**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) =**(ii)**f(x) =

**Solution.**

**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**

**Find**

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

**Solution.**

**Question 5. Find the range of each of the following functions.**

**(i)**f(x) = 2 – 3x, x ∈ R, x>0.**(ii)**f(x)=x^{2}+ 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) = x^{2} + 2, x is a real number

We know x^{2}≥ 0 ⇒ x^{2} + 2 ≥ 0 + 2

⇒ x^{2} + 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.