# RD Sharma Solutions for Class 12 Maths Chapter 2 Functions

## RD Sharma Solutions for Class 12 Maths Chapter 2 – Free PDF Download Updated for (2021-22)

RD Sharma Solutions Class 12 Maths Chapter 2 Functions help students who aspire to obtain a good academic score in the exam. The solutions are designed by experts to boost confidence among students in understanding the concepts covered in this chapter and methods to solve problems in a shorter period. Students can access RD Sharma Solutions while practising problems for step-by-step guidance and instant doubt clearance.

Chapter 2 Functions of Class 12 RD Sharma Solutions explains the function, domain and codomain of functions. It has four exercises. RD Sharma Solutions for Class 12 resources are prepared based on the Class 12 CBSE syllabus by our subject experts, taking into consideration, the types of questions asked in CBSE Class 12 board exam. Students can easily access answers to the problems present in RD Sharma Solutions for Class 12. Let us have a look at some of the important concepts that are discussed in this chapter.

• Classification of functions
• Types of functions
• Constant function
• Identity function
• Modulus function
• Integer function
• Exponential function
• Logarithmic function
• Reciprocal function
• Square root function
• Operations on real functions
• Kinds of functions
• One-one function
• On-to function
• Many one function
• In to function
• Bijection
• Composition of functions
• Properties of the composition of functions
• Composition of real function
• Inverse of a function
• Inverse of an element
• Relation between graphs of a function and its inverse

## Download the PDF of RD Sharma Solutions For Class 12 Maths Chapter 2 Functions

### Exercise 2.1 Page No: 2.31

1. Give an example of a functionÂ

(i) Which is one-one but not onto.

(ii) Which is not one-one but onto.

(iii) Which is neither one-one nor onto.

Solution:

(i) Let f:Â ZÂ â†’Â ZÂ given byÂ f(x) = 3xÂ + 2

Let us check one-one condition on f(x) = 3xÂ + 2

Injectivity:

LetÂ xÂ andÂ yÂ be any two elements in the domain (Z), such thatÂ f(x) =Â f(y).

fÂ (x) =Â f(y)

â‡’ 3xÂ + 2 =3yÂ + 2

â‡’Â 3xÂ = 3y

â‡’Â xÂ =Â y

â‡’Â f(x) =Â f(y)

â‡’Â xÂ =Â y

So,Â fÂ is one-one.

Surjectivity:

LetÂ yÂ be any element in the co-domain (Z), such thatÂ f(x) =Â yÂ for some elementÂ xÂ inÂ Z(domain).

Let f(x) =Â y

â‡’ 3xÂ + 2 =Â y

â‡’Â 3xÂ =Â yÂ – 2

â‡’Â x =Â (y – 2)/3.Â ItÂ mayÂ notÂ beÂ inÂ theÂ domainÂ (Z)

BecauseÂ ifÂ weÂ takeÂ yÂ =Â 3,

x = (y – 2)/3 = (3-2)/3 = 1/3 âˆ‰Â domainÂ Z.

So, for every element in the co domain there need not be any element in the domain such thatÂ f(x) =Â y.

Thus,Â fÂ is not onto.

(ii) Example for the function which is not one-one but onto

Let f:Â ZÂ â†’Â NÂ âˆª {0} given byÂ f(x) = |x|

Injectivity:

LetÂ xÂ andÂ yÂ be any two elements in the domain (Z),

Such thatÂ f(x) =Â f(y).

â‡’ |x| = |y|

â‡’Â x =Â Â± y

So, different elements of domainÂ fÂ may give the same image.

So,Â fÂ is not one-one.

Surjectivity:

LetÂ yÂ be any element in the co domain (Z), such thatÂ f(x) =Â yÂ for some elementÂ xÂ inÂ Z (domain).

f(x) =Â y

â‡’ |x| =Â y

â‡’ xÂ = Â± y

Which is an element inÂ ZÂ (domain).

So, for every element in the co-domain, there exists a pre-image in the domain.

Thus,Â fÂ is onto.

(iii) Example for the function which is neither one-one nor onto.

Let f:Â ZÂ â†’Â ZÂ given byÂ f(x) = 2x2Â + 1

Injectivity:

LetÂ xÂ andÂ yÂ be any two elements in the domain (Z), such thatÂ f(x) =Â f(y).

f(x)Â =Â f(y)

â‡’Â 2x2+1Â =Â 2y2+1

â‡’Â 2x2Â =Â 2y2

â‡’Â x2Â =Â y2

â‡’Â xÂ =Â Â±Â y

So, different elements of domainÂ fÂ may give the same image.

Thus,Â fÂ is not one-one.

Surjectivity:

LetÂ yÂ be any element in the co-domain (Z), such thatÂ f(x) =Â yÂ for some elementÂ xÂ inÂ Z (domain).

f (x) =Â y

â‡’Â 2x2+1=y

â‡’Â 2x2=Â yÂ âˆ’Â 1

â‡’Â x2 = (y-1)/2

â‡’Â x = âˆš ((y-1)/2)Â âˆ‰ Z always.

ForÂ example,Â ifÂ weÂ take,Â yÂ =Â 4,

x = Â± âˆš ((y-1)/2)

= Â± âˆš ((4-1)/2)

= Â± âˆš (3/2) âˆ‰ Z

So,Â xÂ mayÂ notÂ beÂ inÂ ZÂ (domain).

Thus,Â fÂ is not onto.

2. Which of the following functions fromÂ AÂ toÂ BÂ are one-one and onto?
(i) f1Â = {(1, 3), (2, 5), (3, 7)};Â AÂ = {1, 2, 3},Â BÂ = {3, 5, 7}

(ii) f2Â = {(2,Â a), (3,Â b), (4,Â c)};Â AÂ = {2, 3, 4},Â BÂ = {a,Â b,Â c}

(iii) f3Â = {(a,Â x), (b,Â x), (c,Â z), (d,Â z)};Â AÂ = {a,Â b,Â c,Â d,},Â BÂ = {x,Â y,Â z}.Â

Solution:

(i) Consider f1Â = {(1, 3), (2, 5), (3, 7)};Â AÂ = {1, 2, 3},Â BÂ = {3, 5, 7}

Injectivity:

f1Â (1) = 3

f1Â (2) = 5

f1Â (3) = 7

â‡’ Every element ofÂ AÂ has different images inÂ B.

So,Â f1Â is one-one.

Surjectivity:

Co-domain ofÂ f1Â = {3, 5, 7}

Range ofÂ f1Â =set of imagesÂ  =Â  {3, 5, 7}

â‡’ Co-domain = range

So,Â f1Â is onto.

(ii) Consider f2Â = {(2,Â a), (3,Â b), (4,Â c)};Â AÂ = {2, 3, 4},Â BÂ = {a,Â b,Â c}

Injectivity:

f2Â (2)Â = a

f2Â (3)Â = b

f2Â (4)Â = c

â‡’ Every element ofÂ AÂ has different images inÂ B.

So,Â f2Â is one-one.

Surjectivity:

Co-domain ofÂ f2Â = {a,Â b,Â c}

Range ofÂ f2Â = set of images = {a,Â b,Â c}

â‡’ Co-domain = range

So,Â f2Â is onto.

(iii) Consider f3Â = {(a,Â x), (b,Â x), (c,Â z), (d,Â z)} ;Â AÂ = {a,Â b,Â c,Â d,},Â BÂ = {x,Â y,Â z}

Injectivity:

f3Â (a) = x

f3Â (b) = x

f3Â (c) = z

f3Â (d) = z

â‡’ aÂ andÂ bÂ have the same imageÂ x.

AlsoÂ cÂ andÂ dÂ have the same imageÂ z

So,Â f3Â is not one-one.

Surjectivity:

Co-domain ofÂ f3Â ={x, y, z}

Range ofÂ f3Â =set of images =Â {x, z}

So, the co-domainÂ  is not same as the range.

So,Â f3Â is not onto.

3. Prove that the functionÂ f:Â NÂ â†’Â N, defined byÂ f(x) =Â x2Â +Â xÂ + 1, is one-one but not onto

Solution:

Given f:Â NÂ â†’Â N, defined byÂ f(x) =Â x2Â +Â xÂ + 1

Now we have to prove that given function is one-one

Injectivity:

LetÂ xÂ andÂ yÂ be any two elements in the domain (N), such thatÂ f(x) = f(y).

