Relations and Functions Test

Result

Relations and Functions Test - 14

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Weekly Quiz Competition

Question 1
1 / -0

Let A be the set of all 50 students of Class X in a school. Let f : A → N be function defined by f (x) = roll number of the student, f is

A
many-one onto

B
Onto

C
one – one and onto

D
one – one but not onto

Solution

Since no two different students of the class can have the same roll number. Therefore , f is one – one,also f(A) = f(A) = Range of f ={1,2,3,4,…50} ≠ N i.e. range of f is not same as its co-domain. So ,f is not onto.
Question 2
1 / -0

If N is the set of integers f : N → N, given by f(x) = 2x is

A
many-one onto

B
one – one but not onto

C
Onto

D
one – one and onto

Solution

Injectivity : For f(x₁) = f(x₂) ⇒ 2x₁= 2x₂ ⇒ x₁= x₂ ⇒ f is one-one, ∀ x₁,x₂∈ N

Surjectivity : y = f(x) ⇒ y = 2x ⇒ x = y/2 ∉ N(domain) ⇒ f is not onto
Question 3
1 / -0

If R is the set of real numbers, a function f:R→R defined as f (x) = x², then f is

A
one – one and onto

B
one – one but not onto

C
neither one – one nor onto

D
many-one onto

Solution

Here , 1 and – 1 ∈ R such that f(-1) = f(1) i.e. there are two distinct elements in R, which have the same image. So , f is not one –one. Since f(x) assumes only non-negative values.So, no negative real number in R( co-domain ) has its pre-image in domain of f i.e. R. There f is not onto.
Question 4
1 / -0

Which of the following functions from Z to Z is a bijection?

A
f (x) = x² + 1

B
f (x) = x³

C
f (x) = 2x + 1

D
f (x) = x + 2

Solution

Injectivity : Let x , y ∈ Z, then , f(x) = f(y) ⇒ x + 2 = y + 2 ⇒x = y.⇒f is one-one. Surjectivity: Let f(x) = y ,

where y ∈ Z, .⇒ x + 2 = y .⇒ x = y – 2 ∈ Z, ⇒ f is onto. Therefore, f is a bijective function.