GSEB Class 11 Maths Solutions Chapter 2 Relations and Functions Exercise 2.3

Question 1. Which of the following relations are functions? Give reasons. If it is a function, determine its domain and range:
1. \( \{ (2, 1), (5, 1), (8, 1), (11, 1), (14, 1), (17, 1) \} \)
2. \( \{ (2, 1), (4, 2), (6, 3), (8, 4), (10, 5), (12, 6), (14, 7) \} \).
3. \( \{ (1, 3), (1, 5), (2, 5) \} \)
Answer:
1. We have: \( f = \{ (2, 1), (5, 1), (8, 1), (11, 1), (14, 1), (17, 1) \} \) No two ordered pairs possess the same first component. As a result, this relation works as 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 notice that no two ordered pairs share 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. We have: \( f = \{ (1, 3), (1, 5), (2, 5) \} \) We can see that '1' appears multiple times as the first component of the ordered pairs in \( f \). Therefore, \( f \) is not a function.
In simple words: A relation is a function if each input (first number in a pair) goes to only one output (second number). If any input repeats with different outputs, it is not a function. The domain is all the inputs, and the range is all the outputs.

Exam Tip: To determine if a relation is a function, check if any element in the domain maps to more than one element in the range. If so, it's not a function.

Question 2. Find the domain and range of the following functions:
(i) \( f(x) = - |x| \)
(ii) \( f(x) = \sqrt{9-x^{2}} \)
Answer:
(i) \( f(x) = - |x| \) Here, \( f(x) \) will always be less than or equal to 0, for all real numbers \( x \). Domain of \( f = R \) (all real numbers) Range of \( f = \{ y \in R, y \le 0 \} \)
(ii) \( f(x) = \sqrt{9-x^{2}} \) For \( f(x) \) to be defined, the expression inside the square root must be non-negative. So, \( 9 - x^{2} \ge 0 \) This means \( x^{2} \le 9 \) Taking the square root of both sides, we get \( |x| \le 3 \). So, \( -3 \le x \le 3 \). Thus, the domain of \( f \) is \( \{ x : x \in R \text{ and } -3 \le x \le 3 \} \). For the range, let \( y = \sqrt{9-x^{2}} \). Since \( \sqrt{} \) always gives a non-negative result, \( y \ge 0 \). Also, square both sides: \( y^{2} = 9 - x^{2} \). This implies \( x^{2} = 9 - y^{2} \). For \( x^{2} \) to be defined, \( 9 - y^{2} \ge 0 \), which means \( y^{2} \le 9 \), so \( |y| \le 3 \). Since we already know \( y \ge 0 \), combining these gives \( 0 \le y \le 3 \). Therefore, the range of \( f \) is \( \{ y : y \in R \text{ and } 0 \le y \le 3 \} \).
In simple words: For \( f(x) = -|x| \), any real number can be an input, but the output will always be zero or a negative number. For \( f(x) = \sqrt{9-x^2} \), the input must be between -3 and 3 (inclusive) for the square root to work, and the output will be between 0 and 3 (inclusive).

Exam Tip: When finding the domain of a function involving a square root, ensure the expression inside the root is greater than or equal to zero. For the range, consider the possible output values based on the domain restrictions and the nature of the function (e.g., square roots are always non-negative).

Question 3. A function \( f \) is defined by \( f(x) = 2x - 5 \). Write down the values of (i) \( f(0) \) (ii) \( f(7) \) (iii) \( f(-3) \).
Answer:The given function is \( f(x) = 2x - 5 \).
(i) To find \( f(0) \), substitute \( x = 0 \) into the function: \( f(0) = 2 \times 0 - 5 = 0 - 5 = -5 \).
(ii) To find \( f(7) \), substitute \( x = 7 \) into the function: \( f(7) = 2 \times 7 - 5 = 14 - 5 = 9 \).
(iii) To find \( f(-3) \), substitute \( x = -3 \) into the function: \( f(-3) = 2 \times (-3) - 5 = -6 - 5 = -11 \).
In simple words: To find the value of a function at a specific number, just replace every 'x' in the function's rule with that number and then do the math.

