﻿ Class 11 NCERT Math Solution
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TOPICS
Exercise - 2.3

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

Solution :-
```(i) {(2,1), (5,1), (8,1), (11,1), (14,1), (17,1)}
R = {2, 5, 8, 11, 14, 17} are the elements of the domain of the given
relation having their unique images, this relation is a function.
So, 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)}
R = {2, 4, 6, 8, 10, 12, 14} are the elements of the domain of the given
relation having their unique images, this relation is a function.
So, Domain = {2, 4, 6, 8, 10, 12, 14} and
Range = {1, 2, 3, 4, 5, 6, 7}

(iii) {(1,3), (1,5), (2,5)}.
This relation is not a function because there is an element 1 which is associated to two
elements 3 and 5.
```

Question-2 :-  Find the domain and range of the following real functions:
(i) f(x) = –|x|
(ii) f(x) = √9 - x²

Solution :-
```(i) f(x) = –|x|, x ∈ R
Since f(x) = –|x| is defined for x ∈ R, the domain of f is R.
It can be observed that the range of f(x) = –|x| is all real numbers except positive real numbers.

(ii) f(x) = √9 - x²
Since √9 - x² is defined for all real numbers that are greater than or equal to –3 and less than or equal to 3,
Domain of f(x) = {x : –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.
Range of f(x) = {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 :-
```   Given that f(x) = 2x – 5.
(i) f(0) = 2 × 0 – 5 = 0 – 5 = –5
(ii) f(7) = 2 × 7 – 5 = 14 – 5 = 9
(iii) 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)   (ii) t(28)   (iii) t(–10)   (iv) The value of C, when t(C) = 212.

Solution :-
```   Give that t(C) = 9C/5 + 32.
(i) t(0) = (9 x 0)/5 + 32 = 0 + 32 = 32
(ii) t(28) = (9 x 28)/5 + 32 = 252/5 + 32 = (252 + 160)/5 = 412/5
(iii) t(-10) = (9 x (-10))/5 + 32 = -18 + 32 = 14
(iv) t(C) = 212
212 = 9C/5 + 32
212 - 32 = 9C/5
180 = 9C/5
C = (180 x 5)/9
C = 20 x 5 = 100
```

Question-5 :-  Find the range of each of the following functions.
(i) f (x) = 2 – 3x, x ∈ R, x > 0.
(ii) f (x) = x² + 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.
Let x > 0
⇒	3x > 0
⇒	2 –3x < 2
⇒	f(x) < 2
Range of f = (–∞, 2)

(ii) f(x) = x² + 2, x is a real number.
Let x be any real number.
Accordingly, x² ≥ 0
⇒	x² + 2 ≥ 0 + 2
⇒	x² + 2 ≥ 2
⇒	f(x) ≥ 2
Range of f = [2,∞)

(iii) f(x)  = x,  x is a real number.
It is clear that the range of f is the set of all real numbers.
Range of f = R
```
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