RD Sharma Solutions for Class 12 Maths Exercise 2.4 Chapter 2 Function

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RD Sharma Solutions for Class 12 Chapter 2 – Function Exercise 2.4

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Exercise 2.4 Page No: 2.68

1. State with reason whether the 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)}

Solution:

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

We have:

f (1) = f (2) = f (3) = f (4) = 10

⇒ f is not one-one.

⇒ f is not a bijection.

So, f does not have an inverse.

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

From the question, it is clear that g (5) = g (7) = 4

⇒ f is not one-one.

⇒ f is not a bijection.

So, f does not have an inverse.

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

Here, different elements of the domain have different images in the co-domain.

⇒ h is one-one.

Also, each element in the co-domain has a pre-image in the domain.

⇒ h is onto.

⇒ h is a bijection.

Therefore h inverse exists.

⇒ h has an inverse, and it is given by

h^-1 = {(7, 2), (9, 3), (11, 4), (13, 5)}

2. Find f ⁻¹ if it exists:  f: A → B, where 

(i) A = {0, −1, −3, 2}; B = {−9, −3, 0, 6} and f(x) = 3 x.

(ii) A = {1, 3, 5, 7, 9}; B = {0, 1, 9, 25, 49, 81} and f(x) = x²

Solution:

(i) Given A = {0, −1, −3, 2}; B = {−9, −3, 0, 6} and f(x) = 3 x.

So, f = {(0, 0), (-1, -3), (-3, -9), (2, 6)}

Here, different elements of the domain have different images in the co-domain.

Clearly, this is one-one.

Range of f = Range of f = B

So, f is a bijection and,

Thus, f ^-1 exists.

Hence, f ^-1= {(0, 0), (-3, -1), (-9, -3), (6, 2)}

(ii) Given A = {1, 3, 5, 7, 9}; B = {0, 1, 9, 25, 49, 81} and f(x) = x²

So, f = {(1, 1), (3, 9), (5, 25), (7, 49), (9, 81)}

Here, different elements of the domain have different images in the co-domain.

Clearly, f is one-one.

But this is not onto because the element 0 in the co-domain (B) has no pre-image in the domain (A)

⇒ f is not a bijection.

So, f ^-1 does not exist.

3. Consider f: {1, 2, 3} → {a, b, c} and g: {a, b, c} → {apple, ball, cat} defined as f (1) = a, f (2) = b, f (3) = c, g (a) = apple, g (b) = ball and g (c) = cat. Show that f, g and gof are invertible. Find f⁻¹, g⁻¹ and gof⁻¹and show that (gof)⁻¹ = f ⁻¹o g⁻¹

Solution:

Given f = {(1, a), (2, b), (c , 3)} and g = {(a , apple) , (b , ball) , (c , cat)} Clearly , f and g are bijections.

So, f and g are invertible.

Now,

f ^-1 = {(a ,1) , (b , 2) , (3,c)} and g^-1 = {(apple, a), (ball , b), (cat , c)}

So, f^-1 o g^-1= {apple, 1), (ball, 2), (cat, 3)}……… (1)

f: {1,2,3,} → {a, b, c} and g: {a, b, c} → {apple, ball, cat}

So, gof: {1, 2, 3} → {apple, ball, cat}

⇒ (gof) (1) = g (f (1)) = g (a) = apple

(gof) (2) = g (f (2))

= g (b)

= ball,

And (gof) (3) = g (f (3))

= g (c)

= cat

∴ gof = {(1, apple), (2, ball), (3, cat)}

Clearly, gof is a bijection.

So, gof is invertible.

(gof)^-1 = {(apple, 1), (ball, 2), (cat, 3)}……. (2)

Form (1) and (2), we get

(gof)^-1 = f^-1 o g ^-1

4. Let A = {1, 2, 3, 4}; B = {3, 5, 7, 9}; C = {7, 23, 47, 79} and f: A → B, g: B → C be defined as f(x) = 2x + 1 and g(x) = x² − 2. Express (gof)⁻¹ and f⁻¹ og⁻¹ as the sets of ordered pairs and verify that (gof)⁻¹ = f⁻¹ og⁻¹.

Solution:

Given that f (x) = 2x + 1

⇒ f= {(1, 2(1) + 1), (2, 2(2) + 1), (3, 2(3) + 1), (4, 2(4) + 1)}

= {(1, 3), (2, 5), (3, 7), (4, 9)}

Also given that g(x) = x²−2

⇒ g = {(3, 3²−2), (5, 5²−2), (7, 7²−2), (9, 9²−2)}

= {(3, 7), (5, 23), (7, 47), (9, 79)}

Clearly, f and g are bijections and, hence, f⁻¹: B→ A and g⁻¹: C→ B exist.