â‡’ x2 + x + 1 = y2 + y + 1

â‡’ (x2 â€“ y2) + (x – y) = 0 

â‡’ (x + y) (x- y ) + (x – y ) = 0

â‡’ (x – y) (x + y + 1) = 0

â‡’ x – y = 0 [x + y + 1Â cannot be zero because x and y are natural numbers

â‡’ x = y

So,Â fÂ is one-one.

Surjectivity:

When x = 1

x2 + x + 1 = 1 + 1 + 1 = 3

â‡’ x2 + x +1 â‰¥ 3,Â for every x in N.

â‡’Â f(x) will not assume the values 1 and 2.

So, f is not onto.

4. LetÂ AÂ = {âˆ’1, 0, 1} andÂ fÂ = {(x,Â x2) :Â xÂ âˆˆÂ A}. Show thatÂ fÂ :Â AÂ â†’Â AÂ is neither one-one nor onto.

Solution:

Given AÂ = {âˆ’1, 0, 1} andÂ fÂ = {(x,Â x2):Â xÂ âˆˆÂ A}

Also given that,Â f(x) = x2

Now we have to prove that given function neither one-one or nor onto.

Injectivity:

Let x = 1

Therefore f(1) = 12=1 and

f(-1)=(-1)2=1

â‡’ 1 and -1 have the same images.

So,Â fÂ is not one-one.

Surjectivity:

Co-domain ofÂ f =Â {-1, 0, 1}

f(1) = 12Â = 1,

f(-1) = (-1)2Â = 1 and

f(0) = 0

â‡’ Range ofÂ fÂ Â = {0, 1}

So, both are not same.

Hence,Â fÂ is not onto

5. Classify the following function as injection, surjection or bijection:

(i)Â f:Â NÂ â†’Â NÂ given byÂ f(x) =Â x2

(ii)Â f:Â ZÂ â†’Â ZÂ given byÂ f(x) =Â x2

(iii) f:Â NÂ â†’Â NÂ given byÂ f(x) =Â x3

(iv) f:Â ZÂ â†’Â ZÂ given byÂ f(x) =Â x3

(v) f:Â RÂ â†’Â R, defined byÂ f(x) = |x|

(vi) f:Â ZÂ â†’Â Z, defined byÂ f(x) =Â x2Â +Â x

(vii) f:Â ZÂ â†’Â Z, defined byÂ f(x) =Â xÂ âˆ’ 5

(viii) f:Â RÂ â†’Â R, defined byÂ f(x) = sin x

(ix) f:Â RÂ â†’Â R, defined byÂ f(x) =Â x3Â + 1

(x) f:Â RÂ â†’Â R, defined byÂ f(x) =Â x3Â âˆ’Â x

(xi) f:Â RÂ â†’Â R, defined byÂ f(x) = sin2xÂ + cos2x

(xii) f:Â QÂ âˆ’ {3} â†’Â Q, defined by f (x) = (2x +3)/(x-3)

(xiii) f:Â QÂ â†’Â Q, defined byÂ f(x) =Â x3Â + 1

(xiv) f:Â RÂ â†’Â R, defined byÂ f(x) = 5x3Â + 4

(xv) f:Â RÂ â†’Â R, defined byÂ f(x) = 5x3Â + 4

(xvi) f:Â RÂ â†’Â R, defined byÂ f(x) = 1 +Â x2

(xvii) f:Â RÂ â†’Â R, defined byÂ f(x) = x/(x2 + 1)

Solution:

(i) Given f:Â NÂ â†’Â N,Â given byÂ f(x) =Â x2

Now we have to check for the given function is injection, surjection and bijection condition.

Injection condition:

LetÂ xÂ andÂ yÂ be any two elements in the domain (N), such thatÂ f(x)Â = f(y).

f(x) = f(y)

x2 = y2

x = yÂ (WeÂ doÂ notÂ getÂ Â±Â becauseÂ xÂ andÂ yÂ areÂ inÂ N that is natural numbers)

So,Â fÂ is an injection.

Surjection condition:

LetÂ yÂ be any element in the co-domain (N),Â such thatÂ f(x)Â = yÂ for some elementÂ xÂ inÂ N (domain).

f(x)Â = y

x2=Â y

x = âˆšy,Â whichÂ mayÂ notÂ beÂ inÂ N.

ForÂ example,Â ifÂ yÂ = 3,

x = âˆš3Â isÂ notÂ inÂ N.

So,Â fÂ is not a surjection.

Also f is not a bijection.

(ii) Given f:Â ZÂ â†’Â Z,Â given byÂ f(x) =Â x2

Now we have to check for the given function is injection, surjection and bijection condition.

Injection condition:

LetÂ xÂ andÂ yÂ be any two elements in the domain (Z), such thatÂ f(x)Â = f(y).

f(x)Â = f(y)

x2 = y2

xÂ =Â Â±y

So,Â fÂ is not an injection.

Surjection test:

LetÂ yÂ be any element in the co-domain (Z),Â such thatÂ f(x)Â = yÂ for some elementÂ xÂ inÂ Z (domain).

f(x)Â = y

x2 =Â y

x = Â± âˆšyÂ whichÂ mayÂ notÂ beÂ inÂ Z.

ForÂ example,Â ifÂ yÂ = 3,

xÂ =Â Â± âˆš 3 isÂ notÂ inÂ Z.

So,Â fÂ is not a surjection.

Also f is not bijection.

(iii) Given f:Â NÂ â†’Â NÂ given byÂ f(x) =Â x3

Now we have to check for the given function is injection, surjection and bijection condition.

Injection condition:

LetÂ xÂ andÂ yÂ be any two elements in the domain (N), such thatÂ f(x)Â = f(y).

f(x)Â = f(y)

x3Â =Â y3

xÂ =Â y

So,Â fÂ is an injection

Surjection condition:

LetÂ yÂ be any element in the co-domain (N),Â such thatÂ f(x)Â = yÂ for some elementÂ xÂ inÂ NÂ (domain).

f(x)Â = y

x3=Â y

x = âˆ›y whichÂ mayÂ notÂ beÂ inÂ N.

ForÂ example,Â ifÂ yÂ = 3,

X = âˆ›3Â isÂ notÂ inÂ N.

So,Â fÂ is not a surjection andÂ fÂ is not a bijection.

(iv) Given f:Â ZÂ â†’Â ZÂ given byÂ f(x) =Â x3

Now we have to check for the given function is injection, surjection and bijection condition.

Injection condition:

LetÂ xÂ andÂ yÂ be any two elements in the domain (Z), such thatÂ f(x)Â = f(y)

f(x)Â = f(y)

x3Â =Â y3

xÂ =Â y

So,Â fÂ is an injection.

Surjection condition:

LetÂ yÂ be any element in the co-domain (Z),Â such thatÂ f(x)Â = yÂ for some elementÂ xÂ inÂ Z (domain).

f(x) = y

x3Â =Â y

x = âˆ›yÂ whichÂ mayÂ notÂ beÂ inÂ Z.

ForÂ example,Â ifÂ yÂ = 3,

x = âˆ›3 isÂ notÂ inÂ Z.

So,Â fÂ is not a surjection andÂ fÂ is not a bijection.

(v) Given f:Â RÂ â†’Â R, defined byÂ f(x) = |x|

Now we have to check for the given function is injection, surjection and bijection condition.

Injection test:

LetÂ xÂ andÂ yÂ be any two elements in the domain (R), such thatÂ f(x)Â = f(y)

f(x)Â = f(y)

|x|=|y|

x = Â±y

So,Â fÂ is not an injection.

Surjection test:

LetÂ yÂ be any element in the co-domain (R), such thatÂ f(x)Â = yÂ for some elementÂ xÂ inÂ RÂ (domain).

f(x)Â = y

|x|=y

xÂ =Â Â±Â yÂ âˆˆÂ Z

So,Â fÂ is a surjection andÂ fÂ is not a bijection.

(vi) Given f:Â ZÂ â†’Â Z, defined byÂ f(x) =Â x2Â +Â x

Now we have to check for the given function is injection, surjection and bijection condition.

Injection test:

LetÂ xÂ andÂ yÂ be any two elements in the domain (Z), such thatÂ f(x)Â = f(y).

f(x)Â = f(y)

x2+Â xÂ =Â y2Â +Â y

Here,Â weÂ cannotÂ sayÂ thatÂ xÂ =Â y.

ForÂ example,Â xÂ =Â 2Â andÂ yÂ =Â –Â 3

Then,

x2 + x = 22 + 2 =Â 6

y2 + y = (âˆ’3)2 â€“ 3 =Â 6

So,Â weÂ haveÂ twoÂ numbersÂ 2Â andÂ -3Â inÂ theÂ domainÂ ZÂ whoseÂ imageÂ isÂ sameÂ asÂ 6.

