# RD Sharma Solutions For Class 12 Maths Exercise 2.4 Chapter 2 Function

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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} â†’ {10} withÂ fÂ = {(1, 10), (2, 10), (3, 10), (4, 10)}

(ii) g: {5, 6, 7, 8} â†’ {1, 2, 3, 4} withÂ gÂ = {(5, 4), (6, 3), (7, 4), (8, 2)}

(iii) h: {2, 3, 4, 5} â†’ {7, 9, 11, 13} withÂ hÂ = {(2, 7), (3, 9), (4, 11), (5, 13)}

Solution:

(i) Given f: {1, 2, 3, 4} â†’ {10} withÂ fÂ = {(1, 10), (2, 10), (3, 10), (4, 10)}

We have:

fÂ (1) =Â fÂ (2) =Â fÂ (3) =Â fÂ (4) = 10

â‡’Â fÂ is not one-one.

â‡’Â f is not a bijection.

So,Â fÂ does not have an inverse.

(ii) Given g: {5, 6, 7, 8} â†’ {1, 2, 3, 4} withÂ gÂ = {(5, 4), (6, 3), (7, 4), (8, 2)}

from the question it is clear that gÂ (5) =Â gÂ (7) = 4

â‡’ fÂ is not one-one.

â‡’ fÂ is not a bijection.

So,Â fÂ does not have an inverse.

(iii) Given h: {2, 3, 4, 5} â†’ {7, 9, 11, 13} withÂ hÂ = {(2, 7), (3, 9), (4, 11), (5, 13)}

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

â‡’ h is one-one.

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

â‡’Â h is onto.

â‡’Â h is a bijection.

Therefore h inverse exists.

â‡’Â h has an inverse and it is given by

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

2. FindÂ fÂ âˆ’1Â if it exists: Â f:Â AÂ â†’Â B,Â whereÂ

(i) AÂ = {0, âˆ’1, âˆ’3, 2};Â BÂ = {âˆ’9, âˆ’3, 0, 6} andÂ f(x) = 3Â x.

(ii) AÂ = {1, 3, 5, 7, 9};Â BÂ = {0, 1, 9, 25, 49, 81} andÂ f(x) =Â x2

Solution:

(i) Given AÂ = {0, âˆ’1, âˆ’3, 2};Â BÂ = {âˆ’9, âˆ’3, 0, 6} andÂ f(x) = 3Â x.

So,Â f =Â {(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 ofÂ f =Â Range ofÂ fÂ = B

so,Â fÂ is a bijection and,

Thus,Â fÂ -1Â exists.

Hence, fÂ -1=Â {(0, 0), (-3, -1), (-9, -3), (6, 2)}

(ii) Given AÂ = {1, 3, 5, 7, 9};Â BÂ = {0, 1, 9, 25, 49, 81} andÂ f(x) =Â x2

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

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

Clearly,Â fÂ 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)

â‡’ fÂ is not a bijection.

So,Â fÂ -1does not exist.

3. ConsiderÂ f: {1, 2, 3} â†’ {a,Â b,Â c} andÂ g: {a,Â b,Â c} â†’ {apple, ball, cat} defined asÂ fÂ (1) =Â a,Â fÂ (2) =Â b,Â fÂ (3) =Â c,Â gÂ (a) = apple,Â gÂ (b) = ball andÂ gÂ (c) = cat. Show thatÂ f,Â gÂ andÂ gofÂ are invertible. FindÂ fâˆ’1,Â gâˆ’1Â andÂ gofâˆ’1and show that (gof)âˆ’1Â =Â fÂ âˆ’1oÂ gâˆ’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,

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

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

f: {1,2,3,}Â â†’ {a, b, c} and g: {a, b, c}Â â†’ {apple, ball, cat}

So, gof: {1, 2, 3}Â â†’ {apple, ball, cat}

â‡’ (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

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

Clearly, gof is a bijection.

So, gof is invertible.

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

Form (1) and (2), we get

(gof)-1Â = f-1Â o gÂ -1

4. LetÂ AÂ = {1, 2, 3, 4};Â BÂ = {3, 5, 7, 9};Â CÂ = {7, 23, 47, 79} andÂ f:Â AÂ â†’Â B,Â g:Â BÂ â†’Â CÂ be defined asÂ f(x) = 2xÂ + 1 andÂ g(x) =Â x2Â âˆ’ 2. Express (gof)âˆ’1Â andÂ fâˆ’1Â ogâˆ’1Â as the sets of ordered pairs and verify that (gof)âˆ’1Â =Â fâˆ’1Â ogâˆ’1.