Exam Tip: Remember to follow the order of operations (PEMDAS/BODMAS) carefully when evaluating function values, especially with negative numbers.

Question 4. The function 't' maps temperature in Celsius into temperature in degree Fahrenheit is \( t(C) = \frac{9C}{5} + 32 \). Find (i) \( t(0) \) (ii) \( t(28) \) (iii) \( t(-10) \) (iv) The value of C, when \( t(C) = 212 \).
Answer:The given function is \( t(C) = \frac{9C}{5} + 32 \).
(i) To find \( t(0) \), substitute \( C = 0 \): \( t(0) = \frac{9 \times 0}{5} + 32 = 0 + 32 = 32 \).
(ii) To find \( t(28) \), substitute \( C = 28 \): \( t(28) = \frac{9 \times 28}{5} + 32 = \frac{252}{5} + 32 \) \( = \frac{252 + (32 \times 5)}{5} = \frac{252 + 160}{5} = \frac{412}{5} \).
(iii) To find \( t(-10) \), substitute \( C = -10 \): \( t(-10) = \frac{9 \times (-10)}{5} + 32 = \frac{-90}{5} + 32 = -18 + 32 = 14 \).
(iv) To find the value of \( C \) when \( t(C) = 212 \): Set the function equal to 212: \( 212 = \frac{9C}{5} + 32 \). Subtract 32 from both sides: \( 212 - 32 = \frac{9C}{5} \) \( 180 = \frac{9C}{5} \). Multiply both sides by 5: \( 180 \times 5 = 9C \) \( 900 = 9C \). Divide by 9: \( C = \frac{900}{9} = 100 \).
In simple words: For the first three parts, replace 'C' in the formula with the given Celsius temperature and calculate to get Fahrenheit. For the last part, set the Fahrenheit formula equal to 212 and solve for 'C' to find the Celsius temperature.

Exam Tip: Pay close attention to fraction calculations and the order of operations. When solving for an unknown variable, perform inverse operations to isolate it correctly.

Question 5. Find the range of each of the following functions.
1. \( f(x) = 2 - 3x, x \in R, x > 0 \).
2. \( f(x) = x^{2} + 2, x \) is a real number.
3. \( f(x) = x, x \) is a real number.
Answer:
1. Given \( f(x) = 2 - 3x \), with \( x \in R \) and \( x > 0 \). Let \( y = 2 - 3x \). To find the range, we need to express \( x \) in terms of \( y \). \( 3x = 2 - y \) \( x = \frac{2 - y}{3} \). Since \( x > 0 \), we must have \( \frac{2 - y}{3} > 0 \). Multiplying by 3, \( 2 - y > 0 \). This means \( 2 > y \), or \( y < 2 \). So, the range of \( f \) is \( \{ y : y \in R \text{ and } y < 2 \} \).
2. Given \( f(x) = x^{2} + 2 \), with \( x \) being a real number. Let \( y = x^{2} + 2 \). Since \( x \) is a real number, \( x^{2} \ge 0 \). Adding 2 to both sides, \( x^{2} + 2 \ge 2 \). Since \( y = x^{2} + 2 \), this means \( y \ge 2 \). So, the range of \( f \) is \( \{ y : y \in R \text{ and } y \ge 2 \} \).
3. Given \( f(x) = x \), with \( x \) being a real number. Let \( y = x \). Since \( x \) can be any real number, \( y \) can also be any real number. So, the range of \( f \) is \( \{ y : y \in R \} \) (all real numbers).
In simple words: The range is all the possible output values of the function. For \( 2-3x \) with \( x>0 \), the output will be less than 2. For \( x^2+2 \), since \( x^2 \) is always at least 0, the output will always be 2 or more. For \( f(x)=x \), the output is the same as the input, so if the input can be any real number, the output can too.

Exam Tip: When determining the range of a function, consider the fundamental properties of the operations involved (e.g., squares are always non-negative, square roots yield non-negative results, linear functions have ranges dependent on their domain). If the domain is restricted, this will also impact the range.