So,Â fÂ is not an injection.

Surjection test:

LetÂ yÂ be any element in the co-domain (Z),

such thatÂ f(x)Â = y for some elementÂ xÂ inÂ ZÂ (domain).

f(x)Â = y

x2Â +Â xÂ =Â y

Here,Â weÂ cannotÂ sayÂ xÂ âˆˆÂ Z.

ForÂ example,Â yÂ = – 4.

x2Â +Â xÂ =Â âˆ’Â 4

x2 +Â xÂ +Â 4Â =Â 0

xÂ = (-1 Â± âˆš-5)/2 = (-1 Â± i âˆš5)/2Â whichÂ isÂ notÂ inÂ Z.

So,Â fÂ is not a surjection andÂ fÂ is not a bijection.

(vii) Given f:Â ZÂ â†’Â Z, defined byÂ f(x) =Â xÂ â€“ 5

Now we have to check for the given function is injection, surjection and bijection condition.

Injection test:

LetÂ xÂ andÂ yÂ be any two elements in the domain (Z), such thatÂ f(x)Â = f(y).

f(x)Â = f(y)

x –Â 5 =Â y –Â 5

x = y

So,Â fÂ is an injection.

Surjection test:

LetÂ yÂ be any element in the co-domain (Z),Â such thatÂ f(x)Â = yÂ for some elementÂ xÂ inÂ ZÂ (domain).

f(x)Â = y

x –Â 5 =Â y

x = yÂ + 5, which is inÂ Z.

So,Â fÂ is a surjection andÂ fÂ is a bijection

(viii) Given f:Â RÂ â†’Â R, defined byÂ f(x) = sin x

Now we have to check for the given function is injection, surjection and bijection condition.

Injection test:

LetÂ xÂ andÂ yÂ be any two elements in the domain (R), such thatÂ f(x)Â = f(y).

f(x)Â = f(y)

SinÂ xÂ =Â sinÂ y

Here,Â xÂ mayÂ notÂ beÂ equalÂ toÂ yÂ becauseÂ sin 0 = sin Ï€.

So,Â 0Â andÂ Ï€Â haveÂ theÂ sameÂ imageÂ 0.

So,Â fÂ is not an injection.

Surjection test:

Range ofÂ fÂ = [-1, 1]

Co-domain ofÂ fÂ =Â R

Both are not same.

So,Â fÂ is not a surjection andÂ fÂ is not a bijection.

(ix) Given f:Â RÂ â†’Â R, defined byÂ f(x) =Â x3Â + 1

Now we have to check for the given function is injection, surjection and bijection condition.

Injection test:

LetÂ xÂ andÂ yÂ be any two elements in the domain (R), such thatÂ f(x)Â = f(y).

f(x)Â = f(y)

x3+1Â =Â y3+Â 1

x3= y3

xÂ =Â y

So,Â fÂ is an injection.

Surjection test:

LetÂ yÂ be any element in the co-domain (R),Â such thatÂ f(x)Â = yÂ for some elementÂ xÂ inÂ RÂ (domain).

f(x) = y

x3+1=y

x = âˆ› (y – 1) âˆˆ R

So,Â fÂ is a surjection.

So,Â fÂ is a bijection.

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

Now we have to check for the given function is injection, surjection and bijection condition.

Injection test:

LetÂ xÂ andÂ yÂ be any two elements in the domain (R), such thatÂ f(x)Â = f(y).

f(x)Â = f(y)

x3 â€“ x = y3 âˆ’ y

Here,Â weÂ cannotÂ sayÂ x = y.

ForÂ example,Â x = 1Â andÂ y = -1

x3 âˆ’ xÂ = 1 âˆ’ 1 =Â 0

y3 â€“ y = (âˆ’1)3âˆ’ (âˆ’1) â€“ 1 + 1 = 0

So,Â 1Â andÂ -1Â haveÂ theÂ sameÂ imageÂ 0.

So,Â fÂ is not an injection.

Surjection test:

LetÂ yÂ be any element in the co-domain (R),Â such thatÂ f(x)Â = yÂ for some elementÂ xÂ inÂ RÂ (domain).

f(x) =Â y

x3Â âˆ’Â xÂ =Â y

ByÂ observationÂ weÂ canÂ sayÂ thatÂ thereÂ existÂ someÂ xÂ inÂ R,Â suchÂ thatÂ x3Â – x = y.

So,Â fÂ is a surjection andÂ fÂ is not a bijection.

(xi) Given f:Â RÂ â†’Â R, defined byÂ f(x) = sin2xÂ + cos2x

Now we have to check for the given function is injection, surjection and bijection condition.

Injection condition:

f(x) = sin2xÂ + cos2x

We know that sin2xÂ + cos2xÂ = 1

So,Â f(x) = 1 for everyÂ xÂ in R.

So, for all elements in the domain, the image is 1.

So,Â fÂ is not an injection.

Surjection condition:

Range ofÂ fÂ = {1}

Co-domain ofÂ fÂ =Â R

Both are not same.

So,Â fÂ is not a surjection andÂ fÂ is not a bijection.

(xii) Given f:Â QÂ âˆ’ {3} â†’Â Q, defined by f (x) = (2x +3)/(x-3)

Now we have to check for the given function is injection, surjection and bijection condition.

Injection test:

LetÂ xÂ andÂ yÂ be any two elements in the domain (QÂ âˆ’Â {3}), such thatÂ f(x)Â = f(y).

f(x)Â =Â f(y)

(2x + 3)/(x – 3) = (2y + 3)/(y – 3)

(2x + 3)Â (y âˆ’ 3) =Â (2y + 3)Â (x âˆ’ 3)

2xyÂ âˆ’Â 6x +Â 3y âˆ’ 9Â =Â 2xyÂ âˆ’Â 6y + 3x âˆ’ 9

9xÂ =Â 9y

xÂ =Â y

So,Â fÂ is an injection.

Surjection test:

LetÂ yÂ be any element in the co-domain (QÂ âˆ’Â {3}),Â such thatÂ f(x)Â = yÂ for some elementÂ xÂ inÂ QÂ (domain).

f(x) =Â y

(2x + 3)/(x – 3) = y

2x + 3 = x y âˆ’ 3y

2x â€“ x y = âˆ’3y âˆ’ 3

xÂ (2âˆ’y) = âˆ’3Â (y + 1)

x = -3(y + 1)/(2 – y)Â whichÂ isÂ notÂ definedÂ atÂ y = 2.

So,Â fÂ is not a surjection andÂ fÂ is not a bijection.

(xiii) Given f:Â QÂ â†’Â Q, defined byÂ f(x) =Â x3Â + 1

Now we have to check for the given function is injection, surjection and bijection condition.

Injection test:

LetÂ xÂ andÂ yÂ be any two elements in the domain (Q), such thatÂ f(x)Â = f(y).

f(x)Â = f(y)

x3 + 1Â =Â y3 +Â 1

x3Â =Â y3

xÂ =Â y

So,Â fÂ is an injection.

Surjection test:

LetÂ yÂ be any element in the co-domain (Q),Â such thatÂ f(x)Â = yÂ for some elementÂ xÂ inÂ Q (domain).

f(x)Â = y

x3+Â 1Â =Â y

x = âˆ›(y-1), whichÂ mayÂ notÂ beÂ inÂ Q.

ForÂ example,Â ifÂ y=Â 8,

x3+Â 1Â =Â Â 8

x3=Â 7

x = âˆ›7, whichÂ isÂ notÂ inÂ Q.

So,Â fÂ is not a surjection andÂ fÂ is not a bijection.

(xiv) Given f:Â RÂ â†’Â R, defined byÂ f(x) = 5x3Â + 4

Now we have to check for the given function is injection, surjection and bijection condition.

Injection test:

LetÂ xÂ andÂ yÂ be any two elements in the domain (R), such thatÂ f(x)Â = f(y).

f(x)Â = f(y)

5x3 + 4Â =Â 5y3 + 4

5x3=Â 5y3

x3 =Â y3

xÂ =Â y

So,Â fÂ is an injection.

Surjection test:

LetÂ yÂ be any element in the co-domain (R),Â such thatÂ f(x)Â = yÂ for some elementÂ xÂ inÂ R (domain).

f(x) = y

5x3+ 4Â =Â y

x3 = (y – 4)/5 âˆˆ R

So,Â fÂ is a surjection andÂ fÂ is a bijection.

(xv) Given f:Â RÂ â†’Â R, defined byÂ f(x) = 5x3Â + 4

Now we have to check for the given function is injection, surjection and bijection condition.

