Relations and Functions Test

Result

Relations and Functions Test - 10

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Question 1
1 / -0

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)}. Then R is

A
Reflexive and transitive but not symmetric

B
An equivalence relation.

C
Reflexive and symmetric but not transitive

D
Symmetric and transitive but not reflexive.

Solution

The relation R is not symmetric, (1,2) ∈ R , but (2,1) ∉ R , (1,3) ∈R ,but (3,1) ∉ R , (3,2) ∈ R, but (2,3) ∉ R.
Question 2
1 / -0

The function f : R → R given by f(x) = cosx ∀x ∈ R is :

A
One – one and onto

B
One – one

C
onto but not one-one

D
Neither One – one nor onto .

Solution

f(0) = cos 0 = 1 ,and f (2π) =cos (2π)= 1. So, different elements in R may have the same image. Hence, f is not an injective function. Also, range of f(x) is not equal to its co-domain. So, f is not surjective.
Question 3
1 / -0

Let A = {1, 2, 3} and B = {2, 3, 4}, then which of the following is a function from A to B?

A
{(0, 3), (2, 4)}

B
{(1, 3), (2, 3), (3, 3)}

C
{(1, 2), (1, 3), (2, 3), (3, 3)}

D
{(1, 2), (2, 3), (3, 4), (3, 2)}

Solution

{(1, 3), (2, 3), (3, 3)} is a function ,because for each x ∈ A , there is a unique y ∈ B such that ( x , y) ∈∈f., i.e. xfy.
Question 4
1 / -0

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

A
{1, 3, 2}

B
1, 3 and 2

C
{3, 5, 6}

D
3, 5 and 6.

Solution

The domain in ordered pair (x,y) is represented by x-co ordinate. Therefore, the domain of the given function is given by: { 1 , 3 , 2 }.
Question 5
1 / -0

The range of the function f(x) =|x−1| is

A
(−∞,0)

B
(−∞,∞)(−∞,∞)

C
[0,∞)

D
(0,∞)

Solution

We have , f(x) =|x − 1|, which always gives non-negative values of f(x) for all x∈R.Therefore range of the given function is all non-negative real numbers i.e. [0,∞) .