Solution:

Given that f (x) = 2x + 1

â‡’Â f=Â {(1,Â 2(1) + 1),Â (2,Â 2(2) + 1),Â (3,Â 2(3) + 1),Â (4,Â 2(4) + 1)}

= {(1,Â 3),Â (2,Â 5),Â (3,Â 7),Â (4,Â 9)}

Also given that g(x) = x2âˆ’2

â‡’Â g =Â {(3,Â 32âˆ’2),Â (5,Â 52âˆ’2),Â (7,Â 72âˆ’2),Â (9,Â 92âˆ’2)}

= {(3,Â 7),Â (5,Â 23),Â (7,Â 47),Â (9,Â 79)}

ClearlyÂ fÂ andÂ gÂ areÂ bijectionsÂ and,Â hence,Â fâˆ’1:Â Bâ†’ AÂ andÂ gâˆ’1:Â Câ†’ BÂ exist.

So,Â fâˆ’1=Â {(3,Â 1),Â (5,Â 2),Â (7,Â 3),Â (9,Â 4)}

AndÂ gâˆ’1=Â {(7,Â 3),Â (23,Â 5),Â (47,Â 7),Â (79,Â 9)}

Now,Â (fâˆ’1Â oÂ gâˆ’1):Â Câ†’ A

fâˆ’1Â oÂ gâˆ’1 = {(7,Â 1),Â (23,Â 2),Â (47,Â 3),Â (79,Â 4)}â€¦â€¦….(1)

Also,Â f:Â Aâ†’BÂ andÂ g:Â BÂ â†’Â C,

â‡’Â gof:Â AÂ â†’Â C,Â (gof)Â âˆ’1Â :Â Câ†’ A

So,Â fâˆ’1Â oÂ gâˆ’1andÂ (gof)âˆ’1Â haveÂ sameÂ domains.

(gof) (x) = gÂ (fÂ (x))

=gÂ (2x + 1)

=(2x +1 )2âˆ’ 2

â‡’Â (gof)Â (x)Â =Â 4x2 +Â 4xÂ + 1 âˆ’ 2

â‡’Â (gof)Â (x)Â =Â 4x2+Â 4xÂ âˆ’1

Then,Â (gof)Â (1)Â =Â gÂ (fÂ (1))

=Â 4 + 4 âˆ’ 1

=7,

(gof) (2) = gÂ (fÂ (2))

= 4(2)2 + 4(2) â€“ 1 = 23,

(gof) (3) = gÂ (fÂ (3))

= 4(3)2 + 4(3) â€“ 1 = 47Â and

(gof) (4) = gÂ (fÂ (4))

= 4(4)2 + 4(4) âˆ’ 1 = 79

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

â‡’ (gof)â€“ 1 = {(7,Â 1),Â (23,Â 2),Â (47,Â 3),Â (79,Â 4)}…… (2)

FromÂ (1)Â andÂ (2),Â weÂ get:

(gof)âˆ’1Â =Â fâˆ’1Â oÂ gâˆ’1

5. Show that the functionÂ f:Â QÂ â†’Â Q,Â defined byÂ f(x) = 3xÂ + 5, is invertible. Also, findÂ fâˆ’1

Solution:

Given functionÂ f:Â QÂ â†’Â Q,Â defined byÂ f(x) = 3xÂ + 5

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

Injection of f:

LetÂ xÂ andÂ yÂ be two elements of the domain (Q),

Such that f(x) = f(y)

â‡’ 3xÂ + 5 = 3yÂ + 5

â‡’ 3xÂ = 3y

â‡’ x = y

so,Â f is one-one.

Surjection ofÂ f:

LetÂ yÂ be in the co-domain (Q),

Such thatÂ f(x) = y

â‡’ 3x +5 = y

â‡’ 3x = y – 5

â‡’ x = (y -5)/3 belongs to Q domain

â‡’ fÂ is onto.

So,Â fÂ is a bijection and, hence, it is invertible.

Now we have to findÂ f-1:

LetÂ f-1(x) = yâ€¦… (1)

â‡’Â x = f(y)

â‡’Â x = 3y + 5

â‡’Â xÂ âˆ’5Â =Â 3y

â‡’ y = (x – 5)/3

Now substituting this value in (1) we get

So, f-1(x) = (x – 5)/3

6. ConsiderÂ f:Â RÂ â†’Â RÂ given byÂ f(x) = 4xÂ + 3. Show thatÂ fÂ is invertible. Find the inverse ofÂ f.

Solution:

Given f:Â RÂ â†’Â RÂ given byÂ f(x) = 4xÂ + 3

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

Consider injection ofÂ f:

LetÂ xÂ andÂ yÂ be two elements of domain (R),

Such that f(x) = f(y)

â‡’ 4xÂ + 3 = 4yÂ + 3

â‡’ 4xÂ = 4y

â‡’ x = y

So,Â fÂ is one-one.

