Gujarat Board GSEB Textbook Solutions Class 12 Maths Chapter 1 Relations and Functions Ex 1.1 Textbook Questions and Answers.

## Gujarat Board Textbook Solutions Class 12 Maths Chapter 1 Relations and Functions Ex 1.2

Question 1.

Show that the function f: R → R, defined by f (x) = \(\frac{1}{x}\) is one-one onto, where R is the set of all non-zero real numbers. Is the result true, if the domain R is replaced by N with co-domain being same as R ?

Solution:

(i) f(x) = \(\frac{1}{x}\), If f(x_{1}) = f(x_{2}) ⇒ \(\frac{1}{x_{1}}\) = \(\frac{1}{x_{2}}\)

⇒ x_{1} = x_{2}

Each x ∈ R, has a unique image in codomain.

⇒ f is one-one.

(ii) For each y belonging codomain, then y = \(\frac{1}{x}\) or x = \(\frac{1}{y}\), there is a unique pre-image of y.

⇒ f is onto.

(b) When domain R, is replaced by N, codomain R, remaining the same, then f : N → R,

If f(x_{1}) = f(x_{2})

⇒ \(\frac{1}{n_{1}}\) = \(\frac{1}{n_{2}}\)

⇒ n_{1} = n_{2} where n_{1}, n_{2} ∈ N.

⇒ f is one-one.

But for every real number belonging to codomain may not have a pre-image in N.

e.g. \(\frac{1}{\frac{2}{3}}\) = \(\frac{3}{2}\) ∉ N

∴ f is not onto.

Question 2.

Check the injectivity and surjectivity of the following functions:

(i) f : N → N is given by f(x) = x^{2}.

(ii) f : Z → Z is given by f(x) = x^{2}.

(iii) f : R → R is given by f(x) = x^{2}.

(iv) f : N → N is given by f(x) = x^{3}.

(v) f : Z → Z is given by f(x) = x^{3}.

Solution:

(i) f : N → N given by f(x) = x^{2}.

(a) f(x_{1}) = f(x_{2}) ⇒ \(x_{1}^{2}\) = \(x_{2}^{2}\) ⇒ x_{1} = x_{2}.

∴ f is one-one, i.e., it is injective.

(b) There are some members of codomain which have no image in domain N.

e.g., 3 ∈ codomain N. But there is no pre-image in domain of f.

⇒ f is not onto, i.e., not surjective.

(ii) f : Z → Z given by f(x) = x^{2}.

(a) f(-1) = f(1) = 1 ⇒ -1 and 1 have the same image.

∴ f is not one-one, i.e., not injective.

(b) There are many elements belonging to codomain have no pre-image in its domain Z.

e.g. 3 ∈ codomain Z but \(\sqrt{3}\) ∉ domain Z of f

∴ f is not onto, i.e., not surjective.

(iii) f : R → R given by f(x) = x^{2}.

(a) f is not one-one since f(- 1) = f(1) = 1

-1 and 1 have the same image, i.e., f is not injective.

(b) – 2 ∈ codomain R of f, but \(\sqrt{-2}\) does not belong to domain R of f.

⇒ f is not onto, i.e., f is not surjective.

(iv) f: N → N given by f(x) = x^{3},

(a) f(x_{1}) = f(x_{2}) ⇒ \(x_{1}^{3}\) = \(x_{2}^{3}\) ⇒ x_{1} = x_{2}

i.e., every x ∈ N has a unique image in its codomain.

∴ f is one-one. It is injective.

(b) There are many members or codomain of f which have no pre-image in its domain.

e.g., 2, 3 etc. These members of codomain have no pre-image in its domain.

∴ f is not onto, i.e., f is not surjective.

(v) f : Z → Z given by f(x) = x^{3}.

(a) Here, also f(x_{1}) = f(x_{2}) ⇒ \(x_{1}^{3}\) = \(x_{2}^{3}\) ⇒ x_{1} = x_{2}

∴ f is one-one, i.e., it is injective.

(b) Many members of codomain of f do not have any pre-image in its domain.

e.g., 2 belonging to its codomain has no pre-image in its domain of Z. Therefore, f is not surjective.

Question 3.

Prove that greatest integer function f : R → R, given by f(x) = [x], is neither one-one nor onto, where [x] denotes the greatest integer less than or equal to x.

Solution:

f : R → R given by f(x) = [x].

(a) f(1.2) = 1, f(1.5) = 1

⇒ f is not one-one.

(b) All the images of x ∈ R belonging to its domain have integers as the images in codomain. But no fraction proper or improper belonging to codomain of f has any pre-image in its domain.

⇒ f is not onto.

Hence, f is neither one-one nor onto.

Question 4.

Show that the Modulus Function f : R → R, given by f(x) = [x], is neither one-one nor onto, where | x | is x, if x is positive and | x | is – x, if x is negative.

Solution:

f : R → R given by f(x) = | x |.

(a) f(-1) = |-1| = 1, f(1) = |1| = 1.

⇒ – 1 and 1 have the same image.

∴ f is not one-one.

(b) No negative value belonging to codomain of f has any pre-image in its domain.

∴ f is not onto.

Hence, f is neither one-one nor onto.

Question 5.

Show that the Signum function f : R → R, given by

is neither one-one nor onto.

