TOPICS

Exercise - 1.2

Relations and Functions

**Question-1 :-** Show that the function f : R∗ → R∗ defined by f(x) = 1/x is one-one and 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∗

Given that f : R∗ → R∗ defined by f(x) = 1/x. ⇒ f(x) = f(y) ⇒ 1/x = 1/y ⇒ x = y ∴f is one-one. Now, f(x) = 1/x ⇒ y = 1/x ⇒ x = 1/y ⇒ f(1/y) = x ∴f is on-to. Again, g : R∗ → R∗ defined by g(x) = 1/x. ⇒ g(x₁) = g(x₂) ⇒ 1/x₁ = 1/x₂ ⇒ x₁ = x₂ and (x₁, x₂) ∉ N ∴g is one-one. Now, g(x) = 1/x But every real number belonging to co-domain may not have a pre-image in N. e.g., 1/3 = 3/2 ∉ N. ∴g is not on-to.

**Question-2 :-** Check the injectivity and surjectivity of the following functions:

(i) f : N → N given by f(x) = x²

(ii) f : Z → Z given by f(x) = x²

(iii) f : R → R given by f(x) = x²

(iv) f : N → N given by f(x) = x³

(v) f : Z → Z given by f(x) = x³

(i) f : N → N given by f(x) = x² ⇒ f(x₁) = f(x₂) ⇒ x₁² = x₂² ⇒ x₁ = x₂ ∴f is one-one. Now, f(x) = x² There are such numbers of co-domain which have no image in domain N. e.g., 3 ∈ co-domain N, but there is no pre-image in domain of f. ∴f is not on-to. Hence, f is injective but not surjective.

(ii) f : Z → Z given by f(x) = x² since, Z = {0, ±1, ±2, ±3...} Therefore, f(-1) = f(1) = 1 ⇒ -1 and 1 have same image. ∴f is not one-one. Now, f(x) = x² There are such numbers of co-domain which have no image in domain Z. e.g., 3 ∈ co-domain Z, but there is no √3 ∉ Z domain of f. ∴f is not on-to. Hence, f is neither injective nor surjective.

(iii) f : R → R given by f(x) = x² since, R = {0, ±1, ±2, ±3...} Therefore, f(-1) = f(1) = 1 ⇒ -1 and 1 have same image. ∴f is not one-one. e.g., 3 ∈ co-domain R, but there is no √-3 ∉ R domain of f. ∴f is not on-to. Hence, f is neither injective nor surjective.

(iv) f : N → N given by f(x) = x³ ⇒ f(x₁) = f(x₂) ⇒ x₁³ = x₂³ ⇒ x₁ = x₂ for every x ∈ N, has a unique image in its co-domain. ∴f is one-one. There are many such members of co-domain of f which do not have pre-image in its domain. e.g., 2, 3 etc. ∴f is not on-to. Hence, f is injective but not surjective.

(v) f : Z → Z given by f(x) = x³ ⇒ f(x₁) = f(x₂) ⇒ x₁³ = x₂³ ⇒ x₁ = x₂ for every x ∈ Z, has a unique image in its co-domain. ∴f is one-one. There are many such members of co-domain of f which do not have pre-image in its domain. ∴f is not on-to. Hence, f is injective but not surjective.

**Question-3 :-** Prove that the 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.

f: → R → R is given by, f(x) = [x] It is seen that f(1.4) = [1.4] = 1, f(1.8) = [1.8] = 1. ∴ f(1.4) = f(1.8), but 1.4 ≠ 1.8. ∴ f is not one-one. Now, consider 0.8 ∈ R. It is known that f(x) = [x] is always an integer. Thus, there does not exist any element x ∈ R such that f(x) = 0.8. ∴ f is not on-to. Hence, the greatest integer function is neither one-one nor on-to.

**Question-4 :-** Show that the Modulus Function f : R → R, given by f(x) = |x|, is neither oneone nor onto, where |x| is x, if x is positive or 0 and |x| is – x, if x is negative.

Modulus Function f : R → R, given by f(x) = |x| f contains (-1,1),(1,1),(-3,3),(3,3) Thus negative integers are not images of any element. ∴f is not one-one. Also second set R contains some negative numbers which are not images of any real number. i.e., f(x) = |x| = -1 ∴f is not on-to. Hence, f is neither one-one nor on-to.

**Question-5 :-** Show that the Signum Function f : R → R, given by
is neither one-one nor onto.

Signum Function f : R → R, given by ⇒ f(1) = f(2) = 1 ⇒ f(x₁) = f(x₂) = 1 for x>1 ⇒ x₁ ≠ x₂ ∴f is not one-one. Now, as f(x) takes only 3 values (1, 0, or −1) for the element −2 in co-domain, there does not exist any x in domain such that f(x) = −2. ∴f is not onto. Hence, f is neither one-one nor on-to.

