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

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

If you're in search of Chapter 2 Relations and Functions Exercise 2.3 NCERT Solutions then you can find on this page. Here you will find NCERT Solutions for Class 11 Maths that will guide every students how to apply formula in right manner and boost their marks in the exams. One can easily solve their doubts by taking help from this page.

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.