# 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`