TOPICS

Exercise - 1.3

Relations and Functions

**Question-1 :-** Let f : {1, 3, 4} → {1, 2, 5} and g : {1, 2, 5} → {1, 3} be given by f = {(1, 2), (3, 5), (4, 1)} and g = {(1, 3), (2, 3), (5, 1)}. Write down gof.

The functions f: {1, 3, 4} → {1, 2, 5} and g: {1, 2, 5} → {1, 3} are defined as f = {(1, 2), (3, 5), (4, 1)} and g = {(1, 3), (2, 3), (5, 1)}. ⇒ gof(1) = f(f(1)) = g(2) = 3 [where f(1) = 2, g(2) = 3] ⇒ gof(3) = f(f(3)) = g(5) = 1 [where f(3) = 5, g(5) = 1] ⇒ gof(4) = f(f(4)) = g(1) = 3 [where f(4) = 1, g(1) = 3] ⇒ gof = {(1,3), (3,1), (4,3)}

**Question-2 :-** Let f, g and h be functions from R to R. Show that (a) (f + g)oh = foh + goh, (b)
(f . g)oh = (foh) . (goh)

(a) (f + g)oh = foh + goh ⇒ L.H.S (Left Hand Side) ⇒ Consider (f + g)oh(x) ⇒ (f+g)(h(x)) ⇒ f(h(x)) + g(h(x)) ⇒ foh(x) + goh(x) ⇒ {(foh) + (goh)}(x) ∴((f+g)oh)(x) = {(foh) + (goh)}(x) for all x ∈ R Hence, (f + g)oh = foh + goh.

(b) (f . g)oh = (foh) . (goh) ⇒ L.H.S (Left Hand Side) ⇒ Consider (f . g)oh(x) ⇒ (f . g)(h(x)) ⇒ f(h(x)) . g(h(x)) ⇒ foh(x) . goh(x) ⇒ {(foh) . (goh)}(x) ∴((f . g)oh)(x) = {(foh) . (goh)}(x) for all x ∈ R Hence, (f . g)oh = (foh) . (goh)

**Question-3 :-** Find gof and fog, if (i) f(x) = |x| and g(x) = | 5x – 2 |,
(ii) f(x) = 8x³ and g(x) = x^{1/3}.

(i) f(x) = |x| and g(x) = | 5x – 2 | gof(x) = g(f(x)) = g(|x|) = | 5 |x| – 2 | fog(x) = f(g(x)) = f(| 5x – 2 |) = || 5x – 2 || = | 5x – 2 |

(ii) f(x) = 8x³ and g(x) = x^{1/3}gof(x) = g(f(x)) = g(8x³) = (8x³)^{1/3}= (2x)^{3 x 1/3}= 2x fog(x) = f(g(x)) = f(x^{1/3}) = 8 (x^{1/3})³ = 8x

**Question-4 :-** If f(x) = (4x+3)/(6x-4), x ≠ 2/3, show that fof(x) = x, for all x ≠ 2/3.
What is the inverse of f ?

Given that f(x) = (4x+3)/(6x-4), x ≠ 2/3 Therefore, fof(x) = x, for all x ≠ 2/3. fof(x) = I Hence, the given function f is invertible and the inverse of f is f itself.

**Question-5 :-**
State with reason whether following functions have inverse

(i) f : {1, 2, 3, 4} → {10} with f = {(1, 10), (2, 10), (3, 10), (4, 10)}

(ii) g : {5, 6, 7, 8} → {1, 2, 3, 4} with g = {(5, 4), (6, 3), (7, 4), (8, 2)}

(iii) h : {2, 3, 4, 5} → {7, 9, 11, 13} with h = {(2, 7), (3, 9), (4, 11), (5, 13)}

(i) f: {1, 2, 3, 4} → {10} defined as: f = {(1, 10), (2, 10), (3, 10), (4, 10)} From the given definition of f, we can see that f is a many one function as: f(1) = f(2) = f(3) = f(4) = 10 ∴f is not one-one. Hence, function f does not have an inverse. (ii) g: {5, 6, 7, 8} → {1, 2, 3, 4} defined as: g = {(5, 4), (6, 3), (7, 4), (8, 2)} From the given definition of g, it is seen that g is a many one function as: g(5) = g(7) = 4. ∴g is not one-one. Hence, function g does not have an inverse. (iii) h: {2, 3, 4, 5} → {7, 9, 11, 13} defined as: h = {(2, 7), (3, 9), (4, 11), (5, 13)} It is seen that all distinct elements of the set {2, 3, 4, 5} have distinct images under h. ∴Function h is one-one. Also, h is onto since for every element y of the set {7, 9, 11, 13}, there exists an element x in the set {2, 3, 4, 5}such that h(x) = y. Thus, h is a one-one and on-to function. Hence, h has an inverse.

