. Range

So, g is a bijection.

3. h(-1) = |-1| = 1

And h(1) = |1| = 1

⇒ -1 and 1 have the same images.

So, h is not a bijection.

4. k(-1) = (-1) = 1

And k(1) = (1) = 1

⇒ -1 and 1 have the same images.

So, k is not a bijection.

Q67. Choose the correct answer from the given four options.

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:

1. Reflexive but not symmetric.

2. Reflexive but not transitive.

3. Symmetric and transitive.

4. Neither symmetric, nor transitive.

Ans: 1. Reflexive but not symmetric.

Solution:

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

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

Hence, R is reflexive.

but

Hence, R is not symmetric.

and

Hence, R is transitive.

Q68. A binary operation * on Z defined by a * b = 3a + b for all a, b ∈ Z, is:

1. Commutative.

2. Associative.

3. Not commutative.

4. Commutative and associative.

Ans: 3. Not commutative.

Solution:

Let

a * b = 3a + b

b * a = 3b + a

Thus, a * b b * a

If a = 1 and b = 2,

1 * 2 = 3(1) + 2

= 5

2 * 1 = 3(2) + 1

= 7

1 * 2 2 * 1

Thus, * is not commutative on Z.

f(x) = x

2

g(x) = sin ( ) πx

2

h(x) = |x|

k(x) = x2

g(x) = sin ( ) πx

2

f = [ , ] ≠ A −1

2

1

2

= [ sin ( ), sin ( )] = [−1, 1] = A −π

2

π

2

2

2

∵ (1, 1), (2, 2), (3, 3) ∈ R

(1, 2) ∈ R (2, 1) ∉ R

(1, 2) ∈ R (2, 3) ∈ R

⇒ (1, 3) ∈ R

a, b ∈ Z

≠

≠

Q69. Let f: R → R be defined as f(x) = 3x. Choose the correct answer.

1. f is one-one onto

2. f is many-one onto

3. f is one-one but not onto

4. f is neither one-one nor onto.

Ans: f: R → R is defined as f(x) = 3x.

Let such that f(x) = f(y).

⇒ 3x = 3y

⇒ x = y

f is one-one.

Also, for any real number (y) in co-domain R, there exists in R such that

f is onto.

Hence, function f is one-one and onto.

The correct answer is A.

1. f is one-one onto.

Q70. The relation ‘R’ in N × N such that (a, b)R(c, d) ⇔ a + d = b + c is:

1. Reflexive but not symmetric.

2. Reflexive and transitive but not symmetric.

3. An equivalence relation.

4. None of the these.

Ans: 3. An equivalence relation.

Solution:

We observe the following properties of relation R.

Reflexivity: Let

So, R is reflexive on N × N.

Symmetry: Let such that (a, b)R(c, d)

So, R is symmetric on N × N.

Transitivity: Let such that (a, b)R(c, d) and (c, d)R(e, f)

⇒ a + d = b + c and c + f = d + e

⇒ a + d + c + f = b + c + d + e

⇒ a + f = b + e

⇒ (a, b)R(e, f)

So, R is transitive on N × N.

Hence, R is an equivalence relation on N.

Q71. The function f : R → R, f(x) = x is:

1. Injective but not surjective.

2. Surjective but not injective.

3. Injective as well as surjective.

4. Neither injective nor surjective.

Ans: 4. Neither injective nor surjective.

Solution:

Given function is f : R → R, f(x) = x

If f(x) = f(y) then

x = y

Hence, it is not one-one or injective.

x, y ∈ R

∴

y

3

f( ) = 3( ) = y y

3

y

3

∴

(a, b) ∈ N × N

⇒ a, b ∈ N

⇒ a + b = b + a

⇒ (a, b) ∈ R

(a, b), (c, d) ∈ N × N

⇒ a + d = b + c

⇒ d + a = c + b

⇒ (d, c), (b, a) ∈ R

(a, b), (c, d), (e, f) ∈ N × N

2

2

2 2

⇒ x ± y

f(x) = y

y = x

But co-domain is R.

