PSEB 11th Class Maths Solutions Chapter 2 Relations and Functions Ex 2.3

Punjab State Board PSEB 11th Class Maths Book Solutions Chapter 2 Relations and Functions Ex 2.3 Textbook Exercise Questions and Answers.

PSEB 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)}
Answer.
(i) {(2, 1), (5,1), (8, 1), (11, 1), (14, 1), (17,1)}
Since 2, 5, 8, 11, 14, and 17 are the elements of the domain of the given relation having their unique images, this relation is a function.
Here, domain={2, 5, 8, 11, 14, 17} and range={l}

(ii) {(2, 1), (4, 2), (6, 3), (8, 4), (10, 5), (12, 6), (14, 7)}
Since 2, 4, 6, 8, 10, 12, and 14 are the elements of the domain of the given relation having their unique images, this relation is a function.
Here, domain = {2, 4, 6, 8, 10, 12, 14} and range = {1, 2, 3, 4, 5, 6, 7}

(iii) {(1, 3), (1, 5), (2, 5)}
Since the same first element i.e., 1 corresponds to two different images i.e.,3 and 5, this relation is not a function.

Question 2.
Find the domain and range of the following real functions:
(i) f(x) = – |x|
(ii) f(x) = \(\sqrt{9-x^{2}}\)
Answer.
(i) f(x) = – |x|, f(x) ≤ 0, ∀ x ∈ R
Domain of f = R
Range of f = {y ∈ R, y < 0}

(ii) f(x) = \(\sqrt{9-x^{2}}\)
All, f is not defined for 9 – x² ≤ 0 or x² ≤ 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 y² = 9 – x²
or x = \(\sqrt{9-x^{2}}\)
Again, x is not defined for
9 – y² < 0 or y² > 9 or y > 3 or y < – 3.
But y cannot be – ve
∴ Range (f) = {y : y ∈ R and 0 < y < 3}

Question 3.
A function fis defined by f(x) = 2x – 5. Write down the values of
(i) f(0)
(ii) f(7)
(iii) f(- 3)
Answer.
The given function is f(x) = 2x – 5. Therefore,
(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 Fahrenheit is defined by t(C) = \(\frac{9 C}{5}\) + 32. Find
(i) f(0)
(ii) t(28)
(iii) t(- 10)
(iv) The value of C, when t(C) = 212.
Answer.
The given function is t (C) = \(\frac{9 C}{5}\) + 32. Therefore,
(i) t(0) = \(\frac{9 \times 0}{5}\) + 32 = 0 + 32 = 32

(ii) t(28) = \(\frac{9 \times 28}{5}\) + 32 = \(\frac{252+160}{5}=\frac{412}{5}\)

(iii) t (- 10) = \(\frac{9 \times(-10)}{5}\) + 32 = 9 × ( – 2) + 32

(iv) It is given that t(C) = 212
∴ 212 = \(\frac{9 C}{5}\) + 32 \(\frac{9 C}{5}\)
= 212 – 32 \(\frac{9 C}{5}\) = 180
9C = 180 × 5 C = \(\frac{180 \times 5}{9}\) = 100.
Thus, the value of t, when t(C) = 212, is 100.

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.
Answer.
(i) Let f(x) = y = 2 – 3x is x = \(\frac{2-y}{3}\)
Now x > 0 is 2 – y > 0 or y < 2
Range (f) = {y : y ∈ R and y < 2}

(ii) Let f(x) = y = x² + 2 is x² = y – 2 or x = \(\sqrt{y-2}\) ⇒ y > 2
Range (f) = {y : y ∈ R and y > 2}

(iii) f(x) = x, x is a real number
x = f(x) = y = a real number
Range (f) = {y : y ∈ R}