Rd sharma solution class 12 chapter 2 functions

Exercise 3.1

Question 1

Give an example of a function
(i) which is one-one but not onto

Sol :

f:Z\rightarrow Z\text{ given by }f(x)=3x+2

Injectivity :

Let x and be any two elements in the domain (Z) , such that 

f(x)=f(y)

\Rightarrow 3x+2=3y+2

\Rightarrow 3x=3y

\Rightarrow x=y

So , f is one-one

 

Surjectivity :

Let be any element in the co-domain (Z) , such that f(x)=y for some element in Z (domain) .

f(x)=y

\Rightarrow 3x+2=y

\Rightarrow 3x=y-2

\Rightarrow x=\dfrac{y-2}{3} it may not be in the domain (Z) because if we take y=3 ,

x=\dfrac{y-2}{3}=\dfrac{3-2}{3} =\dfrac{1}{3}\not\in\text{domain (Z)}

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

Thus , f is not onto .

 

(ii) which is not one-one but onto

f:Z\rightarrow N\cup \left\{0\right\}\text{ given by }f(x)=|x|

Injectivity :

Let x and be any two elements in the domain (Z) , such that 

f(x)=f(y)

\Rightarrow |x|=|y|

\Rightarrow x=\pm y

So ,different element of domain f may give the same image .

So , f is not one-one .

 

Surjectivity :

Let be any element in the co-domain (Z) , such that f(x)=y for some element in Z (domain) .

f(x)=y

\Rightarrow |x|=y

\Rightarrow x=\pm y , which is an element in Z (domain) .

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

Thus , f is onto .

(iii) which is neither one-one nor onto

f:Z\rightarrow Z\text{ given by }f(x)=2x^2+1

Injectivity :

Let x and be any two elements in the domain (Z) , such that 

f(x)=f(y)

\Rightarrow 2x^2+1=2y^2+1

\Rightarrow 2x^2=2y^2

\Rightarrow x^2=y^2

\Rightarrow x=\pm y

So , different elements of domain f may give the same image .

So , f is not one-one

 

Surjectivity :

Let be any element in the co-domain (Z) , such that f(x)=y for some element in Z (domain) .

f(x)=y

\Rightarrow 2x^2+1=y

\Rightarrow 2x^2=y-1

\Rightarrow x^2=\dfrac{y-1}{2}

\Rightarrow x=\pm \sqrt{\dfrac{y-1}{2}}\notin always .

For example if we take , y=4

x=\pm \sqrt{\dfrac{y-1}{2}} =\pm \sqrt{\dfrac{4-1}{2}} =\pm \sqrt{\dfrac{3}{2}}\notin Z

So , may not be in Z (domain)

Thus , f is not onto .

 

Question 2

Which of the following functions from to are one-one and onto ?

(i) f_1=\left\{(1,3),(2,5),(3,7)\right\} ; A=\left\{1,2,3\right\} ; B=\left\{3,5,7\right\}

Sol :

Injectivity :

f_1(1)=3

f_1(2)=5

f_1(3)=7

\Rightarrow Every element of has different images in .

So , f_1 is one-one .

 

Surjectivity :

Co-domain of f_1=\left\{3,5,7\right\}

Range of f_1 = set of images =\left\{3,5,7\right\}

\Rightarrow co-domain = range

So , f_1 is onto .

 

 

(ii) f_2=\left\{(2,a),(3,b),(4,c)\right\} ; A=\left\{2,3,4\right\} ; B=\left\{a,b,c\right\}

Sol :

Injectivity :

f_2(2)=a

f_2(3)=b

f_2(4)=c

\Rightarrow Every element of has different images in .

So , f_2 is one-one .

 

Surjectivity :

Co-domain of f_2=\left\{a,b,c\right\}

Range of f_2 = set of images =\left\{a,b,c\right\}

\Rightarrow co-domain = range

So , f_2 is onto .

 

 

(iii) f_3=\left\{(a,x),(b,x),(c,z),(d,z)\right\} ; A=\left\{a,b,c,d\right\} ; B=\left\{x,y,z\right\}

Sol :

Injectivity :

f_3(a)=x

f_3(b)=x

f_3(c)=z

f_3(d)=z

\Rightarrow and have same image . Also , and have same image z .

So , f_3 is not one-one .

 

Surjectivity :

Co-domain of f_3=\left\{x,y,z\right\}

Range of f_3 = set of images =\left\{x,z\right\}

So , the co-domain is not equals to range .

So , f_3 is  not onto .

 

 

Question 3

Prove that the function f:N\rightarrow N ,\text{ defined by }f(x)=x^2+x+1 , is one-one but not onto .

Sol :

f:N\rightarrow N ,\text{ defined by }f(x)=x^2+x+1

Injectivity :

Let and be any two elements in the domain (N) , such that 

f(x)=f(y)

\Rightarrow x^2+x+1=y^2+y+1

\Rightarrow x^2+x=y^2+y

\Rightarrow (x^2-y^2)+(x-y)=0

\Rightarrow (x-y)(x+y)+(x-y)=0

\Rightarrow (x-y)[x+y+1]=0 taking common (x – y)

\Rightarrow x-y=0 [(x+y+1) can not be zero because x and y are natural numbers]

\Rightarrow x=y

So , f is one-one

 

Surjectivity :

The minimum number in is 1 .

When = 1 ,

\Rightarrow x^2+x+1=1+1+1=3

\Rightarrow x^2+x+1\geq 3 , for every x in N

\Rightarrow f(x) will not assume the values 1 and 2 

So , f is  not onto

 

 

Question 4

Let A=\left\{-1,0,1\right\} and f=\left\{(x,x^2):x\in A\right\} . Show that f:A\rightarrow A is neither one-one nor onto .

Sol :

Given , A=\left\{-1,0,1\right\} , f=\left\{(x,x^2):x\in A\right\} and f(x)=x^2

Injectivity :

f(x)=x^2

f(1)=1^2=1

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

\Rightarrow 1 and -1 have same images .

So , f is not one-one

 

Surjectivity :

Co-domain of f=\left\{-1,0,1\right\}

f(1)=1^2=1

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

f(0)=0

\Rightarrow Range of f=\left\{0,1\right\}

So , co-domain is not same as range

Hence , f is not onto

 

 

Question 5

Classify the following functions as injection, surjection or bijection :

(i) f:N\rightarrow N \text{ given by } f(x)=x^2

Sol :

Injection :

Let x and y be any elements in the doamin (N) such that 

f(x)=f(y)

x^2=y^2

x = y we do not get \pm because x and y are natural numbers

So , f is an injection

 

Surjection :

Let y be any element in co-domain (N) such that f(x)=y for some element x in (N) domain 

f(x)=y

x^2=y

x=\sqrt{y} , which may not be in N

For example , if y=3 ,

x=\sqrt{3} is not in N

So , f is not a surjection

So , f is not a bijection

 

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