**Question-6 :-** Show that f : [–1, 1] → R, given by f(x) = x /(x+2) is one-one. Find the inverse of the function f : [–1, 1] → Range f.
(Hint: For y ∈ Range f, y = f(x) = x /(x+2), for some x in [–1, 1], i.e., x = 2y/(1-y))

f: [−1, 1] → R is given as f(x) = x /(x+2), Let f(x) = f(y). ⇒ x/(x+2) = y/(y+2) ⇒ x(y+2) = y(x+2) ⇒ xy + 2x = xy + 2y ⇒ 2x = 2y ⇒ x = y ∴ f is a one-one function. It is clear that f: [−1, 1] → Range f is on-to. ∴ f: [−1, 1] → Range f is one-one and on-to and therefore, the inverse of the function: f: [−1, 1] → Range f exists. Let g: Range f → [−1, 1] be the inverse of f. Let y be an arbitrary element of range f. Since f: [−1, 1] → Range f is on-to, we have: y = f(x) for same x ∈ [-1,1] ⇒ y = x/(x+2) ⇒ y(x+2) = x ⇒ xy + 2y = x ⇒ x = 2y/(1-y), y ≠ 1 Now, let us define g: Range f → [−1, 1] as g(y) = 2y/(1-y), y ≠ 1 f^{-1}(y) = 2y/(1-y), y ≠ 1

**Question-7 :-** Consider f: R → R given by f(x) = 4x + 3. Show that f is invertible. Find the inverse of f.

Given that f: R → R given by f(x) = 4x + 3 Let f(x) = f(y). ⇒ 4x + 3 = 4y + 3 ⇒ 4x = 4y ⇒ x = y ∴f is one-one. For y ∈ R , let y = 4x + 3. ⇒ 4x = y - 3 ⇒ x = (y-3)/4 ∈ R Therefore, for any y ∈ R, there exists x = (y-3)/4 ∈ R such that ⇒ f(x) = 4((y-3)/4) + 3 ⇒ f(x) = y - 3 + 3 ⇒ f(x) = y ∴f is on-to. Thus, f is one-one and on-to and therefore, f^{-1}exists. Let us define g: R → R by g(x) = (y-3)/4 gof = fog = I Hence, f is invertible and the inverse of f is given by f^{-1}(y) = g(y) = (y-3)/4

**Question-8 :-**
Consider f : R+ → [4, ∞) given by f(x) = x² + 4. Show that f is invertible with the inverse f^{-1} of f given by f^{-1}(y) = √y - 4 ,
where R+ is the set of all non-negative real numbers.

f : R+ → [4, ∞) given by f(x) = x² + 4. Let f(x) = f(y). ⇒ x² + 4 = y² + 4 ⇒ x² = y² ⇒ x = y ∴ f is a one-one function. f : R+ → [4, ∞), Let y = x² + 4. ⇒ x² = y - 4 ⇒ x = √y - 4 > 0 ⇒ f(√y - 4) = (√y - 4)² + 4 = y ∴f is on-to. Thus, f is one-one and on-to and therefore, f^{-1}exists. Let us define f : R+ → [4, ∞) given by g(y) = √y - 4 gof = fog = I Hence, f is invertible and the inverse of f is given by f^{-1}(y) = g(y) = √y - 4