**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.

Given that A = {1, 2, 3}, B = {4, 5, 6, 7}. f: A → B is defined as f = {(1, 4), (2, 5), (3, 6)}. ∴f (1) = 4, f (2) = 5, f (3) = 6 It is seen that the images of distinct elements of A under f are distinct. Hence, function 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 + x²

(i) f : R → R defined by f(x) = 3 – 4x ⇒ f(x₁) = f(x₂) ⇒ 3 – 4x₁ = 3 – 4x₂ ⇒ x₁ = x₂ ∴f is one-one. Now, f(x) = 3 – 4x, Let f(x) = y ⇒ y = 3 – 4x ⇒ x = (3 - y)/4 ∈ R ⇒ f((3 - y)/4 ) = 3 - 4[(3 - y)/4 ] = 3 - 3 + y = y ∴f is on-to. Hence, f is injective and surjective or f is bijective function.

(ii) f : R → R defined by f(x) = 1 + x² ⇒ f(x₁) = f(x₂) ⇒ 1 + x₁² = 1 + x₂² ⇒ x₁² = x₂² ⇒ x₁ = ± x₂ ∴f is not one-one. Now, f(x) = 1 + x², Let f(x) = y ⇒ y = 1 + x² ⇒ x = ± √y - 1 ⇒ f(√y - 1) = 1 + (1-y) = 2 - y ≠ y ∴f is not on-to. Hence, f is neither injective nor surjective so, f is not bijective function.

**Question-8 :-** Let A and B be sets. Show that f : A × B → B × A such that f(a, b) = (b, a) is bijective function

f : A × B → B × A such that f(a, b) = (b, a) ⇒ Let (a₁, b₁) & (a₂, b₂) ∈ A x B such that f(a₁, b₁) = f(a₂, b₂) ⇒ (b₁, a₁) = (b₂, a₂) ⇒ a₁ = a₂, b₁ = b₂ ⇒ (a₁, b₁) = (a₂, b₂) ⇒ a₁ = a₂, b₁ = b₂ So, for all (a₁, b₁),(a₂, b₂) ∈ A x B ∴f is one-one. Now, let (b, a) ∈ B × A be any element. Then, there exists (a, b) ∈ A × B such that f(a, b) = (b, a). ∴f is on-to. Hence, f is injective and surjective so, f is bijective function.

**Question-9 :-** Let f : N → N be defined by
State whether the function f is bijective. Justify your answer.

Let f : N → N be defined by (a) f(1) = (n+1)/2 = (1+1)/2 = 2/2 = 1 and f(2) = n/2 = 2/2 = 1 The elements 1, 2, belonging to domain of f have the same image 1 in its co-domain. ∴f is one-one. (b) Every number of co-domain has pre-image in its domain e.g., 1 has two pre-images 1 and 2. ∴f is on-to. Hence, f is not injective but surjective. So, f is not bijective function.

**Question-10 :-** Let A = R – {3} and B = R – {1}. Consider the function f : A → B defined by
f(x) = (x-2)/(x-3). Is f one-one and onto? Justify your answer

Let A = R – {3} and B = R – {1} and the function f : A → B defined by f(x) = (x-2)/(x-3). ⇒ f(x₁) = f(x₂) ⇒ (x₁-2)/(x₁-3) = (x₂-2)/(x₂-3) ⇒ (x₁-2)(x₂-3) = (x₁-3)(x₂-2) ⇒ x₁x₂ - 3x₁ - 2x₂ + 6 = x₁x₂ - 2x₁ - 3x₂ + 6 ⇒ - 3x₁ - 2x₂ = - 2x₁ - 3x₂ ⇒ - 3x₁ + 2x₁ = - 3x₂ + 2x₂ ⇒ -x₁ = -x₂ ⇒ x₁ = x₂ ∴f is one-one. Now, f(x) = (x-2)/(x-3) Let f(x) = y Then y = (x-2)/(x-3) ⇒ y(x-3) = x - 2 ⇒ xy - 3y = x - 2 ⇒ xy - x = 3y - 2 ⇒ x(y-1) = 3y - 2 ⇒ x = (3y-2)/(y-1) ⇒ f(x) = y ∴f is on-to. Hence, f is injective and surjective so, f is bijective 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.

Let f : R → R be defined as f(x) = x⁴ ⇒ f(x₁) = f(x₂) ⇒ x₁⁴ = x₂⁴ ⇒ ± x₁ = ± x₂ ∴f is not one-one. Now, f(x) = x⁴ Let f(x) = y Then y = x⁴ ⇒ x = ± (y)^{1/4}f(x) = (- y^{1/4})⁴ ⇒ f(x) = y ∴f is not on-to. Hence, f is neither one-one nor on-to. Therefore, option (D) is correct.

**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.

Let f : R → R be defined as f(x) = 3x ⇒ f(x₁) = f(x₂) ⇒ 3x₁ = 3x₂ ⇒ x₁ = x₂ ∴f is one-one. Now, f(x) = 3x Let f(x) = y Then y = 3x x = y/3 f(x) = 3 x (y/3) = y ∴f is not on-to. Hence, f is one-one and on-to. Therefore, option (A) is correct.

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