Gujarat Board GSEB Textbook Solutions Class 11 Maths Chapter 2 Relations and Functions Ex 2.3 Textbook Questions and Answers.

## Gujarat Board Textbook Solutions 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:

- {(2, 1), (5, 1), (8, 1), (11, 1), (14, 1), (17, 1)}
- {(2, 1), (4, 2), (6, 3), (8, 4), (10, 5), (12, 6), (14, 7)}.
- {1,3}, {1,5}, {2, 5}

Solution:

1. We have : f= {(2, 1), (5, 1), (8, 1), (11, 1), (14, 1), (17, 1)}

No two ordered pairs have the same first component. Therefore, this relation is a function.

Domain (f) = {2, 5, 8, 11, 14, 17}

Range (f) = {1}

2. We have : f = {(2, 1), (4, 2), (6, 3), (8, 4), (10, 5), (12, 6), (14, 7)}

We observe that no two ordered pairs have the same first component.

So, f is a function.

Domain (f) = {2, 4, 6, 8, 10, 12, 14}.

Range (f) = {1, 2, 3, 4, 5, 6, 7}.

3. f = (1, 3), (1, 5), (2, 5)

We observe that 1 has appeared more than once as first component of the ordered pairs in f.

Therefore, f is not a function.

Question 2.

Rind the domain and range of the following functions:

(i) f(x) = – |x|

(ii) f(x) = \(\sqrt{9-x^{2}}\)

Solution:

(i) f(x) = – |x|, f(x) ≤ 0, ∀ x ∈ R

Domain of f = R

Range of f = {y ∈ R, y ≤ 0}

(ii) f(x) = \(\sqrt{9-x^{2}}\)

Here, f is not defined for 9 – x^{2} < 0 or x^{2} > 9 or when x > 3 or x < – 3.

Also, for each real number x lying between – 3 and 3 or for x = – 3, 3; f(x) is unique.

∴ Domain (f) = {x : x ∈ R and – 3 ≤ x ≤ 3}

Further, y = \(\sqrt{9-x^{2}}\) or y^{2} = 9 – x^{2}

or x = \(\sqrt{9-x^{2}}\).

Again, x is not defined for 9 – y^{2} < 0 or y^{2} > 9 or y > 3 or y < – 3. Buty cannot be – ve. ∴ Range (f) – {y : y ∈R and 0 ≤ y ≤ 3}

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:

f(x) = 2x – 5

(i) f(0) = 2 × 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’ maps temperature in Celsius into temperature in degree Fahrenheit is defined 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: t(C) = \(\frac{9C}{5}\) + 32

(i) When C = 0, t(0) = 0 + 32 = 32

(ii) t(C) = \(\frac{9C}{5}\) + 32 When C = 28, t(28) = \(\frac{9×28}{5}\) + 32 = \(\frac{252}{5}\) + 32 = \(\frac{252+160}{5}\) = \(\frac{412}{5}\).

(iii) When C = – 10 t(- 10) = (- 10) \(\frac{9}{5}\) + 32 = – 18 + 32 = 14.

(iv) t(C) = 212 ⇒ 212 = \(\frac{9C}{5}\) + 32 or \(\frac{9C}{5}\) = 212 – 32 = 180 or C = \(\frac{180×5}{9}\) = 100.

Question 5.

Find the range of each of the following functions.

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

Solution:

1. Let f(x) = y = 2 – 3x, i.e; x = \(\frac{2-y}{3}\)

Now, x > 0, i.e; 2 – y > 0 or y < 2.

∴ Range (f) = {y : y ∈ R and y < 2}.

2. Let f(x) = y = x^{2} + 2, i.e; x^{2} = y – 2 or x = \(\sqrt{y-2}\) ⇒ y > 2.

∴ Range (f) = {y : y ∈ R and y ≥ 2}.

3. f(x) = x, x is a real number

x = f(x) = y = a real number

∴ Range (f) = {y : y ∈ R}