Injection condition:

LetÂ xÂ andÂ yÂ be any two elements in the domain (R), such thatÂ f(x)Â = f(y).

f(x)Â = f(y)

5x3 + 4Â =Â 5y3 + 4

5x3 =Â 5y3

x3 =Â y3

xÂ =Â y

So,Â fÂ is an injection.

Surjection test:

LetÂ yÂ be any element in the co-domain (R),Â such thatÂ f(x)Â = yÂ for some elementÂ xÂ inÂ R (domain).

f(x) = y

5x3 + 4Â =Â y

x3 = (y – 4)/5 âˆˆ R

So,Â fÂ is a surjection andÂ fÂ is a bijection.

(xvi) Given f:Â RÂ â†’Â R, defined byÂ f(x) = 1 +Â x2

Now we have to check for the given function is injection, surjection and bijection condition.

Injection condition:

LetÂ xÂ andÂ yÂ be any two elements in the domain (R), such thatÂ f(x)Â = f(y).

f(x)Â = f(y)

1Â +Â x2 = 1Â +Â y2

x2Â =Â y2

xÂ =Â Â±Â y

So,Â fÂ is not an injection.

Surjection test:

LetÂ yÂ be any element in the co-domain (R),Â such thatÂ f(x)Â = yÂ for some elementÂ xÂ inÂ R (domain).

f(x)Â = y

1Â +Â x2 =Â y

x2 =Â yÂ âˆ’Â 1

x = Â± âˆš-1 = Â± iÂ isÂ notÂ inÂ R.

So,Â fÂ is not a surjection andÂ fÂ is not a bijection.

(xvii) Given f:Â RÂ â†’Â R, defined byÂ f(x) = x/(x2 + 1)

Now we have to check for the given function is injection, surjection and bijection condition.

Injection condition:

LetÂ xÂ andÂ yÂ be any two elements in the domain (R), such thatÂ f(x)Â = f(y).

f(x)Â = f(y)

x /(x2 + 1) = y /(y2 + 1)

x y2+Â xÂ =Â x2yÂ +Â y

xy2 âˆ’ x2yÂ +Â xÂ âˆ’ yÂ =Â 0

âˆ’x yÂ (âˆ’y + x) +Â 1Â (x âˆ’ y)Â =Â 0

(x âˆ’ y)Â (1 â€“ x y)Â =Â 0

xÂ =Â yÂ orÂ xÂ = 1/y

So,Â fÂ is not an injection.

Surjection test:

LetÂ yÂ be any element in the co-domainÂ (R),Â such thatÂ f(x)Â = yÂ for some elementÂ xÂ inÂ R (domain).

f(x)Â = y

x /(x2 + 1) = y

y x2 â€“ x + y = 0

x = (-(-1) Â± âˆš (1-4y2))/(2y)Â ifÂ yÂ â‰ Â 0

= (1 Â± âˆš (1-4y2))/ (2y), whichÂ mayÂ notÂ beÂ inÂ R

ForÂ example,Â ifÂ y=1,Â then

(1 Â± âˆš (1-4)) / (2y) = (1 Â± i âˆš3)/2, whichÂ isÂ notÂ inÂ R

So,Â fÂ isÂ notÂ surjectionÂ andÂ fÂ isÂ notÂ bijection.

6. IfÂ f:Â AÂ â†’Â BÂ is an injection, such that range ofÂ fÂ = {a}, determine the number of elements inÂ A.

Solution:

Given f:Â AÂ â†’Â BÂ is an injection

And also given that range ofÂ fÂ = {a}

So, the number of images ofÂ Â fÂ = 1

Since,Â fÂ Â is an injection, there will be exactly one image for each element ofÂ fÂ .

So, number of elements inÂ AÂ = 1.

7. Show that the functionÂ f:Â RÂ âˆ’ {3} â†’Â RÂ âˆ’ {2} given byÂ f(x) = (x-2)/(x-3)Â is a bijection.

Solution:

Given that f:Â RÂ âˆ’ {3} â†’Â RÂ âˆ’ {2} given by f (x) = (x-2)/(x-3)

Now we have to show that the given function is one-one and on-to

Injectivity:

LetÂ xÂ andÂ yÂ be any two elements in the domain (RÂ âˆ’ {3}), such thatÂ f(x) = f(y).

f(x) = f(y)

â‡’ (x – 2) /(x – 3) = (y – 2) /(y – 3)

â‡’ (x – 2) (y – 3) =Â (y – 2) (x – 3)

â‡’ x y â€“ 3 x â€“ 2 y +Â 6 = x y – 3y – 2x + 6

â‡’ x = y

So,Â fÂ is one-one.

Surjectivity:

LetÂ yÂ be any element in the co-domain (RÂ âˆ’ {2}), such thatÂ f(x) = yÂ for some elementÂ x inÂ RÂ âˆ’ {3}Â (domain).

f(x) = y

â‡’ (x – 2) /(x – 3) = y

â‡’ x – 2 = x y – 3y

â‡’ x y – x = 3y – 2

â‡’ x ( y – 1 ) = 3y – 2

â‡’Â x = (3y – 2)/ (y – 1), whichÂ isÂ inÂ R – {3}

So, for every element in the co-domain, there exists some pre-image in the domain.

â‡’Â fÂ is onto.

Since,Â fÂ is both one-one and onto, it is a bijection.

8. LetÂ AÂ = [-1, 1]. Then, discuss whether the following function fromÂ AÂ to itself is one-one,Â onto or bijective:

(i) f (x) = x/2

(ii) g (x) = |x|

(iii) h (x) = x2

Solution:

(i) Given f: AÂ â†’ A, givenÂ by f (x) = x/2

Now we have to show that the given function is one-one and on-to

Injection test:

LetÂ xÂ andÂ yÂ be any two elements in the domain (A), such thatÂ f(x) =Â f(y).

f(x) =Â f(y)

x/2 = y/2

x = y

So,Â fÂ is one-one.

Surjection test:

LetÂ yÂ be any element in the co-domain (A), such thatÂ f(x) =Â yÂ for some elementÂ xÂ inÂ A (domain)

f(x) =Â y

x/2 = y

xÂ = 2y, which may not be inÂ A.

For example, ifÂ yÂ = 1, then

xÂ = 2, which is not inÂ A.

So,Â fÂ is not onto.

So,Â fÂ is not bijective.

(ii) Given g: AÂ â†’ A, givenÂ by g (x) = |x|

Now we have to show that the given function is one-one and on-to

Injection test:

LetÂ xÂ andÂ yÂ be any two elements in the domain (A), such thatÂ f(x) =Â f(y).

g(x) =Â g(y)

|x| = |y|

xÂ =Â Â± y

So,Â fÂ is not one-one.

Surjection test:

ForÂ yÂ = -1, there is no value ofÂ xÂ inÂ A.

So,Â gÂ is not onto.

So,Â gÂ is not bijective.

(iii) Given h: AÂ â†’ A, givenÂ by h (x) = x2

Now we have to show that the given function is one-one and on-to

Injection test:

LetÂ xÂ andÂ yÂ be any two elements in the domain (A), such thatÂ h(x) =Â h(y).

h(x) =Â h(y)

x2Â =Â y2

xÂ = Â±y

So,Â fÂ is not one-one.

Surjection test:

ForÂ yÂ = – 1, there is no value ofÂ xÂ inÂ A.

So,Â hÂ is not onto.

So,Â hÂ is not bijective.

9. Are the following set of ordered pair ofÂ a function? If so, examine whether the mapping isÂ injective or surjective:

(i) {(x,Â y):Â xÂ is a person,Â yÂ is the mother ofÂ x}

(ii) {(a,Â b):Â aÂ is a person,Â bÂ is an ancestor ofÂ a}Â

Solution:

Let fÂ = {(x,Â y):Â xÂ is a person,Â yÂ is the mother ofÂ x}

As, for each elementÂ xÂ in domain set, there is a unique related elementÂ yÂ in co-domain set.

So,Â fÂ is the function.

Injection test:

As,Â yÂ can be mother of two or more persons

So,Â fÂ is not injective.

Surjection test:

For every motherÂ yÂ defined by (x,Â y), there exists a personÂ xÂ for whomÂ yÂ is mother.

So,Â fÂ is surjective.

Therefore,Â fÂ is surjective function.

(ii) Let gÂ = {(a,Â b):Â aÂ is a person,Â bÂ is an ancestor ofÂ a}

Since, the ordered map (a,Â b) does not map ‘a’ – a person to a living person.