-1, if x< 0

Solution:

f : R → R, given by

(a) f(1) = f(2) = 1

∴ 1 and 2 have the same image.

i.e. f(x_{1}) = f(x_{2}) = 1 for x > 0

⇒ x_{1} ≠ x_{2}.

Similarly, f(x_{1}) = f(x_{2}) = —1, for x < 0,

where x_{1} ≠ x_{2}

⇒ f is not one-one.

(b) Except – 1, 0 and 1, no other members of codomain of f has any pre-image in its domain.

f is not onto.

⇒ f is neither one-one nor onto.

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:

A = (1, 2, 3}, B = {4, 5, 6, 7} and f = {(1, 4), (2, 5), (3, 6)}.

Every member of A has a unique image in B as shown in the figure.

∴ f is one-one.

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 defined by f(x) = 3 – 4x.

(ii) f : R → R defined by f(x) = 1 + x^{2}.

Solution:

(i) f : R → R defined by f(x) = 3 – 4x.

f(x_{1}) = 3 – 4x_{1}, f(x_{2}) = 3 – 4x_{2}

(a) f(x_{1}) = f(x_{2}) ⇒ 3 – 4x_{1} = 3 – 4x_{2}

∴ x_{1} = x_{2}

This shows that f is one-one.

(b) f(x) = y = 3 – 4x

∴ x = \(\frac{3-y}{4}\)

For every value of y belonging to its codomain, there is a pre-image in its domain.

⇒ f is onto.

Hence, f is one-one and onto.

(ii) f : R → R given by f(x) = 1 + x^{2}

(a) f(1) = 1 + 1 = 2, f(- 1) = 1 + 1 = 2.

∴ f(- 1) = f(1) = 2

i. e., – 1 and 1 have the same image 2.

⇒ f is not one-one.

(b) No negative number belonging to its codomain has its pre-image in its domain.

⇒ f is not onto.

Thus, f is neither one-one nor onto.

Question 8.

Let A and B be two sets. Show that f : A × B → B × A such that f(a, b) = (b, a) is bijective function.

Solution:

We have f : (A × B) → (B × A) such that f(a, b) = (b, a).

(a) ∴ f{a_{1}, b_{1}) = (b_{1}, a_{1}), f(a_{2}, b_{2}) = (b_{2}, a_{2}) and

f(a_{1}, b_{1}) = f(a_{2}, b_{2}) ⇒ (b_{1}, _{1}) = (b_{2}, a_{2})

⇒ b_{1} = b_{2} and a_{1} = a_{2}

∴ f is onto

Thus, f is one-one and onto, ie. it is bijective.

Question 9.

Let f: N → N be defined by

State whether the function f is bijective. Justify your answer.

Solution:

f: N → N be defined by

(a) f(1) = \(\frac{n+1}{2}\) = \(\frac{1+1}{2}\) = \(\frac{2}{2}\) = 1, f(2) = \(\frac{n}{2}\) = \(\frac{2}{2}\) = 1

The elements 1 and 2 belonging to domain of / have the same image in its codomain.

⇒ f is not one-one.

Therefore, it is not injective:

(b) Every member of codomain has pre-image in its domain e.g., 1 has two pre-images 1 and 2.

⇒ f is onto.

Thus, f is not one-one but it is onto. ⇒ f is not bijective.

Question 10.

Let A = R-{3} and B = R-{1}. consider the function f: A → B defined by f (x) = \(\frac{x-2}{x-3}\). Is f one-one and onto? Justify your answer.

Solution:

f : A → B, where A = R- {3}, B = R – {1}, f is defined by

or (x_{1} – 2)(x_{2} – 3) = (x_{2} – 2)(x_{1} – 3)

or x_{1}x_{2} – 3x_{1} – 2x_{2} + 6 = x_{1}x_{2} – 2x_{1} – 3x_{2} + 6

i.e. – 3x_{1} – 2x_{2} = -2x_{1} – 3x_{2}

or -x_{1} = -x_{2}

⇒ x_{1} = x_{2}

∴ f is one-one.

(b) Let y = \(\frac{x-2}{x-3}\) ⇒ xy – 3y = x – 2

or x(y – 1) = 3y – 2

⇒ x = \(\frac{3 y-2}{y-1}\)

⇒ For every value of y, except y = 1, there is a pre-image

⇒ x = \(\frac{3 y-2}{y-1}\)

⇒ f is onto.

Thus, f is one-one and onto.

Question 11.

Let f: R → R be defined as f(x) = x^{4}.

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:

f(-1) = (-1)^{4} = 1, f(1) = 1,sup>4 = 1

∴ – 1 and 1 have the same image 1.

⇒ f is not one-one.

Further – 2 in the codomain of f has no pre-image in its domain.

∴ f is not onto.

i.e., f is neither one-one nor onto.

Part (D) is the correct answer.

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 not onto.

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

Solution:

f : R → R is defined by f(x) = 3x.

(a) f(x_{1}) = 3x_{1}, f(x_{2}) = 3x_{2}

f(x_{1}) = f(x_{2})

⇒ f is one-one.

(b) For every member y belonging to co-domain has pre-image x in domain of f.

Since y = 3x, ⇒ x = \(\frac{y}{3}\)

∴ f is onto.

∴ f is one-one and onto.

Part (A) is the correct answer.