Hence, it is not onto or surjective.

Q72. Let be a function defined as The inverse of f is the map g: Range

given by:

1.

2.

3.

4.

Ans: Given:

Now, Range of

Let

Therefore, option (B) is correct.

2.

Q73. Let f : Z → Z be given by Then, f is:

1. Onto but not one-one.

2. One-one but not onto.

3. One-one and onto.

4. Neither one-one nor onto.

Ans: 1. Onto but not one-one.

Solution:

Given function is

if x is even

= 0 if x is odd

For f(3) = 0 and f(4) = 0

⇒ f(3) = f(4)

But,

Hence, it is not one-one.

Here, Domain = range of f

Hence, it is onto.

Q74. Let R be a relation on N defined by x + 2y = 8. The domain of R is:

1. {2, 4, 8}

2. {2, 4, 6, 8}

3. {2, 4, 6}

4. {1, 2, 3, 4}

Ans: 3. {2, 4, 6}

Solution:

The relation R is defined as R = x, y: and x + 2y = 8

⇒ R = x, y: and

Domain of R is all values of satisfying the relation R.

Also, there are only three values of x that result in y, which is a natural number.

These are {2, 6, 4}.

2

x = ±√y

f : R − { − } → R 4

3 f(x) = . 4x

3x+4

f : R − { − } → R 4

3

g(y) = 3y

3−4y

g(y) = 4y

4−3y

g(y) = 4y

3−4y

g(y) = . 3y

4−3y

f : R − { − } → R and f(x) = 4

3

4x

3x+4

f → R − { − } 4

3

y = f(x) ∴ y = ⇒ 3xy + 4y = 4x 4x

3x+4

⇒ x(4 − 3y) = 4y ⇒ x = 4y

4−3y

∴ f−1(y) = g(y) = 4y

3−4y

g(y) = . 4y

4−3y

f(x) = { , if x is even

0, if x is odd

.

x

2

f(x) = x

2

3 ≠ 4

x ∈ R ⇒ y ∈ R

x, y ∈ N

x, y ∈ N y = 8−x

2

x ∈ N

Q75. Consider a non-empty set consisting of children in a family and a relation R defined as aRb if a is brother of b.

Then, R is:

1. Symmetric but not transitive.

2. Transitive but not symmetric.

3. Neither symmetric nor transitive.

4. Both symmetric and transitive.

Ans: 4. Transitive but not symmetric.

Solution:

We have,

R = {(a, b): a is brother of b}

Let Then,

a is brother of b.

but b is not necessary brother of a (As, b can be sister of a)

So, R is not symmetric.

Also,

Let and

⇒ a is brother of b and b is brother of c

⇒ a is brother of c

So, R is transitive.

Q76. Let R be a relation on the set N given by R = {(a, b): a = b – 2, b > 6}. Then,

1. (2, 4) ∈ R

2. (3, 8) ∈ R

3. (6, 8) ∈ R

4. (8, 7) ∈ R

Ans: 1. (6, 8) ∈ R

Solution:

a = b – 2 ⇒ 6 = 8 – 2 and b = 8 > 6

Hence, (6, 8) ∈ R

Q77. The law a + b = b + a is called:

1. Closure law.

2. Associative law.

3. Commutative law.

4. Distributive law.

Ans: 3. Commutative law.

Solution:

The law a + b = b + a is commutative.

Q78. Let R be the relation on the set A = {1, 2, 3, 4} given by R = {(1, 2), (2, 2), (1, 1), (4, 4), (1, 3), (3, 3), (3, 2)}. Then,

1. R is reflexive and symmetric but not transitive.

2. R is reflexive and transitive but not symmetric.

3. R is symmetric and transitive but not reflexive.

4. R is an equivalence relation.

Ans: 2. R is reflexive and transitive but not symmetric.