So,Â gÂ is not a function.

10. LetÂ AÂ = {1, 2, 3}. Write all one-one fromÂ AÂ to itself.

Solution:

Given A = {1, 2, 3}

Number of elements inÂ Â AÂ = 3

Number of one-one functions = number of ways of arranging 3 elements = 3! = 6

(i) {(1, 1), (2, 2), (3, 3)}

(ii) {(1, 1), (2, 3), (3, 2)}

(iii) {(1, 2 ), (2, 2), (3, 3 )}

(iv) {(1, 2), (2, 1), (3, 3)}

(v) {(1, 3), (2, 2), (3, 1)}

(vi) {(1, 3), (2, 1), (3,2 )}

11. IfÂ f:Â RÂ â†’Â RÂ be the function defined byÂ f(x) = 4x3Â + 7, show thatÂ fÂ is a bijection.

Â

Solution:

Given f:Â RÂ â†’Â RÂ is a function defined byÂ f(x) = 4x3Â + 7

Injectivity:

LetÂ xÂ andÂ yÂ be any two elements in the domain (R), such thatÂ f(x) = f(y)

â‡’Â 4x3 + 7Â =Â 4y3 +Â 7

â‡’Â 4x3 =Â 4y3

â‡’Â x3 =Â y3

â‡’Â xÂ =Â y

So,Â fÂ is one-one.

Surjectivity:

LetÂ yÂ be any element in the co-domainÂ (R),Â such thatÂ f(x) = yÂ for some elementÂ xÂ inÂ R (domain)

f(x) = y

â‡’Â 4x3 + 7Â =Â y

â‡’Â 4x3 =Â yÂ âˆ’ 7

â‡’ x3 = (y – 7)/4

â‡’ x = âˆ›(y-7)/4 in R

So, for every element in the co-domain, there exists some pre-image in the domain. fÂ is onto.

Since,Â fÂ is both one-to-one and onto, it is a bijection.

### Exercise 2.2 Page No: 2.46

1. FindÂ gofÂ andÂ fogÂ whenÂ f:Â RÂ â†’Â RÂ andÂ gÂ :Â RÂ â†’Â RÂ is defined byÂ

(i) f(x) = 2xÂ + 3 and Â g(x) =Â x2Â + 5.

(ii) f(x) = 2xÂ +Â x2Â andÂ Â g(x) =Â x3

(iii) fÂ (x)Â =Â x2 + 8Â and g(x)Â = 3x3Â + 1

(iv) f (x) =Â xÂ andÂ g(x) = |x|Â

(v) f(x) =Â x2Â + 2xÂ âˆ’ 3 andÂ Â g(x) = 3xÂ âˆ’ 4Â

(vi) f(x) = 8x3Â andÂ Â g(x) =Â x1/3

Solution:

(i) Given,Â f:Â RÂ â†’Â RÂ andÂ g:Â RÂ â†’Â R

So,Â gof:Â RÂ â†’Â RÂ andÂ fog:Â RÂ â†’Â R

Also given that f(x) = 2xÂ + 3 andÂ g(x) =Â x2Â + 5

Now, (gof) (x) = g (f (x))

= gÂ (2xÂ +3)

=Â (2xÂ + 3)2Â + 5

=Â 4x2+ 9 + 12xÂ +5

=4x2+Â 12xÂ + 14

Now, (fog) (x) = f (g (x))

= f (x2Â + 5)

=Â 2Â (x2Â + 5) +3

=Â 2Â x2+ 10 + 3

=Â 2x2Â + 13

(ii) Given,Â f:Â RÂ â†’Â RÂ andÂ g:Â RÂ â†’Â R

so,Â gof:Â RÂ â†’Â RÂ andÂ fog:Â RÂ â†’Â R

f(x) = 2xÂ +Â x2Â andÂ g(x) =Â x3

(gof)Â (x)=Â gÂ (fÂ (x))

=Â gÂ (2x+x2)

=Â (2x+x2)3

Now, (fog)Â (x) =Â fÂ (gÂ (x))

=Â fÂ (x3)

=Â 2Â (x3) + (x3)2

= 2x3 + x6

(iii) Given,Â f:Â RÂ â†’Â RÂ andÂ g:Â RÂ â†’Â R

So,Â gof:Â RÂ â†’Â R andÂ fog:Â RÂ â†’Â R

f(x) =Â x2Â + 8Â  andÂ g(x) = 3x3Â + 1

(gof)Â (x) =Â gÂ (fÂ (x))

=Â gÂ (x2Â +Â 8)

=Â 3Â (x2+8)3Â +Â 1

Now, (fog)Â (x) =Â fÂ (gÂ (x))

=Â fÂ (3x3Â +Â 1)

=Â (3x3+1)2Â +Â 8

=Â 9x6Â +Â 6x3 +Â 1 +Â 8

= 9x6 + 6x3 + 9

(iv) Given,Â f:Â RÂ â†’Â RÂ andÂ g:Â RÂ â†’Â R

So,Â gof:Â RÂ â†’Â RÂ andÂ fog:Â RÂ â†’Â R

f(x) =Â xÂ andÂ g(x) = |x|

(gof)Â (x) =Â gÂ (fÂ (x))

=Â gÂ (x)

=Â |x|

Now (fog)Â (x) =Â fÂ (gÂ (x))

=Â fÂ (|x|)

=Â |x|

(v) Given,Â f:Â RÂ â†’Â RÂ andÂ g:Â RÂ â†’Â R

So,Â gof:Â RÂ â†’Â RÂ andÂ fog:Â RÂ â†’Â R

f(x) =Â x2Â + 2xÂ âˆ’ 3 andÂ g(x) = 3xÂ âˆ’ 4

(gof)Â (x) =Â gÂ (f(x))

=Â gÂ (x2 + 2x âˆ’ 3)

=Â 3Â (x2 + 2x âˆ’ 3)Â âˆ’ 4

=Â 3x2 +Â 6xÂ âˆ’Â 9Â âˆ’Â 4

=Â 3x2 + 6x âˆ’ 13

Now, (fog)Â (x) =Â fÂ (gÂ (x))

=Â fÂ (3x âˆ’ 4)

=Â (3xÂ âˆ’Â 4)2 + 2Â (3xÂ âˆ’Â 4)Â âˆ’3

=Â 9x2 + 16 âˆ’ 24x + 6x â€“ 8 âˆ’ 3

=Â 9x2 âˆ’ 18xÂ +Â 5

(vi) Given,Â f:Â RÂ â†’Â RÂ andÂ g:Â RÂ â†’Â R

So,Â gof:Â RÂ â†’Â RÂ andÂ fog:Â RÂ â†’Â R

f(x) = 8x3Â andÂ g(x) =Â x1/3

(gof)Â (x) =Â gÂ (fÂ (x))

=Â gÂ (8x3)

= (8x3)1/3

= [(2x)3]1/3

=Â 2x

Now, (fog)Â (x) =Â fÂ (gÂ (x))

= f (x1/3)

= 8 (x1/3)3

=Â 8x

2. LetÂ fÂ = {(3, 1), (9, 3), (12, 4)} andÂ gÂ = {(1, 3), (3, 3) (4, 9) (5, 9)}. Show thatÂ gofÂ andÂ fog are both defined. Also, findÂ fogÂ andÂ gof.

Solution:

Given f = {(3, 1), (9, 3), (12, 4)} andÂ gÂ = {(1, 3), (3, 3) (4, 9) (5, 9)}

fÂ : {3, 9, 12} â†’ {1, 3, 4} andÂ gÂ : {1, 3, 4, 5} â†’ {3, 9}

Co-domain ofÂ fÂ is a subset of the domain ofÂ g.

So,Â gofÂ exists andÂ gof: {3, 9, 12} â†’ {3, 9}

(gof)Â (3) = gÂ (fÂ (3)) = gÂ (1)Â = 3

(gof)Â (9) = gÂ (fÂ (9)) = gÂ (3) = 3

(gof)Â (12) = gÂ (fÂ (12)) = gÂ (4) = 9

â‡’Â gofÂ = {(3,Â 3),Â (9,Â 3),Â (12,Â 9)}

Co-domain ofÂ gÂ is a subset of the domain ofÂ f.

