NCERT Solutions for Class 11 Maths Chapter 2 Relations and Functions Exercise 2.3

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NCERT Solutions for Class 11 Maths Chapter 2 Relations and Functions Exercise 2.3

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)}

Answer

(i) It is a function because to each first element of the ordered pair, there corresponds exactly one second element.
Domain = {2, 5, 8, 11, 14, 17}
Range = {1}.

(ii) It is a function because to each first element of the ordered pair, there corresponds exactly one second element.
Domain = {2, 4, 6, 8, 10, 12, 14}
Range = {1, 2, 3, 4, 5, 6, 7}.

(iii) It is not a function because the two ordered pairs (1, 3) and (1, 5) have the same first component.

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

Answer

(i) We have f (x) = – | x | 
Clearly, f = – | x | is ≤ 0, for all x ∈ R 
∴ Domain = R and Range 
= {x ∈ R: x ≤ 0} = R-
(ii) We have f (x) = √(9 - x2)
Clearly, f is not defined when (9 - x2)<0 
i.e. x2 > 9 
⇒ when x > 3 or x < – 3 
Also, for each real number x lying between –3 and 3. 
f(x) = √(9 – x2) = is unique. 
So, Domain = {x ∈ R: – 3 ≤ x ≤ 3} 
Further, y = √(9 – x2) ⇒ y2 = (9 – x2)
⇒ x = √(9 – y2)
Clearly, x is not defined when (9 – y2) < 0 
But (9 – y2) < 0 
⇒ y2 > 9 ⇒ y > 3 or y < – 3 
∴ Range = {y ∈ R: – 3 ≤ y ≤ 3}.

3. A function f is defined by f (x) = 2x – 5. Write down the value of: (i) f (0), (ii) f (7), (iii) f (– 3).

Answer

f is defined by f (x) = 2x – 5
∴ f (0) = 2(0) – 5 = – 5
f (7) = 2(7) – 5 = 9
f (– 3) = 2(– 3) – 5 = – 11.

4. The function ‘t’ maps temperature in degree Celsius into temperature in Fahrenheit. It 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. 

Answer

We have t(C) = 9c/5 + 32
∴ (i) t(0) = 9(0)/5 + 32 = 32
(ii) t(28) = 9(28)/5 + 32 = 252/5 + 32
= 412/5 = 82.4
(iii) t(-10) = 9(-10)/5 + 32
= -90/5 + 32 = -18 + 32 = 14
(iv) when t(C) = 212, i.e., 212 = 9C/5 + 32
212 = (9C + 160)/5 => 1060 = 9C + 160
=> 9C = 900 => C = 100
∴ The value of C is 100

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.

Answer

(i) Let y = 2 – 3x, x ∈ R, x > 0
y – 2 = – 3x ⇒ 3x = 2 – y
⇒ x = (2 - y)/3
As x > 0, we have
(2 - y)/3 > 0 or 2 - y> 0
⇒ y < 2
Hence, Range = (- ∞,2)

(ii) Let y = x2 + 2, x is a real number.
⇒ x2 = y – 2 Þ x = y – 2
As x is real, we have
y – 2 > 0 ⇒ √y > 2
Hence, Range = (2,∞)

(iii) Let y = x, x is a real number Range = R.
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