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`