10 Maths Chapter 1 || Ex 1.4 ||

Exercise 1.4 Solutions

1. Determine whether the graphs given below represent functions. Give reason for your answers concerning each graph using Vertical Line Test.

(i) Curve opening rightwards (Parabola)

Reason: A vertical line intersects the curve at two points. Hence, it is NOT a function.

(ii) S-shaped curve

Reason: Any vertical line intersects the curve at at most one point. Hence, it IS a function.

(iii) S-shaped curve along y-axis

Reason: A vertical line intersects the curve at three points. Hence, it is NOT a function.

(iv) Straight line passing through origin

Reason: Any vertical line intersects the line at exactly one point. Hence, it IS a function.

2. Let \( f: A \to B \) be a function defined by \( f(x) = \frac{x}{2} - 1 \), where \( A = \{2, 4, 6, 10, 12\} \), \( B = \{0, 1, 2, 4, 5, 9\} \). Represent \( f \) by:

Calculations:

• \( f(2) = \frac{2}{2} - 1 = 1 - 1 = 0 \)
• \( f(4) = \frac{4}{2} - 1 = 2 - 1 = 1 \)
• \( f(6) = \frac{6}{2} - 1 = 3 - 1 = 2 \)
• \( f(10) = \frac{10}{2} - 1 = 5 - 1 = 4 \)
• \( f(12) = \frac{12}{2} - 1 = 6 - 1 = 5 \)

(i) Set of ordered pairs

\( f = \mathbf{\{(2, 0), (4, 1), (6, 2), (10, 4), (12, 5)\}} \)

(ii) A table

\( x \)	2	4	6	10	12
\( f(x) \)	0	1	2	4	5

(iii) An arrow diagram

Draw two ovals for set \( A = \{2, 4, 6, 10, 12\} \) and set \( B = \{0, 1, 2, 4, 5, 9\} \). Connect with arrows: \( 2 \to 0 \), \( 4 \to 1 \), \( 6 \to 2 \), \( 10 \to 4 \), \( 12 \to 5 \).

(iv) A graph

Plot the points \( (2, 0) \), \( (4, 1) \), \( (6, 2) \), \( (10, 4) \), and \( (12, 5) \) on a Cartesian plane.

3. Represent the function \( f = \{(1, 2), (2, 2), (3, 2), (4, 3), (5, 4)\} \) through:

(i) An arrow diagram

Set \( A = \{1, 2, 3, 4, 5\} \), Set \( B = \{2, 3, 4\} \). Draw arrows: \( 1 \to 2 \), \( 2 \to 2 \), \( 3 \to 2 \), \( 4 \to 3 \), \( 5 \to 4 \).

(ii) A table form

\( x \)	1	2	3	4	5
\( f(x) \)	2	2	2	3	4

(iii) A graph

Plot the points \( (1, 2), (2, 2), (3, 2), (4, 3) \), and \( (5, 4) \) on an \( xy \)-plane.

4. Show that the function \( f: \mathbb{N} \to \mathbb{N} \) defined by \( f(x) = 2x - 1 \) is one-one but not onto.

1. One-One: Let \( f(x_1) = f(x_2) \)

\( 2x_1 - 1 = 2x_2 - 1 \implies 2x_1 = 2x_2 \implies x_1 = x_2 \)

Since distinct elements have distinct images, it is a One-One function.

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2. Onto: Range of \( f \) for \( x \in \{1, 2, 3, \dots\} \):

\( f(1) = 1, f(2) = 3, f(3) = 5, \dots \)

\( \text{Range} = \{1, 3, 5, 7, \dots\} \) (Set of odd natural numbers)

\( \text{Co-domain} = \mathbb{N} = \{1, 2, 3, 4, 5, \dots\} \)

Here, \( \text{Range} \neq \text{Co-domain} \) (e.g., even numbers like 2, 4 in co-domain have no pre-images in domain).

Therefore, it is NOT an onto function.