Solution:

Reflexivity: Clearly,

So, R is reflexive on A.

Symmetry: Since, but R is not symmetric on A.

Transitivity: Since, and R is transitive on A.

Q79. Let f(x) = x be a function with domain {0, 1, 2, 3}. Then domain of f is:

1. {3, 2, 1, 0}

(a, b) ∈ R.

⇒ (b, a) ∉ R

(a, b) ∈ R (b, c) ∈ R

⇒ (a, c) ∈ R

(a, a) ∈ R ∀ a ∈ A

1, 2 ∈ R, 2, 1 ∉ R,

1, 3, 3, 2 ∈ R 1, 2 ∈ R,

3 -1

2. {0, -1, -2, -3}

3. {0, 1, 8, 27}

4. {0, -1, -8, -27}

Ans: 3. {0, 1, 8, 27}

Solution:

Given function is f(x) = x be a function with domain {0, 1, 2, 3}.

Range = {0, 1 , 2 , 3 } = {0, 1, 8, 27}

f can be written as

{(0, 0), (1, 1), (2, 8), (3, 27)}

Hence, f can be written as

{(0, 0), (1, 1), (8, 2), (27, 3)}

Domain of f is {0, 1, 8, 27}

Q80. An operation * is defined on the set Z of non-zero integers by a * b = ab for all a, b ∈ Z. Then the property

satisfied is:

1. Closure.

2. Commutative.

3. Associative.

4. None of these.

Ans: 4. None of these.

Solution:

* is not clouser because when a = 1 and b = 2,

* is not commutative because when a = 1, b = 2 and c = 3,

Thus,

Q81. Let R be the relation over the set of all straight lines in a plane such that Then, R is:

1. Symmetric.

2. Reflexive.

3. Transitive.

4. An equivalence relation.

Ans: 1. Symmetric.

Solution:

Given R is the relation over the set of all straight lines in a plane such that

It is symmetric relation as we can say either or

Q82. The maximum number of equivalence relations on the set A = {1, 2, 3} is:

1. 1

2. 2

3. 3

4. 5

Ans: 4. 5

Solution:

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

R = {(1, 1)}

R = {(2, 2)}

3

3 3 3

-1

-1

a ∗ b = a = ∈ Z

b

1

2

1 ∗ (2 ∗ 3) = 1 ∗ ( 2 )

3

= 1

( ) 2

3

= 3

2

(1 ∗ 2) ∗ 3 = 1 ∗ 3

2

=

( ) 1

2

3

= 1

6

1 ∗ (2 ∗ 3) ≠ (1 ∗ 2) ∗ 3

l1Rl2 ⇔ l1⊥l2.

l1Rl2 ⇔ l1⊥l2.

l2⊥l1.

1

2

R = {(3, 3)}

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

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

The maximum number of equivalence relations on the set A = {1, 2, 3} is 5.

Q83. Let f : R → R be given by Then, f (1) is:

1.

2.

3. Does not exist.

4. None of these.

Ans: 2.

Solution:

We have, f : R → R is given by

Q84. Let f(x) = x and g(x) = 2 . Then, the solution set of the equation fog(x) = gof(x) is:

1. R

2. {0}

3. {0, 2}

4. None of these.

Ans: 3. {0, 2}

Solution:

Since (fog)(x) = (gof)(x),

f(g(x)) = g(f(x))

Q85. Q is the set of all positive rational numbers with the binary operation * defined by . The

inverse of an element is:

1.

2.

3.

4.

Ans: 4.

Solution:

Let e be the identity element in Q with respect to * such that

a * e = a = e * a,

a * e = a and e * a = a,

and ,

Thus, 2 is the identity element in Q with respect to *.

Let and be the inverse of a.

