Exercise 2.3 Relations and Functions
Question and Answers
Class 11 – Maths
| Class | Class 11 |
| Subject | Mathematics |
| Chapter Name | Relations and Functions |
| Chapter No. | Chapter 2 |
| Exercise | Exercise 2.3 |
| Category | Class 11 Maths NCERT Solutions |
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)}
(i) {(2, 1), (5, 1), (8, 1), (11, 1), (14, 1), (17, 1)}
As 2, 5, 8, 11, 14, and 17 are the elements of the domain of the given relation having their unique images, this relation can be called a function.
Here, domain = {2, 5, 8, 11, 14, 17} and range = {1}
(ii) {(2,1), (4,2), (6,3), (8,4), (10,5), (12,6), (14,7)}
(ii) {(2, 1), (4, 2), (6, 3), (8, 4), (10, 5), (12, 6), (14, 7)}
As 2, 4, 6, 8, 10, 12, and 14 are the elements of the domain of the given relation having their unique images, this relation can be called a function.
Here, domain = {2, 4, 6, 8, 10, 12, 14} and range = {1, 2, 3, 4, 5, 6, 7}
(iii) {(1,3), (1,5), (2,5)}.
(iii) {(1, 3), (1, 5), (2, 5)}
It’s seen that the same first element, i.e., 1, corresponds to two different images, i.e., 3 and 5; this relation cannot be called a function.
Question 2 Find the domain and range of the following real functions:
(i) f(x) = – x
(i) We have the given function as, f (x) = − |x|
It is also know that, | x | = x , if x ≥ 0
−x ,if x < 0
Thus, f(x) = − | x | = − x ,if x ≥ 0
x ,if x < 0
As f(x) is defined for x ∈ R, the domain of f is R.
It is also seen that the range of f(x) = –|x| is all real numbers except positive real numbers.
Therefore, the range of f is given by (–∞, 0].
(ii) f(x) = √9 − x2
(ii) As √(9 – x2) is defined for all real numbers that are greater than or equal to –3 and less than or equal to 3, for 9 – x2 ≥ 0.
So, the domain of f(x) is {x: –3 ≤ x ≤ 3} or [–3, 3]
For any value of x in the range [–3, 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)
(i) Function, f(x) = 2x – 5
f (0) = 2 × 0 – 5 = 0 – 5 = –5
(ii) f (7)
(ii) Function, f(x) = 2x – 5
f(7) = 2 × 7 – 5 = 14 – 5 = 9
(iii) f (–3)
(iii) Function, f(x) = 2x – 5
f(–3) = 2 × (–3) – 5 = – 6 – 5 = –11
Question 4 The function ‘t’ which maps temperature in degree Celsius into temperature in degree Fahrenheit is defined by t(C) = 9C/ 5 + 32. Find
(i) t(0)
(i) We have our given function as, t(C)= (9C/5) +32
To find our needed values of the function we just have to put the values in the given function and simplify it.
So, we get now,
t (0) =[ (9 × o ) /5 ] + 32
t (0)= 32
(ii) t(28)
(ii) We have our given function as, t(C)= (9C/5) +32
To find our needed values of the function we just have to put the values in the given function and simplify it.
So, we get now,
t (28) =[ (9 × 28 ) /5 ] + 32
t (28 ) =( 252 + 160) /5
t (28 ) =(412 /5 ]
t (28 ) = 82.4
(iii) t(–10)
(iii) We have our given function as, t(C)= (9C/5) +32
To find our needed values of the function we just have to put the values in the given function and simplify it.
So, we get now,
t ( -10 ) = [ (9 × (-10) ) /5 ] + 32
t (-10) =−18+32
t (-10) = 14
(iv) The value of C, when t(C) = 212.
(iv) We have our given function as, t(C)= (9C/5) +32
we are given that, t(C)=212
212 = (9C/5) +32
9C/5 = 212−32
9C/5 = 180
C = 900 / 9 = 100
Thus , it can be said that, for t(C)=212 the value of t is 100
Question 5 Find the range of each of the following functions.
(i) f (x) = 2 – 3x, x ∈ R, x > 0.
(i) f(x) = 2 – 3x, x ∈ R, x > 0
|
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 |
… |
We can now see, it can be seen that the elements of the range is less than 2.
So, the range will be, f = (−∞ ,2 )
x > 0, So , 3x > 0
-3x < 0 [Multiplying by -1 on both sides, the inequality sign changes]
2 – 3x < 2
Therefore, the value of 2 – 3x is less than 2.
Hence, Range = (–∞, 2)
(ii) f (x) = x 2 + 2, x is a real number.
(ii) f(x) = x2 + 2, x is a real number
|
x |
0 |
±0.3 |
±0.8 |
±1 |
±2 |
±3 |
… |
|
f(x) |
2 |
2.09 |
2.64 |
3 |
6 |
11 |
… |
So, we see that the range of the function f is the set of all numbers which are greater than or equal to 2.
Thus we can conclude that the range of the function is, [2,∞]
x2 ≥ 0
So,
x2 + 2 ≥ 2 [Adding 2 on both sides]
Therefore, the value of x2 + 2 is always greater or equal to 2, for x is a real number.
Hence, Range = [2, ∞)
(iii) f (x) = x, x is a real number.
(iii) f(x) = x, x is a real number
Clearly, the range of f is the set of all real numbers.
Thus,
Range of f = R