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.
Solution :- 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)
Solution :-(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) = x1/3.
Solution :-(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) = x1/3
gof(x) = g(f(x))
= g(8x³)
= (8x³)1/3
= (2x)3 x 1/3
= 2x
fog(x) = f(g(x))
= f(x1/3)
= 8 (x1/3)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 ?
Solution :- 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))
Solution :- 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.
Solution :- 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.
Solution :- 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
Solution :- 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).
Solution :- 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.
Solution :- 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.
Solution :- 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-1of = 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) x1/3
(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.
