Exercise – 2.3

1. Which of the following relations are functions? Give reasons. If it is a function determine its domain & range.

(i) {(2, 1), (5, 1), (8, 1), (11, 1), (14, 1), (17, 1)}

It is a function. Because every element has unique image

Domain = {2, 5, 8, 11, 14, 17}

Range = {1}

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

It is a function. Because every element has unique image

Domain = {2, 4, 6, 8, 10, 12, 14}

Range = {1, 2, 3, 4, 5, 6, 7}

(iii) {(1, 3), (1, 5), (2, 5)}

It is not a function. Because every element has not unique image

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

(i) f(x) = -|x|

Domain = All Real

For range:

Let y = f(x)

|x| ≥ 0

-|x| ≤ 0

f(x) ≤ 0

y ≤ 0

∴ Range = (-∞, 0)

(ii) f(x) = (9 – x²)^1/2

We know that

9 – x² ≥ 0

9 ≥ x²

x² ≤ 9

-3 ≤ x ≤ 3

∴ Domain = [-3, 3]

For range:

Let y = y(x) = (9 – x²)^1/2

y is always positive

y = (9 – x²)^1/2

sq. on both sides

y² = 9 – x²

x² = 9 – y²

x = ±(9 – y²)^1/2

we know that

 9 – y² ≥ 0

9 ≥ y²

y² ≤ 9

-3 ≤ y ≤ 3

∴ Range = [-3, 3]

But,

As, y is always positive. So, -3 is rejected.

∴ Range = [0, 3]

3. As function f is defined by f(x) = 2x – 5. Write down the values of:

(i) f(0)

As, f(x) = 2x – 5

∴ f(x) = 2 X 0 – 5 = -5

(ii) f(7)

As f(x) = 2x – 5

∴ f(7) = 2 X 7 – 5 = 9

(iii) f(3)

As f(x) = 2x – 5

∴ f(-3) = 2 X (-3) – 5 = -11

4. The function ‘t’ which maps temperature in degree calcium into temperature degree. Fahrenheit is defined by t (C) = (9C/5) + 32. Find:

(i) t(0)

As t (C) = (9C/5) + 32

∴ t(0) = (9*0/5) + 32 = 32. 

(ii) t(28)

As t (C) = (9C/5) + 32

∴ t (28) = (9*28/5) + 32 = (252/5) + 32 = 412/5

(iii) t(-10)

As t (C) = (9C/5) + 32

∴ t(-10) = (9 * -10/5) + 32 = (-90 + 160/5) = 14

(iv) The value of C, when t(C) = 212

As t (C) = (9C/5) + 32

∴ t (C) = (9C/5 + 32)

212 = (9C/5 + 32)

212 – 32 = 9C/5

C = 100

5. Find the range of each of the following functions

(i) f(x) = 2 – 3x, x Є R, x > 0

Let f(x) = y = 2 – 3x

It is given that

X > 0

Multiply both sides by 3

3x > 0

-3x < 0

Now, Add 2 on both sides

2 – 3x < 2

Y < 2

∴ Range = (-∞)

(ii) f(x) = x² + 2, x is a real number

Let f(x) = y = x² + 2

We know that

x² ≥ 0

Add 2 on both sides

x² + 2 ≥ 2

y ≥ 2

∴ Range = [2, ∞]

(iii) f(x) = x, x is real number

Range = All real