So, f⁻¹= {(3, 1), (5, 2), (7, 3), (9, 4)}

And g⁻¹= {(7, 3), (23, 5), (47, 7), (79, 9)}

Now, (f⁻¹ o g⁻¹): C→ A

f⁻¹ o g⁻¹ = {(7, 1), (23, 2), (47, 3), (79, 4)}……….(1)

Also, f: A→B and g: B → C,

⇒ gof: A → C, (gof) ⁻¹ : C→ A

So, f⁻¹ o g⁻¹and (gof)⁻¹ have the same domains.

(gof) (x) = g (f (x))

=g (2x + 1)

=(2x +1 )²− 2

⇒ (gof) (x) = 4x² + 4x + 1 − 2

⇒ (gof) (x) = 4x²+ 4x −1

Then, (gof) (1) = g (f (1))

= 4 + 4 − 1

=7,

(gof) (2) = g (f (2))

= 4(2)² + 4(2) – 1 = 23,

(gof) (3) = g (f (3))

= 4(3)² + 4(3) – 1 = 47 and

(gof) (4) = g (f (4))

= 4(4)² + 4(4) − 1 = 79

So, gof = {(1, 7), (2, 23), (3, 47), (4, 79)}

⇒ (gof)^{– 1} = {(7, 1), (23, 2), (47, 3), (79, 4)}…… (2)

From (1) and (2), we get:

(gof)⁻¹ = f⁻¹ o g⁻¹

5. Show that the function f: Q → Q, defined by f(x) = 3x + 5, is invertible. Also, find f⁻¹

Solution:

Given function f: Q → Q, defined by f(x) = 3x + 5

Now we have to show that the given function is invertible.

Injection of f:

Let x and y be two elements of the domain (Q),

Such that f(x) = f(y)

⇒ 3x + 5 = 3y + 5

⇒ 3x = 3y

⇒ x = y

so, f is one-one.

Surjection of f:

Let y be in the co-domain (Q),

Such that f(x) = y

⇒ 3x +5 = y

⇒ 3x = y – 5

⇒ x = (y -5)/3 belongs to Q domain

⇒ f is onto.

So, f is a bijection, and hence, it is invertible.

Now we have to find f^-1:

Let f^-1(x) = y…… (1)

⇒ x = f(y)

⇒ x = 3y + 5

⇒ x −5 = 3y

⇒ y = (x – 5)/3

Now substituting this value in (1) we get

So, f^-1(x) = (x – 5)/3

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

Solution:

Given f: R → R given by f(x) = 4x + 3

Now we have to show that the given function is invertible.

Consider injection of f:

Let x and y be two elements of the domain (R),

Such that f(x) = f(y)

⇒ 4x + 3 = 4y + 3

⇒ 4x = 4y

⇒ x = y

So, f is one-one.

Now surjection of f:

Let y be in the co-domain (R),

Such that f(x) = y.

⇒ 4x + 3 = y

⇒ 4x = y -3

⇒ x = (y-3)/4 in R (domain)

⇒ f is onto.

So, f is a bijection, and hence, it is invertible.

Now we have to find f ^-1

Let f^-1(x) = y……. (1)

⇒ x = f (y)

⇒ x = 4y + 3

⇒ x − 3 = 4y

⇒ y = (x -3)/4

Now substituting this value in (1), we get

So, f^-1(x) = (x-3)/4

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

Solution:

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

Now we have to show that f is invertible,

Consider injection of f:

Let x and y be two elements of the domain (Q),

Such that f(x) = f(y)

⇒ x² + 4 = y² + 4

⇒ x² = y²

⇒ x = y      (as co-domain as R+)

So, f is one-one

Now surjection of f:

Let y be in the co-domain (Q),

Such that f(x) = y

⇒ x² + 4 = y

⇒ x² = y – 4

⇒ x = √ (y-4) in R

⇒ f is onto.

So, f is a bijection, and hence, it is invertible.

