NCERT SOLUTIONS FOR CLASS 11 MATHS CHAPTER 2 / RELATIONS AND FUNCTIONS EX 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) We have a relation R = {(2, 1), (5, 1), (8, 1), (11, 1), (14, 1), (17, 1)} Since 2, 5, 8, 11, 14, 17 are the elements of domain of R having their unique images.

The given relation is a function.

Hence domain = {2, 5, 8, 11, 14, 17) and Range = {1}.

(ii) We have a relation
R = {(2,1), (4, 2), (6, 3), (8, 4), (10, 5), (12, 6), (14, 7)}
Since 2, 4, 6, 8, 10, 12, 14 are the elements of domain of R having their unique images.
The given relation is a function.
Hence domain = {2, 4, 6, 8, 10, 12, 14} and Range = {1, 2, 3, 4, 5, 6, 7}.

(iii) We have a relation R = {(1, 3), (1, 5), (2, 5)}
Since the distinct ordered pairs (1, 3) and (1, 5) have the same first element i.e., 1 does not have a unique image under R.
It is not a function.

Question 2. Find the domain and range of the following real functions:

  • (i) f(x) = -\left| x \right|
  • (ii) f(x) = \sqrt { 9-{ x }^{ 2 } }

Solution.

NCERT Solutions for Class 11 Maths Chapter 2 Relations and Functions Ex 2.3 1

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.

We are given 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 by

t(C)=\frac { 9C }{ 5 } +32

Find

  • (i) t (0)
  • (ii) t (28)
  • (iii) t (-10)
  • (iv) The value of C, when t (C = 212

Solution.

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

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) Given f (x) = 2 – 3x, x ∈ R, x > 0

∵ x > 0 ⇒ -3x < 0 ⇒ 2 – 3x < 2 + 0 ⇒ f (x) < 2

∴ The range of f (x) is (-2).

(ii) Given f (x) = x2 + 2, x is a real number
We know x2≥ 0 ⇒ x2 + 2 ≥ 0 + 2
⇒ x2 + 2 > 2 ∴ f (x) ≥ 2
∴ The range of f (x) is [2, ∞).

(iii) Given f (x) = x, x is a real number.
Let y =f (x) = x ⇒ y = x
∴ Range of f (x) = Domain of f (x)
∴ Range of f (x) is R.

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