Question

Raji visited the Exhibition along with her family. The Exhibition had a huge swing, which attracted many children. Raji found that the swing traced the path of a Parabola as given by y = x² .

Answer the following questions using the above information.

1. Let f : R → R be defined by f(x) = x² is________

a. Neither Surjective nor Injective

b. Surjective

c. Injective

d. Bijective

2. Let f : N → N be defined by f(x) = x² is ________

a. Surjective but not Injective

b. Surjective

c. Injective

d. Bijective

3. Let f : {1,2,3,….}→{1,4,9,….} be defined by f(x) = x² is _________

a. Bijective

b. Surjective but not Injective

c. Injective but Surjective

d. Neither Surjective nor Injective

4. Let : N→ R be defined by f(x) = x² . Range of the function among the following is _________

a. {1, 4, 9, 16,…}

b. {1, 4, 8, 9, 10,…}

c. {1, 4, 9, 15, 16,…}

d. {1, 4, 8, 16,…}

5. The function f: Z→Z defined by f(x) = x² is__________ 

a. Neither Injective nor Surjective

b. Injective

c. Surjective

d. Bijective

Answer 1

1. (a) Neither Surjective nor Injective

Since x1 does not have unique images,

It is not one-one

Note that y is real number, so it can be negative also

which is not possible as root of negative number is not real

Hence, x is not real

so, f is not onto

Thus, f(x) is neither Surjective nor injective.

2. (C) Injective

But x₁ cannot be negative as x₁ and x₂ are natural numbers

So, only option is

When f(x₁) = f(x₂), then x₁ = X₂

Hence, f is one-one

As x is natural number, it will be positive

∴ x = \(\sqrt{y}\)

Also, since y is a natural number, let's put y = 2

Putting y = 2

x = \(\sqrt{2}\)

which is not possible as is a natural number

So, f is not onto

Thus, f(x) is only injective.

3. (a) Bijective

So, only option is

When f(x₁) = f(x₂), then x₁ = X₂

Hence, f is one-one

Check onto

f(x) = x²

Let f(x) = y, such that y E {1, 4, 9,....}

Also, since y ∈ {1, 4, 9,....}

For all values of y, we will get a value of x,

Where x ∈ {1, 2, 3,....}

So, f is onto

Thus, f(x) is one-one and onto

∴ f(x) is Bijective

4. (a) {1, 4, 9, 16,…}

For f(x) = x², where f: N → R

Range will be = {1², 2², 3², 4², 5², 6², ...}

= {1, 4, 9, 16, 25, 36, ...)

5. (a) Neither Injective nor Surjective

f(x) = x², where f: Z→ Z

Checking one-one

f(x₁) = (x₁)²

f(x₂) = (x₂)²

Putting f (x₁) = f (x₂)

(x₁)² = (x₂)²

X₁ = X₂ or x₁ = -X₂

Since x, does not have unique image,

It is not one-one

Note that y is an integer, let's put y = 5

Putting y = 5

x = \(\pm\sqrt{5}\)

But this is not possible as \(\pm\sqrt{5}\) is not an integer

So, f is not onto

Thus, f(x) is neither Surjective nor Injective.

Answer 2

1. (a) Neither Surjective nor Injective

2. (C) Injective

3. (a) Bijective

4. (a) {1, 4, 9, 16,…}

5. (a) Neither Injective nor Surjective