So,Â fogÂ exists andÂ fog: {1, 3, 4, 5} â†’ {3, 9, 12}

(fog)Â (1) = fÂ (gÂ (1)) = fÂ (3) = 1

(fog)Â (3) = fÂ (gÂ (3)) = fÂ (3) = 1

(fog)Â (4) = fÂ (gÂ (4)) = fÂ (9) = 3

(fog)Â (5) = fÂ (gÂ (5)) = fÂ (9) = 3

â‡’Â fog = {(1,Â 1),Â (3,Â 1),Â (4,Â 3),Â (5,Â 3)}

3. Â LetÂ fÂ = {(1, âˆ’1), (4, âˆ’2), (9, âˆ’3), (16, 4)} andÂ gÂ = {(âˆ’1, âˆ’2), (âˆ’2, âˆ’4), (âˆ’3, âˆ’6), (4, 8)}. Show thatÂ gofÂ is defined whileÂ fogÂ is not defined. Also, findÂ gof.

Solution:

Given fÂ = {(1, âˆ’1), (4, âˆ’2), (9, âˆ’3), (16, 4)} andÂ gÂ = {(âˆ’1, âˆ’2), (âˆ’2, âˆ’4), (âˆ’3, âˆ’6), (4, 8)}

f: {1, 4, 9, 16} â†’ {-1, -2, -3, 4} andÂ g: {-1, -2, -3, 4} â†’ {-2, -4, -6, 8}

Co-domain ofÂ fÂ = domain ofÂ g

So,Â gofÂ exists andÂ gof: {1, 4, 9, 16} â†’ {-2, -4, -6, 8}

(gof)Â (1)Â =Â gÂ (fÂ (1))Â =Â gÂ (âˆ’1)Â =Â âˆ’2

(gof)Â (4)Â =Â gÂ (fÂ (4)) = gÂ (âˆ’2)Â =Â âˆ’4

(gof)Â (9)Â =Â gÂ (fÂ (9))Â =Â gÂ (âˆ’3)Â =Â âˆ’6

(gof)Â (16)Â = gÂ (fÂ (16))Â = gÂ (4)Â =Â 8

So,Â gofÂ =Â {(1,Â âˆ’2),Â (4,Â âˆ’4),Â (9,Â âˆ’6),Â (16,Â 8)}

But the co-domain ofÂ gÂ is not same as the domain ofÂ f.

So,Â fogÂ does not exist.

4. LetÂ AÂ = {a,Â b,Â c},Â BÂ = {u,Â v,Â w} and letÂ fÂ andÂ gÂ be two functions fromÂ AÂ toÂ BÂ and fromÂ BÂ toÂ A,Â respectively, defined as: fÂ = {(a,Â v), (b,Â u), (c,Â w)},Â gÂ = {(u,Â b), (v,Â a), (w,Â c)}.
Show thatÂ fÂ andÂ gÂ both are bijections and findÂ fogÂ andÂ gof.

Â

Solution:

Given fÂ = {(a,Â v), (b,Â u), (c,Â w)},Â gÂ = {(u,Â b), (v,Â a), (w,Â c)}.

Also given that AÂ = {a,Â b,Â c},Â BÂ = {u,Â v,Â w}

Now we have to show f and g both are bijective.

Consider fÂ = {(a,Â v), (b,Â u), (c,Â w)} andÂ f:Â A â†’ B

Injectivity ofÂ f: No two elements ofÂ AÂ have the same image in B.

So,Â fÂ is one-one.

Surjectivity ofÂ f: Co-domain ofÂ fÂ = {u,Â v,Â w}

Range ofÂ fÂ = {u,Â v,Â w}

Both are same.

So,Â fÂ is onto.

Hence,Â fÂ is a bijection.

Now consider gÂ = {(u,Â b), (v,Â a), (w,Â c)} andÂ g:Â B â†’ A

Injectivity ofÂ g: No two elements ofÂ BÂ Â have the same image in A.

So,Â gÂ is one-one.

Surjectivity of g: Co-domain ofÂ gÂ = {a,Â b,Â c}

Range ofÂ gÂ = {a,Â b,Â c}

Both are the same.

So,Â gÂ is onto.

Hence,Â gÂ is a bijection.

Now we have to find fog,

we know that Co-domain ofÂ gÂ is same as the domain ofÂ f.

So,Â fogÂ exists andÂ fog: {uÂ v,Â w}Â â†’Â {u,Â v,Â w}

(fog)Â (u)Â =Â fÂ (gÂ (u))Â =Â fÂ (b)Â =Â u

(fog)Â (v)Â =Â fÂ (gÂ (v))Â =Â fÂ (a)Â =Â v

(fog)Â (w)Â =Â fÂ (gÂ (w))Â =Â fÂ (c)Â =Â w

So,Â fogÂ =Â {(u,Â u),Â (v,Â v),Â (w,Â w)}

Now we have to find gof,

Co-domain ofÂ fÂ is same as the domain ofÂ g.

So,Â fogÂ exists andÂ gof: {a,Â b,Â c}Â â†’Â {a,Â b,Â c}

(gof)Â (a)Â =Â gÂ (fÂ (a))Â =Â gÂ (v)Â =Â a

(gof)Â (b)Â =Â gÂ (fÂ (b))Â =Â gÂ (u)Â =Â b

(gof)Â (c)Â =Â gÂ (fÂ (c))Â =Â gÂ (w)Â =Â c

So,Â gofÂ =Â {(a,Â a),Â (b,Â b),Â (c,Â c)}

5. FindÂ fogÂ (2) andÂ gofÂ (1) when f:Â RÂ â†’Â R;Â f(x) =Â x2Â + 8 andÂ g:Â RÂ â†’Â R;Â g(x) = 3x3Â + 1.

Solution:

Given f:Â RÂ â†’Â R;Â f(x) =Â x2Â + 8 andÂ g:Â RÂ â†’Â R;Â g(x) = 3x3Â + 1.

Consider (fog)Â (2)Â =Â fÂ (gÂ (2))

=Â fÂ (3 Ã— 23 + 1)

=Â f(3 Ã— 8 + 1)

= f (25)

= 252 + 8

=Â 633

(gof)Â (1)Â =Â gÂ (fÂ (1))

=Â gÂ (12 + 8)

=Â gÂ (9)

=Â 3 Ã— 93 + 1

=Â 2188

6. LetÂ R+Â be the set of all non-negative real numbers. IfÂ f:Â R+Â â†’Â R+Â andÂ gÂ :Â R+Â â†’Â R+Â are defined as f(x)=x2 and g(x)=+ âˆšx, find fogÂ andÂ gof. Are they equal functions.

Solution:

Given f:Â R+Â â†’Â R+Â andÂ g:Â R+Â â†’Â R+

So,Â fog:Â R+Â â†’Â R+Â andÂ gof:Â R+Â â†’Â R+

Domains ofÂ fogÂ andÂ gofÂ are the same.

Now we have to find fog and gof also we have to check whether they are equal or not,

Consider (fog)Â (x)Â =Â fÂ (g (x))

= f (âˆšx)

= âˆšx2

= x

Now consider (gof)Â (x)Â = g (f (x))

= g (x2)

= âˆšx2

= x

So,Â (fog)Â (x)Â =Â (gof)Â (x),Â âˆ€xÂ âˆˆ R+

Hence,Â fogÂ =Â gof

7. LetÂ f:Â RÂ â†’Â RÂ andÂ g:Â RÂ â†’Â RÂ be defined byÂ f(x) =Â x2Â andÂ g(x) =Â xÂ + 1. Show thatÂ fogÂ â‰ Â gof.

Solution:

Given f:Â RÂ â†’Â RÂ andÂ g:Â RÂ â†’ R.

So, the domains ofÂ fÂ andÂ gÂ are the same.

Consider (fog)Â (x)Â =Â fÂ (gÂ (x))

=Â fÂ (x + 1) =Â (x + 1)2

=Â x2 + 1 + 2x

Again consider (gof)Â (x)Â =Â gÂ (fÂ (x))

=Â gÂ (x2) = x2 + 1

So,Â fog â‰  gof

### Exercise 2.3 Page No: 2.54

1. FindÂ fogÂ andÂ gof,Â if Â

(i) Â fÂ (x)Â =Â ex,Â g (x)Â =Â loge x

(ii) fÂ (x)Â =Â x2,Â g (x)Â =Â cosÂ x

(iii) fÂ (x)Â =Â |x|,Â gÂ (x)Â =Â sinÂ x

(iv) f (x) = x+1, g(x) = ex

(v) f (x)Â =Â sinâˆ’1Â x,Â g(x)Â =Â x2

(vi) fÂ (x)Â =Â x+1,Â gÂ (x)Â =Â sinÂ x

(vii) f(x)=Â xÂ +Â 1,Â gÂ (x)Â =Â 2xÂ +Â 3

(viii) f(x) = c, c âˆˆ R, g(x) = sin x2

(ix) f(x) = x2 + 2 , g (x) = 1 âˆ’ 1/ (1-x)

Solution:

(i) Given fÂ (x) = ex,Â g(x)Â =Â logeÂ x

Let f:Â RÂ â†’Â (0,Â âˆž);Â and g:Â (0,Â âˆž)Â â†’Â R

Now we have to calculate fog,

Clearly,Â theÂ rangeÂ ofÂ gÂ isÂ aÂ subsetÂ ofÂ theÂ domainÂ ofÂ f.

fog:Â (Â 0,Â âˆž)Â â†’Â R

(fog)Â (x)Â =Â fÂ (gÂ (x))

=Â fÂ (logeÂ x)

=Â logeÂ ex

=Â x

Now we have to calculate gof,

Clearly,Â theÂ rangeÂ ofÂ fÂ isÂ aÂ subsetÂ ofÂ theÂ domainÂ ofÂ g.