Now surjection ofÂ f:

LetÂ yÂ be in the co-domain (R),

Such thatÂ f(x) = y.

â‡’ 4x + 3 = y

â‡’ 4x = y -3

â‡’ x = (y-3)/4 in RÂ (domain)

â‡’ fÂ is onto.

So,Â fÂ is a bijection and, hence, it is invertible.

Now we have to findÂ fÂ -1

Let f-1(x) = y……. (1)

â‡’Â xÂ =Â fÂ (y)

â‡’Â xÂ =Â 4yÂ +Â 3

â‡’Â xÂ âˆ’Â 3Â =Â 4y

â‡’ y = (x -3)/4

Now substituting this value in (1) we get

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

7. ConsiderÂ f:Â RÂ â†’Â R+Â â†’ [4, âˆž) given byÂ f(x) =Â x2Â + 4. Show thatÂ fÂ is invertible with inverseÂ fâˆ’1Â ofÂ fÂ given by fâˆ’1(x) = âˆš (x-4) where R+Â is the set of all non-negative real numbers.

Solution:

Given f:Â RÂ â†’Â R+Â â†’ [4, âˆž) given byÂ f(x) =Â x2Â + 4.

Now we have to show that f is invertible,

Consider injection ofÂ f:

LetÂ xÂ andÂ yÂ be two elements of the domain (Q),

Such that f(x) = f(y)

â‡’Â x2 + 4 = y2 + 4

â‡’Â x2 = y2

â‡’Â xÂ =Â yÂ  Â  Â Â (asÂ co-domainÂ asÂ R+)

So,Â fÂ is one-one

Now surjection ofÂ f:

LetÂ yÂ be in the co-domain (Q),

Such thatÂ f(x) = y

â‡’ x2Â + 4 = y

â‡’ x2Â = y – 4

â‡’ x = âˆš (y-4) in R

â‡’Â fÂ is onto.

So,Â fÂ is a bijection and, hence, it is invertible.

Now we have to findÂ f-1:

LetÂ fâˆ’1Â (x)Â =Â yâ€¦… (1)

â‡’Â xÂ =Â fÂ (y)

â‡’Â xÂ =Â y2Â +Â 4

â‡’Â xÂ âˆ’Â 4Â =Â y2

â‡’ y = âˆš (x-4)

So, f-1(x) = âˆš (x-4)

Now substituting this value in (1) we get,

So, f-1(x) = âˆš (x-4)

8. If f(x) = (4x + 3)/ (6x – 4), x â‰  (2/3)Â show thatÂ fof(x) =Â x, for all xÂ â‰  (2/3). What is the inverse ofÂ f?

Solution:

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

Now we have to show fof(x) =Â 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 â‰  2/3

=> fofÂ = 1

Hence, the given functionÂ fÂ is invertible and the inverse ofÂ fÂ isÂ fÂ itself.

9. ConsiderÂ f:Â R+Â â†’ [âˆ’5, âˆž) given byÂ f(x) = 9x2Â + 6xÂ âˆ’ 5. Show thatÂ fÂ is invertible with

f-1(x) = (âˆš(x +6)-1)/3Â

Solution:

Given f:Â R+Â â†’ [âˆ’5, âˆž) given byÂ f(x) = 9x2Â + 6xÂ â€“ 5

We have to show that f is invertible.

Injectivity ofÂ f:

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

Such that f(x) = f(y)

â‡’Â 9x2 + 6x â€“ 5 = 9y2 +Â 6yÂ âˆ’Â 5

â‡’Â 9x2 + 6x = 9y2 + 6y

â‡’ x = y (As, x, y âˆˆ R+)

So,Â fÂ is one-one.

Surjectivity ofÂ f:

LetÂ yÂ is in the co domain (Q)

Such thatÂ f(x) = y

â‡’ 9x2Â + 6x – 5 = y

â‡’ 9x2Â + 6x = y + 5

â‡’ 9x2Â + 6x +1 = y + 6 (By adding 1 on both sides)

â‡’ (3x + 1)2Â = y + 6

â‡’ 3x + 1 = âˆš(y + 6)

â‡’ 3x = âˆš (y + 6) – 1

â‡’ x = (âˆš (y + 6)-1)/3 in R+ (domain)

fÂ is onto.