**Question-9 :-**
Consider f : R+ → [– 5, ∞) given by f(x) = 9x² + 6x – 5. Show that f is invertible with f^{-1}(y) = (√y+6-1)/3

f: R+ → [−5, ∞) is given as f(x) = 9x² + 6x − 5. Let y be an arbitrary element of [−5, ∞). Let y = 9x² + 6x − 5. f(x) = f(y) 9x² + 6x − 5 = 9y² + 6y − 5 ⇒ 9x² + 6x = 9y² + 6y ⇒ 9x² - 9y²= 6y - 6x ⇒ 9(x²-y²) = -6(x-y) ⇒ 9(x+y)(x-y) - 6(x-y) = 0 ⇒ (x-y)[9(x+y) - 6] = 0 ⇒ x - y = 0 ⇒ x = y ∴f is one-one. Now, f: R+ → [−5, ∞) is given as f(x) = 9x² + 6x − 5 Let f(x) = y y = 9x² + 6x − 5 ⇒ 9x² + 6x - (y + 5) = 0 By discrimination method values find of x ⇒ a = 9, b = 6, c = -(y + 5) ⇒ d = b² - 4ac ⇒ d = 6² - 4 x 9 x (-(y + 5)) ⇒ d = 36 + 36y + 180 ⇒ d = 216 + 36y ⇒ d = 36(6 + y) ⇒ d = 36(y + 6) ⇒ x = (-b ± √d)/2a = (-6 ± √36(y+6))/(2 x 9) ⇒ x = (-6 ± 6√(y+6))/18 ⇒ x = (√y+6-1)/3 Therefore, ∴f is on-to.

**Question-10 :-** Let f : X → Y be an invertible function. Show that f has unique inverse.
(Hint: suppose g₁ and g₂ are two inverses of f. Then for all y ∈ Y, fog₁(y) = I(y) = fog₂(y). Use one-one ness of f).

Given f : X → Y be an invertible function. Also, suppose f has two inverses (g₁ and g₂) Then for all y ∈ Y, fog₁(y) = I(y) = fog₂(y) ⇒ f(g₁(y)) = f(g₂(y)) ⇒ g₁(y) = g₂(y) ⇒ g₁ = g₂ The inverse is unique and hence f has a unique inverse.

**Question-11 :-**
Consider f : {1, 2, 3} → {a, b, c} given by f(1) = a, f(2) = b and f(3) = c. Find f^{-1} and show that (f^{-1})^{-1} = f.

Function f: {1, 2, 3} → {a, b, c} is given by, f(1) = a, f(2) = b, and f(3) = c If we define g: {a, b, c} → {1, 2, 3} as g(a) = 1, g(b) = 2, g(c) = 3, then we have: fog(a) = f(g(a)) = f(1) = a fog(b) = f(g(b)) = f(2) = b fog(c) = f(g(c)) = f(3) = c and gof(1) = g(f(1)) = f(a) = 1 gof(2) = g(f(2)) = f(b) = 2 gof(3) = g(f(3)) = f(c) = 3 gof = I, fog = I f = {(1,a),(2,b),(3,c)} Therefore, f^{-1}= {(1,a),(2,b),(3,c)} (f^{-1})^{-1}= {(1,a),(2,b),(3,c)} = f Hence, (f^{-1})^{-1}= f

**Question-12 :-**
Let f: X → Y be an invertible function. Show that the inverse of f^{-1} is f, i.e., (f^{-1})^{-1} = f.

Let f: X → Y be an invertible function. Then, there exists a function g: Y → X such that gof = IX and fog = IY. Here, f^{-1}= g. Now, gof = IX and fog = IY ⇒ f^{-1}of = IX and fof^{-1}= IY Hence, f^{-1}: Y → X is invertible and f is the inverse of f^{-1}i.e., (f^{-1})^{-1}= f.

**Question-13 :-** If f: R → R be given by f(x) = (3 - x³)^{1/3}, then fof(x) is

(A) x^{1/3}

(B) x³

(C) x

(D) (3 – x³).

Given f: R → R be given by f(x) = (3 - x³)^{1/3}The correct answer is C.

**Question-14 :-**
Let f: R-{-4/3} → R be a function defined as f(x) = 4x/(3x + 4). The inverse of f is the map g: Range f → R-{-4/3} given by

(A) g(y) = 3y/(3-4y)

(B) g(y) = 4y/(4-3y)

(C) g(y) = 4y/(3-4y)

(D) g(y) = 3y/(4-3y)

Given f: R-{-4/3} → R be a function defined as f(x) = 4x/(3x + 4). Now, Range of f → R-{-4/3} Let y = f(x) y = 4x/(3x + 4) ⇒ y(3x + 4) = 4x ⇒ 3xy + 4y = 4x ⇒ 3xy - 4x = -4y ⇒ x(3y-4) = -4y ⇒ x = 4y/(4-3y) f^{-1}(y) = g(y) = 4y/(3-4y) Therefore, option (B) is correct.

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