Then,

a * e = a = e * a

a * b = e and b * a = e

and

3

4

5

f(x) = tan x. -1

π

4

{nπ + : n ∈ Z} π

4

{nπ + : n ∈ Z} π

4

f(x) = tan x

⇒ f−1(x) = tan−1 x

∴ f−1(1) = tan−1 1 = {nπ + : n ∈ Z} π

4

2 x

⇒ f(2x) = g(x2)

⇒ (2x)2

= 2×2

⇒ 22x = 2×2

⇒ x2 = 2x

⇒ x2 − 2x = 0

⇒ x(x − 2) = 0

⇒ x = 0, 2

⇒ x ∈ {0, 2}

+ a ∗ b = ab ∀ a, b ∈ Q+

2

a ∈ Q+

a

1

a2

a4

a

4

a

+

∀ a ∈ Q+

∀ a ∈ Q+

= a ae

2 = a ea

2 ∀ a ∈ Q+

e = 2 ∈ Q+, ∀ a ∈ Q+

+

a ∈ Q+ b ∈ Q+

= 2 ab

2 = 2 ba

2

b = 4 ∈ Q+

a

Thus, is the inverse of .

Q86. If the binary operation * on Z is defined by a * b = a − b + ab + 4, then value of (2 * 3) * 4 is:

1. 233

2. 33

3. 55

4. −55

Ans: 2. 33

Solution:

Given that a * b = a – b + ab + 4

So,

2 * 3

= 2 – 3 + 2.3 + 4

= 4 – 9 + 6 + 4

= 5

Now,

(2 * 3) * 4

= 5 * 4

= 5 – 4 + 5.4 + 4

= 25- 16 + 20 + 4

= 33

Q87. If A = {1, 2, 3}, B = {1, 4, 6, 9} and R is a relation from A to B defined by ‘x is greater than y’. The range of R is:

1. {1, 4, 6, 9}

2. {4, 6, 9}

3. {1}

4. None of these.

Ans: 3. {1}

Solution:

Here, and ⇒ R = 2, 1, 3, 1

Thus, Range of R = {1}

Q88. Let * be a binary operation on R defined by a * b = ab + 1. Then, * is:

1. Commutative but not associative.

2. Associative but not commutative.

3. Neither commutative nor associative.

4. Both commutative and associative.

Ans: 1. Commutative but not associative.

Solution:

Commutativity:

Let

a * b = ab + 1

= ba + 1

= b * a

Therefore,

a * b = b * a,

Therefore, * is commutative on R.

Associativity:

Let

a * (b * c) = a * (bc + 1)

= a(bc + 1) + 1

= abc + a + 1

(a * b) * c = (ab + 1) * c

= (ab + 1)c + 1

= abc + c + 1

a * (b * c) (a * b) * c

4

a a ∈ Q+

2 2

2 2

2 2

2 2

R = x, y : x ∈ A y ∈ B : x > y

a, b ∈ R

∀ a, b ∈ R

a, b, c ∈ R

∴ ≠

For example: a = 1, b = 2 and c = 3 [which belong to R]

Now,

1 * (2 * 3) = 1 * (6 + 1)

= 1 * 7

= 7 + 1

= 8

(1 * 2) * 3 = (2 + 1) * 3

= 3 * 3

= 9 + 1

= 10

⇒ 1 * (2 * 3) (1 * 2) * 3

Therefore, a = 1, b = 2 and c = 3 which belong to R such that

a * (b * c) (a * b) * c

Hence, * is not associative on R.

Q89. The function f : A → B defined by f(x) = -x + 6x- 8 is a bijection if,

1. and

2. and

3. and

4. and

Ans: 1. and

Solution:

f(x) = -x + 6x – 8, is a polynomial function and the domain of polynomial function is real number.

f(x) = -x + 6x – 8

= -(x – 6x + 8)

= -(x – 6x + 9 – 1)

= -(x – 3) + 1

Maximum value of -(x – 3) woud be 0

Maximum value of -(x – 3) + 1 woud be 1

We can see from the given graph that function is symmetrical about x = 3 and the given function is bijective.

