# Relations and Functions

## NCERT Solution

### Exercise 3

Question 1: Which of the following relations are functions? Give reasons. If it is a function determine its domain and range.

(i) {(2,1), (5,1), (8,1), (11, 1), (14,1), (17,1)}

Solution: Because to each of first element of the ordered pair and there corresponds exactly one second element, hence it is a function.

Domain = {2, 5, 8, 11, 14, 17}

Range = {1}

(ii) {(2,1), (4,2), (6,3), (8,4), (10,5), (12,6), (14,7)}

Solution: Because to each of first element of the ordered pair and there corresponds exactly one second element, hence it is a function.

Domain = {2, 4, 6, 8, 10, 12, 14}

Range = {1}

(iii) {(1,3), (1,5), (2,5)}

Solution: As the two ordered pairs (1, 3) and (1, 5) have the same first component, hence it is not a function.

Question 2: Find the domain and range of the following functions:

(i) f (x) = - |x|

Solution: Given, f (x) = - |x|

Therefore, f (x)= - |x| is ≤ 0, for all x ∈ R

Therefore, Domain = R

As the range of f(x) = - |x| is all real numbers except positive real numbers.

Therefore, rage of f is {- ∞, 0}

(ii) f(x)=sqrt(9-x^2)

Solution: Given,

f(x)=sqrt(9-x^2)

Here, f is not defined

When (9-x^2)<0

Therefore,

x^2>9

Or, x>3 or x<-3

Also for each number x is lying between – 3 and 3

As f(x)=sqrt(9-x^2) is unique

Hence, Domain ={x&isn;R: - 3 ≤ x ≤ 3} or [ - 3, 3]

For any value of x such that - 3 ≤ x ≤ 3 the value of f(x) will lie between 0 and 3

Therefore, the range of f (x) is { x: - 0 ≤ x ≤ 3} or [ 0, 3]

Question 3: A function f is defined by f (x) = 2x – 5. Write down the values of

(i) f ( 0 ), (ii) f (7), (iii) f ( - 3)

Solution: (i) Here f is defined by f (x) = 2x – 5

Therefore, f(0)= 2 × 0 -5= -5

f(7)= 2 xx 7 -5=9

f( - 3)=2 xx ( - 3) - 5= - 6 - 5= - 11

Question 4: The function ‘t’ which maps temperature in degree Celsius into temperature in degree Fahrenheit is defiend by t(C)=(9C)/(5)+32

Find (i) t(0) (ii) t(28) (iii) t(-10), (iv) The value of C, when t(C) = 212

Solution: Given, t(C)=(9C)/(5)+32

(i) t(o)=(9xx0)/(5)+32=32

(ii) t(28)=(9xx28)/(5)+32

=(252)/(2)+32

=(252+160)/5

=(412)/(5)=82.4

(iii) t(-10)=(9xx(-10))/(5)+32

=-(90)/(5)+32

=-18+32=14

(iv) Given t(C)=212

Or, 212=(9C)/(5)+32

Or, (9C)/(5)=212-32=180

Or, 9C=180xx5=900

Or, C=900÷9=100

Question 5: Find the range of each of the following functions.

(i) f(x)=2-3x, x∈R, x>0

(ii) f(x)=x^2+2, x is a real number

(iii) f(x)=x, x is a real number

Solution: (i) f(x)=2-3x, x∈R, x>0

The values of f(x) for different values of real numbers x > 0 can be written in the tabular form which is as follows:

 x 0.01 0.1 0.9 1 2 2.5 4 5 f(x) 1.97 1.7 -0.7 -1 -4 -5.5 -10 -13

After the observation of above table it is clear that the range of f is the set of all real numbers less than 2.

Therefore, range of f=(-∞,2)

Alernate Method:

Let x > 0

⇒ 3x > 0

⇒ 2 – 3x < 2

Therefore, Range of f = (-∞, 2)

(ii) f(x)=x^2+2, x is a real number

The values of f(x) for different values of real numbers x can be written in the tabular form which is as follows:

 x 0 ± 0.3 ± 0.8 ± 1 ± 2 ± 3 f(x) 2 2.09 2.64 3 6 11

From the above table it is clear that the range of f is the set of all real numbers greater than 2.

Therefore, range of f = [2, ∞)

Alternate Method:

Let y=x^2+2, x is a real number.

Therefore, x^2=y-2

Or, x=sqrt(y-2)

Here, it is given that x is real number,

Therefore,

y – 2 > 0

⇒ y > 0

Hence, Range = [2, ∞)

(iii) f(x)= x, x is a real number

It is clear that the range f is the set of all real numbers.

Therefore, Range of f = R