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QUESTION: 1

Let A = {1, 2, 3} and let R = {(1, 1), (2, 2), (3, 3), (1, 3), (3, 2), (1, 2)}. Then R is

Solution:

R is reflexive and transitive but not symmetric

QUESTION: 2

Let A {a, b, c} and let R = {(a, a)(a, b), (b, a)}. Then, R is

Solution:

R is symmetric and transitive but not reflexive

QUESTION: 3

Let A = {1, 2, 3} then total number of element in A x A is

Solution:

QUESTION: 4

Let S be the set of all straight lines in a plane. Let R be a relation on S defined by a R b ⇔ a ⊥ b. then, R is

Solution:

a ⊥ a is not true. So, R is not reflexive

a ⊥ b and b ⊥ c does not imply a ⊥ c. So, R is not transitive

But, a ⊥ b ⇒ b ⊥ a is always true.

QUESTION: 5

Let S be the set of all real numbers and let R be a relation on S, defined by a R b ⇔ |a – b| __<__ 1. Then, R is

Solution:

(i) |a – a| = 0 __<__ 1 is always true

(ii) a R b ⇒ |a – b| __<__ 1 ⇒ |-(a – b)| __<__ ⇒ |b – a| __<__ 1 ⇒ b R a.

(iii) 2R 1 and

But, 2 is not related to 1/2. So, R is not transitive.

QUESTION: 6

Domain of f (x)

Solution:

Domain of sin ^{-1}x is [-1,1]

QUESTION: 7

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

Solution:

R = {(a, b): a = b − 2, b > 6}

Now, since b > 6, (2, 4) ∉ R

Also, as 3 ≠ 8 − 2, (3, 8) ∉ R

And, as 8 ≠ 7 − 2

∴ (8, 7) ∉ R

Now, consider (6, 8).

We have 8 > 6 and also, 6 = 8 − 2.

∴ (6, 8) ∈ R

QUESTION: 8

Let R = {(3, 3), (6, 6), (9, 9), (3,6), (3, 9), (9, 12), (3,12), (6, 12), (12, 12)}, be a relation on the set A = {3, 6, 9, 12} Then the relation is

Solution:

R is reflexive

∴ (3, 3), (6,6), (9, 9), (12, 12) ∈ R

again ∴ (6, 12) ∈ R but (12, 6) ∉ R ⇒ R is not symmetric

R is transitive

[∴ (3, 6) ∈ R, (6, 12) ∈ R, (3, 12) ∈ R others are clear

QUESTION: 9

If f(x) = (a – x^{n})^{1/n}. then f(f(x)) =

Solution:

f(x) = (a – x^{n})^{1/n} = y

QUESTION: 10

Solution:

QUESTION: 11

f: N → N : f(x) = 2x is

Solution:

2x = 3 ⇒ x = 3/2 ∉ N. So, f is in to many-one and into

many-one and into

QUESTION: 12

f: R → R : f(x) = x^{2} is

Solution:

Given, function f : R→R such that f(x) = x^{2},

Let A and B be two sets of real numbers.

Let x_{1}, x_{2} ∈ A such that f(x_{1}) = f(x_{2}).

⇒1 + (x_{1})^{2} = 1 + (x_{2})^{2}

⇒ (x_{1})^{2} − (x_{2})^{2} = 0

⇒(x_{1 }− x_{2})(x_{1} + x_{2})=0

⇒ x_{1 }= ± x_{2}.

Thus f(x_{1}) = f(x_{2}) does not imply that x_{1} = x_{2}.

For instance, f(1) = f(−1) = 1, i.e. , two elements (1, -1) of A have the same image in B. So, f is many-one function.

QUESTION: 13

f : R → R : f(x) = x^{3} is

Solution:

⇒ x^{2 }= x^{2}

∴ f is one-one

Let y∈R and let y=x^{3}. then x=y^{⅓} ∈ R

Thus for each y in the codomain R there exists y^{1/3} in R such that

f(y^{13})=(y^{13})^{3}=y

∴ f is onto.

Hence f is one -one onto.

QUESTION: 14

Solution:

f(x) = sinx

f(x) = sinx is a one-one function.

codomain = range

therefore, f(x) = sinx is an onto function.

QUESTION: 15

f : R → R : f(x) = cos x is

Solution:

cos(2π - θ) = cosq ⇒ f is many-one.

Range (f) = [-1,1] ⊂ R ⇒ f is in to.