So, x would be either

The correct option which satisfy A and B both is:

and

Q90. If a relation R is defined on the set Z of integers as follows: (a, b) ∈ R ⇔ a + b = 25. Then, domain (R) is:

1. {3, 4, 5}

2. {0, 3, 4, 5}

3.

4. None of these.

Ans: 3.

Solution:

As aRb ⇔ a < b

does not satisfy reflexive and symmetric relation.

Q91. Let f : R → R be a function defined by Then,

1. f is a bijection.

2. f is an injection only.

3. f is surjection on only.

4. f is neither an injection nor a surjection.

≠

∃

≠

2

A = (−∞, 3] B = (−∞, 1]

A = [−3,∞) B = (−∞, 1]

A = (−∞, 3] B = [1,∞)

A = [3,∞) B = [1,∞)

A = (−∞, 3] B = (−∞, 1]

2

∴ x ∈ R

2

2

2

2

2

∴ 2

∴ f(x) ∈ (−∞, 1]

(−∞, 3] or [3,∞)

A = (−∞, 3] B = (−∞, 1]

2 2

{0,±3,±4,±5}

{0,±3,±4,±5}

f(x) = . e|x|−e−x

ex+e−x

Ans: 4. f is neither an injection nor a surjection.

Solution:

f : R → R

For x = -2 and -3

Hence, for different values of x we are getting same values of f(x)

That means, the given function is many one.

Therefore, this function is not injective.

For x < 0

f(x) = 0

For x > 0

The value of is always positive.

Therefore, the value of f(x) is always less than 1.

Numbers more than 1 are not included in the range but they are included in co-domain.

As the codomain is R.

Hence, the given function is not onto.

Therefore, this function is not surjective.

Q92. The function f : R → R defined by f(x) = 6 + 6 is:

1. One-one and onto.

2. Many one and onto.

3. One-one and into.

4. Many one and into.

Ans: 4. Many one and into.

Solution:

Graph of the given function is as follows:

A line parallel to X-axis is cutting the graph at two different values.

Therefore, for two different values of x we are getting the same value of y.

That means it is many one function.

From the given graph we can see that the range is and R is the co-domain of the given function.

Hence, Co-dornain = Range

Therefore, the given function is into.

Q93. Let A = {1, 2, ……., n} and B = {a, b}. Then the number of subjections from A into B is:

1.

2.

3.

4.

Ans: 2.

Solution:

The number of functions from a set with n number of elements into a set with 2 number of elements = 2

But two functions can be many-one into function.

f(x) = e|x|−e−x

ex+e−x

∈ R

f(-2) = e|−2|−e2

e−2+e2

= e2−e2

e−2+e2

= 0

f(x) = ex−e−x

ex+e−x

= − ex+e−x

ex+e−x

2e−x

ex+e−x

= 1 − 2e−x

ex+e−x

2e−x

ex+e−x

∴ Co-domain ≠ Range

x |x|

[2,∞)

nP2

2n − 2

2n − 1

nC2

2n − 2

n

Q94. For the binary operation * defined on R − {1} by the rule a * b = a + b + ab for all a, b ∈ R − {1}, the inverse of a is:

1.

2.

3.

4.

Ans: 2.

Solution:

Let e be the identity element in R – {1} with respect to * such that

a * e = a = e * a,

a * e = a and e * a = a,

Then,

a + e + ae = a and e + a + ea = a,

e(1 + a) = 0,

Thus, 0 is the identity element in R – {1} with respect to *.

Let and be the inverse of a. Then,

a * b = e = b * a

a * b = e and b * a = e

⇒ a + b + ab = 0 and b + a + ba = 0

Thus, is the inverse of .

Q95. Let Then, for what value of is f(f(x)) = x?

1.

2.

