Relations and Functions NCERT Solutions

NCERT Exercise 1.2 Q: 6 - 12

Question: 6. Let A = {1, 2, 3,}, B = {4, 5, 6, 7 } and let f = {(1, 4), (2, 5), (3, 6)} be a function from A to B. Show that f is one-one.

Solution:

Here, A = {1, 2, 3} and B = {4, 5, 6, 7}

F: A →B = {(1, 4), (2, 5), (2, 6)}

F (1) = 4, f(2) = 5 and f(3) = 6

⇒ f(x₁) ≠ f(x₂) for every x₁ ≠ x₂

So, f is one-one function.

Question: 7. In each of the following cases state whether the function is one-one, onto or bijective. Justify your answer.

(i) f :R → R define by f(x) = 3 – 4x

(ii) f : R → R defined by f(x) = 1 + x²

Solution:

(i) Let x₁ and x₂ ∈ R

Here f(x) = 3 – 4x

Then f(x₁) = 3 – 4x₁

and f (x₂) = 3 – 4x₂

Now f(x₁) = f (x₂)

⇒ 3 – 4x₁ = 3 – 4x₂

⇒ – 4x₁ = –4x₂

⇒ x₁ = x₂

∴ f(x₁) = f(x₂)

& x₁ = x₂

So, f is one-one function.

Let f(x) = y ∈ R

So, f is onto function.

(ii) Let x₁ = 2 and x₂ = –2 ∈ R

Here f(x) = 1 + x²

Then f(x₁) = f (2) = 1 + (2) ² = 5

and f (x₂) = f(–2) = 1 + (–2) ² = 5

∴ f (x₁) = f(x₂) but x₁ ≠ x₂

So, f is not one-one function.

Let f(x) = – 2 ⇒R

Then 1 + x² = –2

⇒ x₂ = –3 = ± √–3 ∈R

So, f is not onto function.

Question: 8. Let A and B sets show that f : A x B → B x A such that f (a, b) = (b, a) is bijective function.

Solution:

Let x₁ = (a, b) and x₂ = (c, d) ∈ A x B

Then f (x₁) = f (a, b) = (b, a)

F(x₂) = f (c, d) = (d, c)

Now f (x₁) = f (x₂)

⇒ (b, a) = (d, c) ⇒b = d and a = c

Now x₁ = (a, b) = (c, d) = x₂

& there4; f(x₁) = f(x₂)⇒ x₁ = x₂

So f is one-one function.

Let y = (p, q) ∈ B x A and x = (a, b) ∈ A x B

∴ f(a, b) = (p, q)

⇒ b = p and a = q

&rArr x = (a, b) = (p, q)

Also f (x) = f (q, p) = (p, q) = y

So, f is onto function.

Thus f is one-one and onto function.

Question: 9. Let f: N → N be defined by f(n) for all n ∈ N. State whether the function f is bijective. Justify your answer.

Solution:

Let n₁ = 3 and n₂ – 4 ∈ N

∴ f(n₁) = f (n₂) but n₁ ≠ n₂

So, f is not one-one function.

Since, f is not one - one, it can't be bijective.

Question: 10. Let A = R – {3} and B = R – {1}. Consider the function f : A → B defined by . Is f one-one and onto? Justify your answer.

Solution:

Let x₁ ≠ 3 and x₂≠ 3 ∈ A

⇒ (x₁ – 2)(x₂ –3) = ((x₁ – 3) ((x₂ – 2)

⇒ x₁ x₂ – 3x₁ – 2x₂ + 6 = x₁x₂ – 2x₁ – 3x₂ + 6

⇒ – 3x₁ – 2x₂ = –2x₁ – 3x₂

⇒ –3x₁ + 2x₁ = –3 x₂ + 2x₂

⇒ – x₁ = –x₂

⇒ x₁ = x₂

∴ f(x₁) = f (x₂)

So, f is one-one function.

Let f(x) = y ≠ 1 ∈ B

⇒ x – 2 = y (x – 3)

⇒ x – 2 = xy – 3y

⇒ x(y – 1) = 3y – 2

Since x is the only value for which f(x) = y (as proved avove), f is onto function. Thus, f is one-one and onto function.

Question: 11. Let f : R → R be defined as f(x) = x⁴ choose the correct answer.

(A) f is one -one onto

(B) f is many -one onto

(C) f is one -one but not onto

(D)f is neither one -one nor onto

Solution:

Let x₁ = 2 and x₂ = 2 ∈ R

Here, f(x) = x⁴

Then, f(x₁) = f(2) = 2⁴ = 16

and f(x₂) = f(–2)= (–2)⁴ = 16

∴ f(x₁) = f(x₂) but x₁ ≠ x₂

So, f is not one-one function.

Let f(x) = – 2 ∈ R

Then x⁴ = –2 which is not possible because there is no value of x corresponding to which x⁴ = –2

So, f is not onto function.

Thus, f is neither one-one nor onto.

So, answer is (D).

Question: 12. Let f : R → R be defined as f(x) = 3x choose the correct answer.

(A) f is one-one onto

(B) f is many-one onto

(C) f is one -one but onto

(D) f is neither one-one nor onto.

Solution:

Let, x₁ and x₂ ∈ R

Here, f(x) = 3x

Then, f (x₁) = 3x₁ and f(x₂) = 3x₂

Now f(x₁) = f(x₂)

⇒ 3 x₁ = 3 x₂

⇒ x₁ = x₂

∴ f(x₁) = f(x₂) implies x₁ = x₂

So, f is one-one function.

Let f(x) = y ∈ R. Then, 3x = y

So, f is onto function.

Thus f is one-one and onto function. So answer is (A).

12-math-home

Reference: