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.

Answer 1:

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) = g[f(1)] = g(2) = 3 [as f(1) = 2 and g(2) = 3]

gof(3) = g[f(3)] = g(5) = 1 [as f(3) = 5 and g(5) = 1]

gof(4) = g[f(4)] = g(1) = 3 [as f(4) = 1 and 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

(f + g)oh = foh + goh

(f.g)oh = (foh).(goh)

Answer 2:

To prove: (f + g)oh = foh + goh

LHS =  [(f + g)oh](x)

= (f + g)[h(x)]

= f [h(x)] + g[h(x)]

= (foh)(x) + (goh)(x)

= {(foh)(x) + (goh)}(x) = RHS

∴ {(f + g)oh}(x) = {(foh)(x) + (goh)}(x) for all x ∈R

Hence, (f + g)oh = foh + goh

To Prove: (f.g)oh = (foh).(goh)

LHS = [(f.g)oh](x)

= (f.g)[h(x)]

= f[h(x)] . g[h(x)]

= (foh)(x) . (goh)(x)

= {(foh).(goh)}(x)  = RHS

∴[(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

 

Answer 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

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 ?

Answer 4:

It is given that f(x) =(4x+3)/(6x-4), x ≠2/3

(fof)(x) = f(f(x)) = f ((4x + 3)/(6x-4))=

= (16x + 12 + 18x-12) / (24x + 18-24x + 16) = (34x / 34)

= x

∴fof(x) = x, for all x ≠2/3.

⇒ fof = I_x

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)}

Answer 5:

(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 onto function.

Hence, h has an inverse.

 

Question 6: Show that f: [−1, 1] → R, given byf(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)

Answer 6:

f: [−1, 1] → R is given as f(x) =x(x+2)

For one – one

Let f(x) = f(y)

⇒(x/x+2) = (y/y+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 onto.

∴ f: [−1, 1] → Range f is one – one and onto 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 onto, we have

y = f(x) for some x ∈ [−1, 1]

⇒ y = (x/x+2)

⇒ xy + 2y = x

⇒ x(1-y) = 2y

⇒ x =(2y/1-y), y ≠ 1

Now, let us define g: Range f → [−1, 1] as

g(y) =(2y/1-y), y ≠ 1

Now,

 

(gof)(x) = g(f(x)) = g (x/(x + 2))=

∴ gof = x = I_[-1,1] and fog = y = I_{Range f}

∴ f^-1 = g

⇒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.

Answer 7:

f: R → R is given by, f(x) = 4x + 3

For one – one

Let f(x) = f(y)

⇒ 4x + 3 = 4y + 3

⇒ 4x = 4y

⇒ x = y

∴ f is a one – one function.

For onto

For y ∈ R, let y = 4x + 3.

⇒ x = ((y-3)/4) ∈ R

Therefore, for any y ∈ R, there exists x = (y-3)/4)) ∈ R, such that

f(x) = f ((y-3)/4)) = 4 ((y-3)/4)) + 3 = y.

∴ f is onto.

Thus,  f is one – one and onto and therefore,  f

−1

exists.

Let us define g: R → R by g(x) =((y-3)/4)

Now,

(gof)(x) = g(f(x)) = g(4x + 3) =((4x + 3)-3) /4 = 4x/4= x

and

(fog)(y) = f(g(y)) = f ((y-3)/4) = 4 ((y-3)/4)) + 3 = y-3 + 3 = y

∴ gof = fog = IR

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⁻¹of given f  byf⁻¹(y) = (y – 4)^1/2, where 𝑹₊ is the set of all nonnegative real numbers.

Answer 8:

f: R+ → [4, ∞) is given as f(x) = x²+ 4.

For one – one

Let f(x) = f(y)

⇒ x² + 4 = y² + 4

⇒ x² = y²

⇒ x = y                [as  x = y ∈ R+]

∴ f is a one – one function.

For onto

For y ∈ [4, ∞), let y = x²+ 4

 ⇒x² = y − 4 ≥ 0                              [as y ≥ 4]

⇒ x = (y-4)^1/2 ≥ 0

Therefore, for any y ∈ [4, ∞), there exists x = (y-4)^1/2 ∈ R+, such that

f(x) = f(y-4)^1/2 = ((y-4)^1/2 )²+ 4 = y-4 + 4 = y

∴ f is onto.

Thus, f is one – one and onto and therefore, f

−1

exists.

Let us define g: [4, ∞) → R+ by g(y) = Thus, f is one – one and onto and therefore, f⁻¹exists.

Let us define g: [4, ∞) → R+ by g(y) = (y-4)^1/2  

Now,

(gof)(x) = g(f(x)) = g(x² + 4) = ((x² + 4)-4))^1/2 = (x2)^1/2 = x

and

(fog)(y) = f(g(y)) = f(y-4 )^1/2 = ((y-4 )^1/2)²+ 4 = y-4 + 4 = y

∴ gof = fog = IR

Hence, f is invertible and the inverse of f is given by f^-1(y) = g(y) = (y-4 )^1/2

 

Question 9: Consider f: R+ → [−5, ∞) given by f(x) = 9x²+ 6x − 5. Show that f is invertible with

Answer 9:

f: R+ → [−5, ∞) is given as f(x) = 9x²+ 6x − 5.

Let y be an arbitrary element of [−5, ∞).

Let y = 9x²+ 6x – 5

⇒ y = (3x + 1)²-1-5 = (3x + 1)²-6

⇒ y + 6 = (3x + 1)²

⇒ 3x + 1 = (y + 6)^1/2        [as y ≥ −5 ⇒ y + 6 > 0]

⇒ x =((y+6 )^1/2-1) / 3)

∴ f is onto, thereby range f = [−5, ∞).

Let us define g: [−5, ∞) → R+ as g(y) =((y+6 )^1/2-1) / 3)

Now,

 

(gof)(x) = g(f(x)) = g(9x² + 6x-5) = g((3x + 1)²-6)

= ((3x + 1)²-6 + 6)^1/2 -1

= (3x + 1-1) /3 = (3x / 3) = x

and

(fog)(y) = f(g(y)) = f

 

= ((y + 6)^1/2)² -6 = y + 6-6 = y

∴ gof = x = I_R and fog = y = I_{Range f}

Hence,  f is invertible and the inverse of f is given by

f⁻¹(y) = g(y) =

 

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(y) = fog₂(y). Use one – one ness of f).

Answer 10:

Let f: X → Y be an invertible function. 

Also, suppose f has two inverses (sayg₁ and g₂)

Then, for all y ∈Y, we have

fog₁(y) = I_Y(y) = fog₂(y)

⇒ f(g₁(y)) = f(g₂(y))

⇒ g₁(y) = g₂(y)                  [as f  is invertible ⇒ f  is one – one]

⇒ g₁ = g₂                           [as g is one – one]

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⁻¹and show that (f⁻¹)⁻¹= f.

Answer 11:

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.

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_X and fog = I_Y, where X = {1, 2, 3} and Y= {a, b, c}.

Thus, the inverse of f exists and f⁻¹= g.

∴ f⁻¹:{a, b, c} → {1, 2, 3} is given by f⁻¹(a) = 1, f−1(b) = 2, f^–1   (c) = 3

Let us now find the inverse of f⁻¹ i.e., find the inverse of g.

If we define h: {1, 2, 3} → {a, b, c} as h(1) = a, h(2) = b, h(3) = c

We have

(goh)(1) = g(h(1)) = g(a) = 1

(goh)(2) = g(h(2)) = g(b) = 2

(goh)(3) = g(h(3)) = g(c) = 3

and

(hog)(a) = h(g(a)) = h(1) = a

(hog)(b) = h(g(b)) = h(2) = b

(hog)(c) = h(g(c)) = h(3) = c

∴ goh = I_X and hog = I_Y, where X = {1, 2, 3} and Y = {a, b, c}.

Thus, the inverse of g exists and g⁻¹= h ⇒ (f⁻¹)⁻¹= h.

It can be noted that h = 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.

Answer 12:

Let f: X → Y be an invertible function. 

Then, there exists a function g: Y → X such that gof = I_X and fog = I_Y. 

Here, f⁻¹= g.

Now, gof = I_X and fog = I_Y

⇒ f⁻¹of = I_X and fof⁻¹= I_Y

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) (1/x3)           (B) x³                   (C) x                     (D) (3 − x³)

Answer 13:

f: R → R be given as f(x) = (3-x³)^1/3

∴ fof(x) = f(f(x)) = f

 

= [3-(3-x³)]^1/3 = (x³)^1/3 = x

∴ fof(x) = x

The correct answer is C

 

Question 14: Let f: R- {-4/3} → R be a function as f(x) =(4x / (3x+4)). The inverse of f is 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))

Answer 14:

It is given that f: R − {−4/3} → R be a function as f(x) = (4x / (3x+4))

Let y be an arbitrary element of Range f.

Then, there exists x ∈ R- {-4/3} such that y = f(x)

⇒ y = (4x / (3x + 4))

⇒ 3xy + 4y = 4x

⇒ x(4-3y) = 4y

⇒ x = (4y / (4-3y))

Let us define g: Range f → R- {-4/3} as g(y) = (4y / (4-3y))

Now,

Now,

gof (x)= g(f(x)) =  

=(16x / (12x + 16 – 12x)) =(16x / 16) = x

and

fog (y)= f(g(y)) =  

=(16y / (12y + 16 – 12y)) =(16y / 16) = y

∴ gof = I_R-{-4/3}and fog = I_{Range f}

Thus, g is the inverse of f i.e., f⁻¹= g.

Hence, the inverse of f is the map g: Range f → R- {-4/3}, which is given by g(y) = (4y / (4-3y)).

The correct answer is B