â‡’Â fog:Â Râ†’Â R

(gof)Â (x)Â =Â gÂ (fÂ (x))

=Â gÂ (ex)

=Â logeÂ ex

=Â x

(ii) fÂ (x)Â =Â x2,Â g(x)Â =Â cosÂ x

f:Â Râ†’Â [0,Â âˆž)Â ;Â g:Â Râ†’[âˆ’1,Â 1]

Now we have to calculate fog,

Clearly,Â theÂ rangeÂ ofÂ gÂ isÂ notÂ aÂ subsetÂ ofÂ theÂ domainÂ ofÂ f.

â‡’Â DomainÂ (fog)Â =Â {x:Â x âˆˆ domainÂ ofÂ gÂ andÂ gÂ (x)Â âˆˆ domainÂ ofÂ f}

â‡’Â DomainÂ (fog)Â =Â x:Â xÂ âˆˆÂ RÂ andÂ cosÂ xÂ âˆˆÂ R}

â‡’Â DomainÂ ofÂ (fog)Â =Â R

(fog):Â Râ†’Â R

(fog)Â (x)Â =Â fÂ (gÂ (x))

=Â fÂ (cosÂ x)

=Â cos2 x

Now we have to calculate gof,

Clearly,Â theÂ rangeÂ ofÂ fÂ isÂ aÂ subsetÂ ofÂ theÂ domainÂ ofÂ g.

â‡’Â fog:Â Râ†’R

(gof)Â (x)Â =Â gÂ (fÂ (x))

=Â gÂ (x2)

=Â cosÂ x2

(iii) Given fÂ (x)Â =Â |x|,Â g(x)Â =Â sinÂ x

f:Â RÂ â†’Â (0,Â âˆž)Â ;Â gÂ :Â Râ†’[âˆ’1,Â 1]

Now we have to calculate fog,

Clearly,Â theÂ rangeÂ ofÂ gÂ isÂ aÂ subsetÂ ofÂ theÂ domainÂ ofÂ f.

â‡’Â fog:Â Râ†’R

(fog)Â (x)Â =Â fÂ (gÂ (x))

=Â fÂ (sinÂ x)

=Â |sinÂ x|

Now we have to calculate gof,

Clearly,Â theÂ rangeÂ ofÂ fÂ isÂ aÂ subsetÂ ofÂ theÂ domainÂ ofÂ g.

â‡’Â fogÂ :Â Râ†’Â R

(gof)Â (x)Â =Â gÂ (fÂ (x))

=Â gÂ (|x|)

=Â sinÂ |x|

(iv) Given fÂ (x)Â =Â x + 1,Â g(x)Â =Â ex

f:Â Râ†’RÂ ;Â g:Â RÂ â†’Â [Â 1,Â âˆž)

Now we have calculate fog:

Clearly,Â rangeÂ ofÂ gÂ isÂ aÂ subsetÂ ofÂ domainÂ ofÂ f.

â‡’Â fog:Â Râ†’R

(fog)Â (x)Â =Â fÂ (gÂ (x))

=Â fÂ (ex)

=Â ex + 1

Now we have to compute gof,

Clearly,Â rangeÂ ofÂ fÂ isÂ aÂ subsetÂ ofÂ domainÂ ofÂ g.

â‡’Â fog:Â Râ†’R

(gof)Â (x)Â =Â gÂ (fÂ (x))

=Â gÂ (x+1)

=Â ex+1

(v) Given fÂ (x)Â =Â sinÂ âˆ’1Â x,Â g(x)Â =Â x2

f:Â [âˆ’1,1]â†’ [(-Ï€)/2 ,Ï€/2];Â gÂ :Â RÂ â†’Â [0,Â âˆž)

Now we have to computeÂ fog:

Clearly,Â theÂ rangeÂ ofÂ gÂ isÂ notÂ aÂ subsetÂ ofÂ theÂ domainÂ ofÂ f.

DomainÂ (fog)Â =Â {x:Â xÂ âˆˆÂ domainÂ ofÂ gÂ andÂ gÂ (x)Â âˆˆÂ domainÂ ofÂ f}

DomainÂ (fog) = {x:Â xÂ âˆˆÂ RÂ andÂ x2Â âˆˆÂ [âˆ’1, 1]}

DomainÂ (fog) = {x:Â xÂ âˆˆÂ RÂ andÂ xÂ âˆˆÂ [âˆ’1, 1]}

DomainÂ ofÂ (fog) =Â [âˆ’1, 1]

fog:Â [âˆ’1,1]Â â†’Â R

(fog)Â (x)Â =Â fÂ (gÂ (x))

=Â fÂ (x2)

=Â sinâˆ’1Â (x2)

Now we have to computeÂ gof:

Clearly,Â theÂ rangeÂ ofÂ fÂ isÂ aÂ subsetÂ ofÂ theÂ domainÂ ofÂ g.

fog:Â [âˆ’1, 1]Â â†’Â R

(gof)Â (x)Â =Â gÂ (fÂ (x))

=Â gÂ (sinâˆ’1Â x)

=Â (sinâˆ’1Â x)2

(vi) Given f(x)Â =Â x+1,Â g(x)Â =Â sinÂ x

f:Â Râ†’RÂ ;Â g:Â Râ†’[âˆ’1,Â 1]

Now we have to computeÂ fog

Clearly,Â theÂ rangeÂ ofÂ gÂ isÂ aÂ subsetÂ ofÂ theÂ domainÂ ofÂ f.

SetÂ ofÂ theÂ domainÂ ofÂ f.

â‡’Â fog:Â Râ†’Â R

(fog)Â (x)Â =Â fÂ (gÂ (x))

=Â fÂ (sinÂ x)

=Â sinÂ xÂ +Â 1

Now we have to compute gof,

Clearly,Â theÂ rangeÂ ofÂ fÂ isÂ aÂ subsetÂ ofÂ theÂ domainÂ ofÂ g.

â‡’Â fog:Â RÂ â†’Â R

(gof)Â (x)Â =Â gÂ (fÂ (x))

=Â gÂ (x+1)

=Â sinÂ (x+1)

(vii) Given fÂ (x)Â =Â x+1,Â gÂ (x)Â =Â 2xÂ +Â 3

f:Â Râ†’RÂ ;Â g:Â RÂ â†’Â R

Now we have to computeÂ fog

Clearly,Â theÂ rangeÂ ofÂ gÂ isÂ aÂ subsetÂ ofÂ theÂ domainÂ ofÂ f.

â‡’Â fog:Â Râ†’Â R

(fog)Â (x)Â =Â fÂ (gÂ (x))

=Â fÂ (2x+3)

=Â 2xÂ +Â 3Â +Â 1

=Â 2xÂ +Â 4

Now we have to computeÂ gof

Clearly,Â theÂ rangeÂ ofÂ fÂ isÂ aÂ subsetÂ ofÂ theÂ domainÂ ofÂ g.

â‡’Â fog:Â RÂ â†’Â R

(gof)Â (x)Â =Â gÂ (fÂ (x))

=Â gÂ (x+1)

=Â 2Â (xÂ +Â 1)Â +Â 3

=Â 2xÂ +Â 5

(viii) Given fÂ (x)Â =Â c,Â gÂ (x)Â =Â sinÂ x2

f:Â RÂ â†’Â {c}Â ;Â g:Â Râ†’Â [Â 0,Â 1Â ]

Now we have to computeÂ fog

Clearly,Â theÂ rangeÂ ofÂ gÂ isÂ aÂ subsetÂ ofÂ theÂ domainÂ ofÂ f.

fog:Â Râ†’R

(fog)Â (x)Â =Â fÂ (gÂ (x))

=Â fÂ (sinÂ x2)

=Â c

Now we have to compute gof,

Clearly,Â theÂ rangeÂ ofÂ fÂ isÂ aÂ subsetÂ ofÂ theÂ domainÂ ofÂ g.