QUESTION: 16

The domain of the function f = {(1, 3), (3, 5), (2, 6)} is

Solution:

Domain = {1, 3, 2}

QUESTION: 17

Let f(x) = x - 1/x +1 , x ≠ -1, then f^{-1} (x) is

Solution:

QUESTION: 18

If f(x) = x/x -1 , x ≠ 1, then f^{-1} (x) is

Solution:

QUESTION: 19

Solution:

QUESTION: 20

If f(x) = cos (log x), then has the value

Solution:

= cos (log x) cos (cos y) [cos(log(x/y))+cos(log(xy))]

= cos (log x) cos (log y) [cos(log x - log y)+cos(log x +log y)]

= cos (log x) cos (log y) [2 cos(log x)cos(log y)] = 0

QUESTION: 21

Solution:

QUESTION: 22

If f(x) = then (fof) (x) = ?

Solution:

(fof) (x) = f[f(x)] – {(3 – x^{3})^{1/3}} = f(y) where y = (3 – x^{3})^{1/3}

= (3 – y^{3})^{1/3} = [3- (3 – x^{3})]^{1/3} = (x^{3})^{1/3} = x

QUESTION: 23

If f(x) = x^{2} – 3x + 2, then (fof) (x) = ?

Solution:

(fof) (x) = f[f(x)] = f(x^{2} - 3x + 2)^{2} - 3(x^{2} – 3x + 2)

= y^{2} – 3y + 2 = (x^{2} – 3x + 2)^{2} – 3(x^{2} – 3x + 2) + 2 = (x^{4} – 6x^{3} + 10x^{2} – 3x)

QUESTION: 24

Let S = {1, 2, 3}. The function f : S → S defined as below have inverse for

Solution:

Since f(2) = f(3) = 1, then inverse does not exists

Inverse exist for (C) f^{-1} = {(3, 1), (2, 3), (1, 2)}.

QUESTION: 25

The relation R defined on the set N of natural numbers by xRy ⇔2x^{2} - 3xy + y^{2} = 0 is

Solution:

(i) xRx ⇔ 2x^{2} - 3x.x + x^{2}

∴ R is reflexive

(ii) For x = 1, y= 2; 2x^{2} - 3xy + y^{2} = 0

∴ 1 R 2 but 2.2^{2} - 3.2.1 + 1^{2} = 3 ≠ 0.

So, 2 is not R-related to 1.

∴ R is not symmetric.

QUESTION: 26

Let R = {(3, 3), (6, 6), (9, 9), (12, 12), (6, 12), (3, 9), (3, 12), (3, 6) be a relation on the set A = {3, 6, 9, 12}. The relation is

Solution:

Since, (3, 3), (6, 6), (9, 9), (12, 12) ∈ R ⇒ R is reflexive relation.

Now, (6, 12)∉R but (12, 6) ∉ R ⇒ R is not a symmetric relation.

Also, (3, 6),(6, 12) ∈ R ⇒ (3, 12) ∈ R

⇒ R is transitive relation.

QUESTION: 27

Let R be a relation on the set A of ordered pairs of positive integers defined by (x, y)R(u, v) if and only if xv = yu, then

Solution:

Clearly, (x, y)R(x, y),

since xy = yx. This shows that R is reflexive.

Further, (x, y)R(u, v) ⇒ xv = yu ⇒ uy vx and hence (u, v)R(x, y). This shows that R is symmetric.

Similarly, (x, y)R(u, v) and (u, v)R(a, b) ⇒ xv = yu and ub = va ⇒ xb = ya and hence (x, y)R(a, b). Thus, R is transitive. Thus, R is an equivalent relation.

QUESTION: 28

If R be a relation defined as aRb iff |a -b|> 0, then the relation is

Solution:

Since, R is a defined as aRb iff |a – b| > 0

For reflexive aRa iff |a – a| > 0

Which is not true, So, R is not reflexive

For symmetric aRb iff |a – b| > 0

Now bRa iff |b – a| > 0

⇒ |a – b| > 0 ⇒ ArB

Thus, R is symmetric.

For transitive aRb iff |a – b| > 0

bRc iff |b – c| > 0

⇒ |a – b + b – c| > 0

⇒ |a – c| > 0 ⇒ | c – a| > 0 ⇒ aRc

∴ R is transitive

QUESTION: 29

If R is an equivalence relation of a set A, then R^{-1} is

Solution:

If R is an equivalence relation, then R^{-1} is also an equivalence relation.

QUESTION: 30

Let r be a relation from R (set of real numbers) to R defined by r = {(a, b)|a, b∈R and a - b + √3 is an irrational number}. The relation r is

Solution:

Given, r = {(a, b)| a, b∈R and as an irrational number

(i) Reflexive which is irrational number.

(ii) Symmetric

Now, which is not an irrational.

Also, which is an irrational.

Which is not symmetric.

(iii) Transitive

∴ It is not transitive.

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