3. 1

4. -1

Ans: 4. -1

Solution: Given function is

Also f(f(x)) = x

Comparing on both sides,

Q96. Choose the correct answer from the given four options.

Let f : N → R be the function defined by and g : Q → R be another function defined by g(x) = x + 2.

Then is:

1.

2.

3.

4.

Ans: 4.

Solution:

We have and g(x) = x + 2

−a

− a

a−1

1a

a2

− a

a−1

∀ a ∈ R − {1}

∀ a ∈ R − {1}

∀ a ∈ R − {1}

∀ a ∈ R − {1}

e = 0 ∈ R − {1}

a ∈ R − {1} b ∈ R − {1}

⇒ b(1 + a) = −a ∈ R − {1}

⇒ b = ∈ R − {1} −a

a−1

−a

a−1 a ∈ R − {1}

f(x) = , x ≠ −1. αx

x+1 α

√2

−√2

f(x) = , x ≠ −1 αx

x+1

f( ) = x αx

x+1

= x

α( ) αx

x+1

+1 αx

x+1

= x α2x

αx+x+1

α2 = αx + x + 1

α2 = (α + 1)x + 1

α + 1 = 0 ⇒ α = −1

f(x) = 2x−1

2

(gof) 3

2

1

1

7

2

None of these.

none of these.

f(x) = 2x−1

2

gof( ) = g(f( )) 3

2

3

2

= g( ) 2× −1 3

2

2

Q97. If R is the largest equivalence relation on a set A and S is any relation on A, then:

1.

2.

3.

4. None of these.

Ans: 2.

Solution:

Given that R is the largest relation on A and S is any relation on A.

We know that R is always subset of A × A.

Hence,

Q98. If A = {a, b, c, d}, then a relation R = {(a, b), (b, a), (a, a)} on A is:

1. Symmetric and transitive only.

2. Reflexive and transitive only.

3. Symmetric only.

4. Transitive only.

Ans: 1. Symmetric and transitive only.

Solution:

Given that A = {a, b, c, d} then a relation R = {(a, b), (b, a), (a, a)} on A.

(a, b), (b, a)

⇒ R is symmetric.

Also for (a, a) R is symmetric.

Q99. Let f: R → R be defined as f(x) = x . Choose the correct answer.

1. f is one-one onto

2. f is many-one onto

3. f is one-one but not onto

4. f is neither one-one nor onto.

Ans: f: R → R is defined as f(x) = x .

Let such that f(x) = f(y).

⇒ x = y

f(x ) = f(x ) does not imply that x = x

For instance,

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

f is not one-one.

Consider an element 2 in co-domain R. It is clear that there does not exist any x in domain R such that f(x) = 2.

f is not onto.

Hence, function f is neither one-one nor onto.

The correct answer is D.

4. f is neither one-one nor onto.

Q100. Let R be the relation in the set N given by R = {(a, b) : a = b – 2, b > 6}. Choose

the correct answer.

1.

2.

3.

4.

Ans: Given: a = b − 2, b > 6

(A) a = 2, b = 4 , Here b > 6 is not true, therefore, this option is incorrect

(B) a = 3, b = 8 and a = b – 2 ⇒ 3 = 8-2 ⇒ 3 = 6, which is false.

Therefore, this option is incorrect

(C) a = 6, b = 8 and b = b – 2 ⇒ 6 = 8 – 2 ⇒ 6 = 6, which is true.

= g(1) = 1 + 2 = 3

R ⊂ S

S ⊂ R

R = S

S ⊂ R

S ⊂ R.

∈ R

4

4

x, y ∈ R

4 4

⇒ x = ±y

∴ 1 2 1 2

∴

∴

(2, 4) ∈ R

(3, 8) ∈ R

(6, 8) ∈ R

(8, 7) ∈ R.

Therefore, this option is correct

(D) a = 8, b = 7 and a = b – 2 ⇒ 8 = 7 – 2 ⇒ 8 = 5, which is false.

3.