5. Show that the function \( f: \mathbb{N} \to \mathbb{N} \) defined by \( f(m) = m^2 + m + 3 \) is a one-one function.

Let \( f(m_1) = f(m_2) \):

\( m_1^2 + m_1 + 3 = m_2^2 + m_2 + 3 \)

\( m_1^2 - m_2^2 + m_1 - m_2 = 0 \)

\( (m_1 - m_2)(m_1 + m_2) + (m_1 - m_2) = 0 \)

\( (m_1 - m_2)(m_1 + m_2 + 1) = 0 \)

Since \( m_1, m_2 \in \mathbb{N} \), \( m_1 + m_2 + 1 > 0 \) (cannot be zero).

Thus, \( m_1 - m_2 = 0 \implies \mathbf{m_1 = m_2} \).

Hence, \( f \) is a One-One function.

6. Let \( A = \{1, 2, 3, 4\} \) and \( B = \mathbb{N} \). Let \( f: A \to B \) be defined by \( f(x) = x^3 \).

\( f(1) = 1^3 = 1 \)

\( f(2) = 2^3 = 8 \)

\( f(3) = 3^3 = 27 \)

\( f(4) = 4^3 = 64 \)

(i) Find the range of \( f \)

Range of \( f \): \(\{1, 8, 27, 64\}\)

(ii) Identify the type of function

Since distinct elements in \( A \) have distinct images in \( B \), and Range \( \neq B \), it is One-One and Into function (or simply One-One).

7. In each case state whether the function is bijective or not. Justify your answer.

(i) \( f: \mathbb{R} \to \mathbb{R} \) defined by \( f(x) = 2x + 1 \)

1. One-one: \( f(x_1) = f(x_2) \implies 2x_1 + 1 = 2x_2 + 1 \implies x_1 = x_2 \). (One-one)

2. Onto: Let \( y = 2x + 1 \implies x = \frac{y - 1}{2} \in \mathbb{R} \). For every \( y \in \mathbb{R} \), there exists pre-image \( x \in \mathbb{R} \). (Onto)

Result: It is a Bijective function.

(ii) \( f: \mathbb{R} \to \mathbb{R} \) defined by \( f(x) = 3 - 4x^2 \)

1. One-one check: \( f(1) = 3 - 4(1)^2 = -1 \), \( f(-1) = 3 - 4(-1)^2 = -1 \).

Distinct elements \( 1 \neq -1 \) have the same image \( -1 \). Thus, it is NOT one-one.

Result: It is NOT a Bijective function.

8. Let \( A = \{-1, 1\} \) and \( B = \{0, 2\} \). If the function \( f: A \to B \) defined by \( f(x) = ax + b \) is an onto function, find \( a \) and \( b \).

Since \( f \) is onto, the range must be \( B = \{0, 2\} \).

Case 1: \( f(-1) = 0 \) and \( f(1) = 2 \)

• \( -a + b = 0 \implies b = a \)
• \( a + b = 2 \implies a + a = 2 \implies 2a = 2 \implies \mathbf{a = 1, b = 1} \)

Case 2: \( f(-1) = 2 \) and \( f(1) = 0 \)

• \( -a + b = 2 \)
• \( a + b = 0 \implies b = -a \)
• \( -a - a = 2 \implies -2a = 2 \implies \mathbf{a = -1, b = 1} \)

Answer: \( (a = 1, b = 1) \) or \( (a = -1, b = 1) \)

9. If the function \( f \) is defined by: \[ f(x) = \begin{cases} x + 2 & x > 1 \\ 2 & -1 \le x \le 1 \\ x - 1 & -3 < x < -1 \end{cases} \] Find the values of:

(i) \( f(3) \)

\( 3 > 1 \implies f(3) = 3 + 2 = \mathbf{5} \)

(ii) \( f(0) \)

\( -1 \le 0 \le 1 \implies f(0) = \mathbf{2} \)

(iii) \( f(-1.5) \)

\( -3 < -1.5 < -1 \implies f(-1.5) = -1.5 - 1 = \mathbf{-2.5} \)