â‡’Â fog:Â Râ†’Â R

(gof)Â (x)Â =Â gÂ (fÂ (x))

=Â gÂ (c)

=Â sinÂ c2

(ix) Given fÂ (x)Â =Â x2+Â 2 and gÂ (x)Â =Â 1 – 1 / (1 – x)

f:Â RÂ â†’Â [Â 2,Â âˆžÂ )

ForÂ domainÂ ofÂ g:Â 1âˆ’Â xÂ â‰ Â 0

â‡’Â xÂ â‰ Â 1

â‡’Â DomainÂ ofÂ g =Â R âˆ’ {1}

gÂ (xÂ )= 1 â€“ [1/(1 – x)] = (1 â€“ x – 1)/ (1 – x) = -x/(1 – x)

ForÂ rangeÂ ofÂ g

y = (- x)/ (1 – x)

â‡’Â yÂ â€“Â x yÂ =Â âˆ’Â x

â‡’Â yÂ =Â x yÂ âˆ’Â x

â‡’Â yÂ =Â xÂ (yâˆ’1)

â‡’ x = y/(y – 1)

RangeÂ ofÂ gÂ = R âˆ’ {1}

So,Â g:Â R âˆ’ {1} â†’ R âˆ’ {1}

Now we have to computeÂ fog

Clearly,Â theÂ rangeÂ ofÂ gÂ isÂ aÂ subsetÂ ofÂ theÂ domainÂ ofÂ f.

â‡’Â fog:Â RÂ âˆ’Â {1} â†’Â R

(fog)Â (x)Â =Â fÂ (gÂ (x))

= f (-x/ (1 – x))

= ((-x)/ (1 – x))2 + 2

= (x2 + 2x2 + 2 – 4x) / (1 – x)2

= (3x2 – 4x + 2)/ (1 – x)2

Now we have to computeÂ gof

Clearly,Â theÂ rangeÂ ofÂ fÂ isÂ aÂ subsetÂ ofÂ theÂ domainÂ ofÂ g.

â‡’Â gof:Â Râ†’R

(gof)Â (x)Â =Â gÂ (fÂ (x))

=Â gÂ (x2Â +Â 2)

= 1 – 1 / (1 – (x2 + 2))

= – 1/ (1 – (x2 + 2))

= (x2 + 2)/ (x2 + 1)

2. Â LetÂ f(x) =Â x2Â +Â xÂ + 1 andÂ g(x) = sinÂ x. Show thatÂ fogÂ â‰ Â gof.

Solution:

Given f(x) =Â x2Â +Â xÂ + 1 andÂ g(x) = sinÂ x

Now we have to prove fogÂ â‰ Â gof

(fog)Â (x)Â =Â fÂ (gÂ (x))

=Â f (sinÂ x)

=Â sin2Â xÂ +Â sinÂ xÂ +Â 1

AndÂ (gof)Â (x)Â =Â gÂ (fÂ (x))

=Â gÂ (x2+Â xÂ +Â 1)

=Â sinÂ (x2+Â xÂ +Â 1)

So,Â fogÂ â‰ Â gof.

3. IfÂ f(x) = |x|, prove thatÂ fofÂ =Â f.

Solution:

Given f(x) = |x|,

Now we have to prove thatÂ fofÂ =Â f.

Consider (fof)Â (x)Â =Â fÂ (fÂ (x))

=Â fÂ (|x|)

=Â ||x||

=Â |x|

=Â fÂ (x)

So,

(fof)Â (x)Â =Â fÂ (x),Â âˆ€xÂ âˆˆÂ R

Hence,Â fofÂ =Â f

4. IfÂ f(x) = 2xÂ + 5 andÂ g(x) =Â x2Â + 1 be two real functions, then describe each of the following functions:
(i) fog
(ii)Â gof
(iii)Â fof
(iv)Â f2
Also, show thatÂ fofÂ â‰ Â f2

Solution:

f(x)Â andÂ g(x)Â are polynomials.

â‡’ f:Â RÂ â†’Â RÂ andÂ g:Â RÂ â†’Â R.

So,Â fog:Â RÂ â†’Â RÂ andÂ gof:Â RÂ â†’Â R.

(i) (fog) (x) = f (g (x))

= f (x2Â + 1)

= 2 (x2 + 1) + 5

=2x2Â + 2 + 5

= 2x2Â +7

(ii) (gof) (x) = g (f (x))

= g (2x +5)

= (2x + 5)2Â + 1

= 4x2Â + 20x + 26

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

= f (2x +5)

= 2 (2x + 5) + 5

= 4x + 10 + 5

= 4x + 15

(iv) f2Â (x) = f (x) x f (x)

= (2x + 5) (2x + 5)

= (2x + 5)2

= 4x2Â + 20x +25

Hence, from (iii) and (iv) clearly fofÂ â‰ Â f2

5. IfÂ f(x) = sinÂ xÂ andÂ g(x) = 2x be two real functions, then describeÂ gofÂ andÂ fog.Â Are these equal functions?

Solution:

Given f(x) = sinÂ xÂ andÂ g(x) = 2x

WeÂ knowÂ that

f:Â Râ†’Â [âˆ’1,Â 1]Â andÂ g:Â Râ†’Â R

Clearly,Â theÂ rangeÂ ofÂ fÂ isÂ aÂ subsetÂ ofÂ theÂ domainÂ ofÂ g.

gof:Â Râ†’Â R

(gof)Â (x)Â =Â gÂ (fÂ (x))

=Â gÂ (sinÂ x)

=Â 2Â sinÂ x

Clearly,Â theÂ rangeÂ ofÂ gÂ isÂ aÂ subsetÂ ofÂ theÂ domainÂ ofÂ f.

fog:Â RÂ â†’Â R

So,Â (fog)Â (x)Â =Â fÂ (gÂ (x))

=Â fÂ (2x)

=Â sinÂ (2x)

Clearly,Â fog â‰  gof

Hence they are not equal functions.

6. LetÂ f,Â g,Â hÂ be real functions given byÂ f(x) = sinÂ x,Â gÂ (x) = 2xÂ andÂ hÂ (x) = cosÂ x. Prove thatÂ fogÂ =Â goÂ (f h).

Solution:

Given that f(x) = sinÂ x,Â gÂ (x) = 2xÂ andÂ hÂ (x) = cosÂ x

WeÂ knowÂ thatÂ f:Â Râ†’ [âˆ’1,Â 1]Â andÂ g:Â Râ†’ R

Clearly,Â theÂ rangeÂ ofÂ gÂ isÂ aÂ subsetÂ ofÂ theÂ domainÂ ofÂ f.

fog:Â RÂ â†’Â R

Now,Â (f h)Â (x) = fÂ (x) hÂ (x)Â =Â (sinÂ x)Â (cosÂ x)Â = Â½ sinÂ (2x)

DomainÂ ofÂ f hÂ isÂ R.

SinceÂ rangeÂ ofÂ sinÂ xÂ isÂ [-1, 1], âˆ’1Â â‰¤Â sinÂ 2xÂ â‰¤Â 1

â‡’ -1/2 â‰¤ sin x/2 â‰¤ 1/2

RangeÂ ofÂ f hÂ = [-1/2, 1/2]

So,Â (f h):Â RÂ â†’ [(-1)/2, 1/2]

Clearly,Â rangeÂ ofÂ f hÂ isÂ aÂ subsetÂ ofÂ g.

â‡’Â goÂ (f h):Â RÂ â†’Â R

â‡’ DomainsÂ ofÂ fogÂ andÂ goÂ (f h)Â areÂ theÂ same.

So,Â (fog)Â (x) = fÂ (gÂ (x))

=Â fÂ (2x)

=Â sinÂ (2x)

AndÂ (goÂ (f h))Â (x)Â =Â gÂ ((f(x). h(x))

=Â gÂ (sin xÂ cosÂ x)

=Â 2sinÂ xÂ cosÂ x

=Â sinÂ (2x)

â‡’Â (fog)Â (x)Â =Â (go (f h))Â (x),Â âˆ€xÂ âˆˆÂ R

Hence,Â fogÂ =Â goÂ (f h)

### 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

### Also, access RD Sharma Solutions for Class 12 Maths Chapter 2 Functions

Exercise 2.1 Solutions

Exercise 2.2 Solutions

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