Now we have to find f^-1:

Let f⁻¹ (x) = y…… (1)

⇒ x = f (y)

⇒ x = y² + 4

⇒ x − 4 = y²

⇒ y = √ (x-4)

So, f^-1(x) = √ (x-4)

Now substituting this value in (1), we get,

So, f^-1(x) = √ (x-4)

8. 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:

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

Now we have to show fof(x) = x

(fof)(x) = f (f(x))

= f ((4x+ 3)/ (6x – 4))

= (4((4x + 3)/ (6x -4)) + 3)/ (6 ((4x +3)/ (6x – 4)) – 4)

= (16x + 12 + 18x – 12)/ (24x + 18 – 24x + 16)

= (34x)/ (34)

= x

Therefore, fof(x) = x for all x ≠ 2/3

=> fof = 1

Hence, the given function f is invertible, and the inverse of f is f itself.

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

f^-1(x) = (√(x +6)-1)/3 

Solution:

Given f: R⁺ → [−5, ∞) given by f(x) = 9x² + 6x – 5

We have to show that f is invertible.

Injectivity of f:

Let x and y be two elements of the domain (R⁺),

Such that f(x) = f(y)

⇒ 9x² + 6x – 5 = 9y² + 6y − 5

⇒ 9x² + 6x = 9y² + 6y

⇒ x = y (As, x, y ∈ R⁺)

So, f is one-one.

Surjectivity of f:

Let y be in the co-domain (Q)

Such that f(x) = y

⇒ 9x² + 6x – 5 = y

⇒ 9x² + 6x = y + 5

⇒ 9x² + 6x +1 = y + 6 (By adding 1 on both sides)

⇒ (3x + 1)² = y + 6

⇒ 3x + 1 = √(y + 6)

⇒ 3x = √ (y + 6) – 1

⇒ x = (√ (y + 6)-1)/3 in R⁺ (domain)

f is onto.

So, f is a bijection, and hence, it is invertible.

Now we have to find f^-1

Let f⁻¹(x) = y….. (1)

⇒ x = f (y)

⇒ x = 9y² + 6y − 5

⇒ x + 5 = 9y² + 6y

⇒ x + 6 = 9y²+ 6y + 1         (adding 1 on both sides)

⇒ x + 6 = (3y + 1)²

⇒ 3y + 1 = √ (x + 6)

⇒ 3y =√(x +6) -1

⇒ y = (√ (x+6)-1)/3

Now substituting this value in (1), we get,

So, f^-1(x) = (√ (x+6)-1)/3

10. If f: R → R be defined by f(x) = x³ −3, then prove that f⁻¹ exists and find a formula for f⁻¹. Hence, find f⁻¹ (24) and f⁻¹ (5).

Solution:

Given f: R → R be defined by f(x) = x³ −3

Now we have to prove that f⁻¹ exists

Injectivity of f:

Let x and y be two elements in the domain (R),

Such that, x³ − 3 = y³ − 3

⇒ x³ = y³

⇒ x = y

So, f is one-one.

Surjectivity of f:

Let y be in the co-domain (R)

Such that f(x) = y

⇒ x³ – 3 = y

⇒ x³ = y + 3

⇒ x = ∛(y+3) in R

⇒ f is onto.

So, f is a bijection, and hence, it is invertible.

Finding f ^-1:

Let f^-1(x) = y…….. (1)

⇒ x= f(y)

⇒ x = y³ − 3

⇒ x + 3 = y³

⇒ y = ∛(x + 3) = f^-1(x)         [from (1)]

So, f^-1(x) = ∛(x + 3)

Now, f^-1(24) = ∛ (24 + 3)

= ∛27

= ∛3³

= 3

And f^-1(5) =∛ (5 + 3)

= ∛8

= ∛2³

= 2

11. A function f: R → R is defined as f(x) = x³ + 4. Is it a bijection or not? In case it is a bijection, find f⁻¹ (3).

Solution:

Given that f: R → R is defined as f(x) = x³ + 4

Injectivity of f:

Let x and y be two elements of the domain (R),

Such that f (x) = f (y)

⇒ x³ + 4 = y³ + 4

⇒ x³ = y³

⇒ x = y

So, f is one-one.

Surjectivity of f:

Let y be in the co-domain (R),

Such that f(x) = y.

⇒ x³ + 4 = y

⇒ x³ = y – 4

⇒ x = ∛ (y – 4) in R (domain)

⇒ f is onto.

So, f is a bijection, and hence, it is invertible.

Finding f^-1:

Let f⁻¹ (x) = y…… (1)

⇒ x = f (y)

⇒ x = y³ + 4

⇒ x − 4 = y³

⇒ y =∛ (x-4)

So, f^-1(x) =∛ (x-4)        [from (1)]

f^-1 (3) = ∛(3 – 4)

= ∛-1

= -1