So,Â fÂ is a bijection and hence, it is invertible.

Now we have to find f-1

LetÂ fâˆ’1(x)Â =Â y….. (1)

â‡’Â xÂ =Â fÂ (y)

â‡’Â xÂ =Â 9y2 +Â 6yÂ âˆ’Â 5

â‡’Â xÂ +Â 5Â =Â 9y2 + 6y

â‡’Â xÂ +Â 6 =Â 9y2+Â 6yÂ +Â 1Â Â  Â  Â  Â Â (addingÂ 1Â onÂ bothÂ sides)

â‡’Â xÂ +Â 6Â =Â (3yÂ +Â 1)2

â‡’ 3y + 1 = âˆš (x + 6)

â‡’Â 3y =âˆš(x +6) -1

â‡’ y = (âˆš (x+6)-1)/3

Now substituting this value in (1) we get,

So, f-1(x)Â = (âˆš (x+6)-1)/3

10. IfÂ f:Â RÂ â†’Â RÂ be defined byÂ f(x) =Â x3Â âˆ’3, then prove thatÂ fâˆ’1Â exists and find a formula forÂ fâˆ’1. Hence, findÂ fâˆ’1 (24) andÂ fâˆ’1Â (5).

Solution:

Given f:Â RÂ â†’Â RÂ be defined byÂ f(x) =Â x3Â âˆ’3

Now we have to prove that fâˆ’1Â exists

Injectivity ofÂ f:

LetÂ xÂ andÂ yÂ be two elements in domain (R),

SuchÂ that,Â x3Â âˆ’Â 3Â =Â y3Â âˆ’Â 3

â‡’Â x3Â =Â y3

â‡’Â xÂ =Â y

So,Â fÂ is one-one.

Surjectivity ofÂ f:

LetÂ yÂ be in the co-domain (R)

Such thatÂ f(x) = y

â‡’ x3Â – 3 = y

â‡’Â x3Â = y + 3

â‡’ x = âˆ›(y+3) in R

â‡’Â fÂ is onto.

So,Â fÂ is a bijection and, hence, it is invertible.

FindingÂ fÂ -1:

LetÂ f-1(x)Â =Â yâ€¦….. (1)

â‡’Â x=Â f(y)

â‡’Â xÂ =Â y3 âˆ’ 3

â‡’Â xÂ +Â 3Â =Â y3

â‡’Â y = âˆ›(x + 3)Â = f-1(x)Â  Â  Â  Â  Â [fromÂ (1)]

So, f-1(x) = âˆ›(x + 3)

Now, f-1(24) = âˆ› (24 + 3)

= âˆ›27

= âˆ›33

= 3

And f-1(5) =âˆ› (5 + 3)

= âˆ›8

= âˆ›23

= 2

11. A functionÂ f:Â RÂ â†’Â RÂ is defined asÂ f(x) =Â x3Â + 4. Is it a bijection or not? In case it is a bijection, findÂ fâˆ’1Â (3).

Solution:

Given that f:Â RÂ â†’Â RÂ is defined asÂ f(x) =Â x3Â + 4

Injectivity ofÂ f:

LetÂ xÂ andÂ yÂ be two elements of domain (R),

Such that fÂ (x)Â =Â fÂ (y)

â‡’Â x3Â +Â 4Â =Â y3Â +Â 4

â‡’Â x3Â =Â y3

â‡’Â xÂ =Â y

So,Â fÂ is one-one.

Surjectivity ofÂ f:

LetÂ yÂ be in the co-domain (R),

Such thatÂ f(x) = y.

â‡’Â x3Â + 4 = y

â‡’ x3Â = y – 4

â‡’ x = âˆ› (y – 4) in R (domain)

â‡’ fÂ is onto.

So,Â fÂ is a bijection and, hence, it is invertible.

FindingÂ f-1:

LetÂ fâˆ’1Â (x)Â =Â yâ€¦… (1)

â‡’Â xÂ =Â fÂ (y)

â‡’Â xÂ =Â y3 +Â 4

â‡’Â xÂ âˆ’Â 4Â =Â y3

â‡’ y =âˆ› (x-4)

So,Â f-1(x)Â =âˆ› (x-4) Â  Â  Â  Â [fromÂ (1)]

f-1 (3) = âˆ›(3 – 4)

= âˆ›-1

= -1