# RD Sharma Solutions For Class 12 Maths Exercise 2.1 Chapter 2 Function

The PDF of RD Sharma Solutions for Class 12 Maths Chapter 2 Function Exercise 2.1 are provided here. Students who are preparing for their exams and who wish to present their answers without any mistakes can refer to RD Sharma Solutions for Class 12. On regular practice, students can speed up the method of solving problems by using shortcut tips to secure high marks in their examination. Students can download the pdf from the links provided below. This exercise deals with basic concepts related to functions. Some of the important concepts of this exercise are listed below.

• 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
• Onto function
• Many one function
• In to function
• Bijection

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Exercise 2.2 Solutions

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### 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.