(iv) \( f(2) + f(-2) \)

\( f(2) = 2 + 2 = 4 \)

\( f(-2) = -2 - 1 = -3 \)

\( f(2) + f(-2) = 4 + (-3) = \mathbf{1} \)

10. A function \( f: [-5, 9] \to \mathbb{R} \) is defined as follows: \[ f(x) = \begin{cases} 6x + 1 & -5 \le x < 2 \\ 5x^2 - 1 & 2 \le x < 6 \\ 3x - 4 & 6 \le x \le 9 \end{cases} \] Find:

(i) \( f(-3) + f(2) \)

\( f(-3) = 6(-3) + 1 = -18 + 1 = -17 \)

\( f(2) = 5(2)^2 - 1 = 5(4) - 1 = 19 \)

\( f(-3) + f(2) = -17 + 19 = \mathbf{2} \)

(ii) \( f(7) - f(1) \)

\( f(7) = 3(7) - 4 = 21 - 4 = 17 \)

\( f(1) = 6(1) + 1 = 7 \)

\( f(7) - f(1) = 17 - 7 = \mathbf{10} \)

(iii) \( 2f(4) + f(8) \)

\( f(4) = 5(4)^2 - 1 = 5(16) - 1 = 79 \)

\( f(8) = 3(8) - 4 = 24 - 4 = 20 \)

\( 2f(4) + f(8) = 2(79) + 20 = 158 + 20 = \mathbf{178} \)

(iv) \( \frac{2f(-2) - f(6)}{f(4) + f(-2)} \)

\( f(-2) = 6(-2) + 1 = -11 \)

\( f(6) = 3(6) - 4 = 14 \)

\( f(4) = 79 \)

\( \frac{2(-11) - 14}{79 + (-11)} = \frac{-22 - 14}{68} = \frac{-36}{68} = \mathbf{-\frac{9}{17}} \)

11. The distance \( S \) an object travels under gravity in time \( t \) seconds is given by \( S(t) = \frac{1}{2}gt^2 + at + b \). Verify whether the function \( S(t) \) is one-one or not.

Let \( S(t_1) = S(t_2) \):

\( \frac{1}{2}gt_1^2 + at_1 + b = \frac{1}{2}gt_2^2 + at_2 + b \)

\( \frac{1}{2}g(t_1^2 - t_2^2) + a(t_1 - t_2) = 0 \)

\( (t_1 - t_2)\left[ \frac{1}{2}g(t_1 + t_2) + a \right] = 0 \)

Since time \( t_1, t_2 > 0 \), gravity \( g > 0 \), and \( a > 0 \), the term \( \left[ \frac{1}{2}g(t_1 + t_2) + a \right] \neq 0 \).

Therefore, \( t_1 - t_2 = 0 \implies \mathbf{t_1 = t_2} \).

Hence, the function \( S(t) \) is One-One.

12. The function '\( t \)' which maps temperature in Celsius (\( C \)) into Fahrenheit (\( F \)) is defined by \( t(C) = F \) where \( F = \frac{9}{5}C + 32 \). Find:

(i) \( t(0) \)

\( t(0) = \frac{9}{5}(0) + 32 = \mathbf{32^\circ F} \)

(ii) \( t(28) \)

\( t(28) = \frac{9}{5}(28) + 32 = 50.4 + 32 = \mathbf{82.4^\circ F} \)

(iii) \( t(-10) \)

\( t(-10) = \frac{9}{5}(-10) + 32 = -18 + 32 = \mathbf{14^\circ F} \)

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

\( \frac{9}{5}C + 32 = 212 \implies \frac{9}{5}C = 180 \implies C = 180 \times \frac{5}{9} = \mathbf{100^\circ C} \)

(v) The temperature when Celsius value is equal to Fahrenheit value

Let \( C = F \):

\( C = \frac{9}{5}C + 32 \implies C - \frac{9}{5}C = 32 \implies -\frac{4}{5}C = 32 \implies C = \mathbf{-40^\circ} \)