Inter 1st Year Maths Relations and Functions Solutions Exercise 2c

Practicing the AP Board Solutions Class 11 Maths and Chapter 2 Inter 1st Year Maths Relations and Functions Solutions Exercise 2c Pdf Download will help students to clear their doubts quickly.

Intermediate 1st Year Maths Relations and Functions Solutions Exercise 2c

Relations and Functions Exercise 2c Solutions

Relations and Functions Class 11 Exercise 2c Solutions – Relations and Functions 2c Exercise Solutions

I.

Question 1.
Find the domain and range of the following real functions:
(i) f(x) = -|x|
(ii) f(x) = \(\sqrt{9-x^2}\)
Solution:
Given, f(x) = -|x|, x ∈ R
we know that,

As f(x) is defined for x ∈ R, the domain of f is R.
It is also observed that the range of f(x) = -|x| includes all real numbers except the positive real numbers.
Therefore, the range of f is given by (-∞, 0).

(ii) f(x) = \(\sqrt{9-x^2}\)
As \(\sqrt{9-x^2}\) is defined for all real numbers that are greater than or equal to -3 and less than or equal to 3 for 9 – x² ≥ 0.
So, the domain of f(x) is (x: -3 ≤ x ≤ 3) or (-3, 3)
Now, for any value of x in the range (-3, 3), the value of f(x) will lie between 0 and 3.
Therefore, the range of f(x) is (x: 0 ≤ x ≤ 3) or (0, 3).

Question 2.
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:
Given, Function 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 3.
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
Here, the values of f(x) for various values of real numbers x > 0 can be given as

It can be observed that the range of f is the set of all real numbers less than 2.
Range of f = (-∞, 2)
We have, x > 0
So, 3x > 0
-3x < 0 (multiplying by -1 on both sides, the inequality sign changes)
2 – 3x < 2
Therefore, the value of 2 – 3x is less than 2.
Hence, Range = (-∞, 2)

(ii) Given, f(x) = x² + 2, x is a real number
Here, the values of f(x) for various values of real numbers x can be given as:

It can be observed that the range of the set of all real numbers greater than 2.
Range of f = (2, ∞)
We know that x² ≥ 0
So, x² + 2 ≥ 2 (Adding 20n both sides)
Therefore, the value of x² + 2 is always greater than or equal to 2 for any real number.
Hence, Range = (2, ∞)

(iii) Given,
f(x) = x, x is a real number
Clearly, the range of the set of all real numbers.

II.

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) {(2, 1), (5, 1), (8, 1), (11, 1), (14, 1), (17, 1)}
As 2, 5, 8, 11, 14, and 17 are the elements of the domain of the given relation having their unique images, this relation can be called a function.
Here, domain = (2, 5, 8, 11, 14, 17) and Range = (1)

(ii) {(2, 1), (4, 2), (6, 3), (8, 4), (10, 5), (12, 6), (14, 7)}
As 2, 4, 6, 8, 10, 12, and 14 are the elements of the domain of the given relation having their unique images, this relation can be called 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)}
It is seen that the same first element
i.e., 1 corresponds to two different images,
i.e., 3 and 5;
This relation cannot be called a function.

Question 2.
The function ‘t’ which maps temperature in degrees Celsius into temperature in degrees 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:
(i) Given function,


Therefore, the value of t when t(C) = 212 is 100.