10 Maths samacheer Kalvi || Ex 1.3 ||

Exercise 1.3 Solutions

1. Let \( f = \{(x, y) \mid x, y \in \mathbb{N} \text{ and } y = 2x\} \) be a relation on \( \mathbb{N} \). Find the domain, co-domain and range. Is this relation a function?

Given: Relation on \( \mathbb{N} \) defined by \( y = 2x \) for \( x, y \in \mathbb{N} = \{1, 2, 3, 4, \dots\} \).

\( f = \{(1, 2), (2, 4), (3, 6), (4, 8), \dots\} \)

• Domain: Set of natural numbers = \( \mathbb{N} = \{1, 2, 3, 4, \dots\} \)
• Co-domain: Set of natural numbers = \( \mathbb{N} = \{1, 2, 3, 4, \dots\} \)
• Range: Set of even natural numbers = \( \{2, 4, 6, 8, \dots\} \)

Is it a function? Yes, every element \( x \in \mathbb{N} \) has a unique image \( y = 2x \in \mathbb{N} \). Hence, it is a function.

2. Let \( X = \{3, 4, 6, 8\} \). Determine whether the relation \( R = \{(x, f(x)) \mid x \in X, f(x) = x^2 + 1\} \) is a function from \( X \) to \( \mathbb{N} \).

Given \( X = \{3, 4, 6, 8\} \) and \( f(x) = x^2 + 1 \):

• \( f(3) = 3^2 + 1 = 9 + 1 = 10 \in \mathbb{N} \)
• \( f(4) = 4^2 + 1 = 16 + 1 = 17 \in \mathbb{N} \)
• \( f(6) = 6^2 + 1 = 36 + 1 = 37 \in \mathbb{N} \)
• \( f(8) = 8^2 + 1 = 64 + 1 = 65 \in \mathbb{N} \)

Ordered pairs: \( R = \{(3, 10), (4, 17), (6, 37), (8, 65)\} \)

Since each element in domain \( X \) has a unique image in \( \mathbb{N} \), \( R \) is a function from \( X \) to \( \mathbb{N} \).

3. Given the function \( f: x \to x^2 - 5x + 6 \), evaluate:

(i) \( f(-1) \)

\( f(-1) = (-1)^2 - 5(-1) + 6 = 1 + 5 + 6 = \mathbf{12} \)

(ii) \( f(2a) \)

\( f(2a) = (2a)^2 - 5(2a) + 6 = \mathbf{4a^2 - 10a + 6} \)

(iii) \( f(2) \)

\( f(2) = (2)^2 - 5(2) + 6 = 4 - 10 + 6 = \mathbf{0} \)

(iv) \( f(x - 1) \)

\( f(x - 1) = (x - 1)^2 - 5(x - 1) + 6 \)

\( = (x^2 - 2x + 1) - 5x + 5 + 6 = \mathbf{x^2 - 7x + 12} \)

4. A graph representing the function \( f(x) \) is given. It is clear that \( f(9) = 2 \).

(i) Find the following values of the function:

(a) \( f(0) = \mathbf{9} \)

(b) \( f(7) = \mathbf{6} \)

(c) \( f(2) = \mathbf{6} \)

(d) \( f(10) = \mathbf{0} \)

(ii) For what value of \( x \) is \( f(x) = 1 \)?

From the graph, when \( y = 1 \), the value of \( x = \mathbf{9.5} \).

(iii) Describe the following: (i) Domain, (ii) Range.

Domain: \( \{x \mid 0 \le x \le 10, x \in \mathbb{R}\} = \mathbf{[0, 10]} \)

Range: \( \{y \mid 0 \le y \le 9, y \in \mathbb{R}\} = \mathbf{[0, 9]} \)

(iv) What is the image of 6 under \( f \)?

From graph, at \( x = 6 \), \( y = 5 \). So the image of 6 is 5.

5. Let \( f(x) = 2x + 5 \). If \( x \neq 0 \), then find \( \frac{f(x + 2) - f(2)}{x} \).

\( f(x) = 2x + 5 \)

\( f(x + 2) = 2(x + 2) + 5 = 2x + 4 + 5 = 2x + 9 \)

\( f(2) = 2(2) + 5 = 4 + 5 = 9 \)

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\( \frac{f(x + 2) - f(2)}{x} = \frac{(2x + 9) - 9}{x} = \frac{2x}{x} = \mathbf{2} \)

6. A function \( f \) is defined by \( f(x) = 2x - 3 \).

(i) Find \( \frac{f(0) + f(1)}{2} \)

\( f(0) = 2(0) - 3 = -3 \)

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

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

(ii) Find \( x \) such that \( f(x) = 0 \)

\( 2x - 3 = 0 \implies 2x = 3 \implies x = \mathbf{\frac{3}{2}} \)

(iii) Find \( x \) such that \( f(x) = x \)

\( 2x - 3 = x \implies 2x - x = 3 \implies x = \mathbf{3} \)

(iv) Find \( x \) such that \( f(x) = f(1 - x) \)

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

\( 2x - 3 = 2 - 2x - 3 \)

\( 4x = 2 \implies x = \frac{2}{4} = \mathbf{\frac{1}{2}} \)

7. An open box is to be made from a square piece of material, 24 cm on a side, by cutting equal squares from the corners and turning up the sides as shown. Express the volume \( V \) of the box as a function of \( x \).

When squares of side \( x \) are cut from four corners:

• Length (\(l\)): \( 24 - 2x \)
• Breadth (\(b\)): \( 24 - 2x \)
• Height (\(h\)): \( x \)

Volume (\(V\)): \( l \times b \times h \)

\( V(x) = (24 - 2x)(24 - 2x)x \)

\( V(x) = (576 - 96x + 4x^2)x \)

\( V(x) = \mathbf{4x^3 - 96x^2 + 576x} \text{ cm}^3 \)

8. A function \( f \) is defined by \( f(x) = 3 - 2x \). Find \( x \) such that \( f(x^2) = (f(x))^2 \).

\( f(x^2) = 3 - 2x^2 \)

\( (f(x))^2 = (3 - 2x)^2 = 9 - 12x + 4x^2 \)

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Given \( f(x^2) = (f(x))^2 \):

\( 3 - 2x^2 = 9 - 12x + 4x^2 \)

\( 6x^2 - 12x + 6 = 0 \)

Divide by 6: \( x^2 - 2x + 1 = 0 \)

\( (x - 1)^2 = 0 \implies x = \mathbf{1} \)

9. A plane is flying at a speed of 500 km per hour. Express the distance '\( d \)' travelled by the plane as function of time '\( t \)' in hours.

Speed: \( 500 \text{ km/h} \)

Time: \( t \text{ hours} \)

Formula: \( \text{Distance} = \text{Speed} \times \text{Time} \)

\( d(t) = 500 \times t \)

Answer: \( d(t) = 500t \)

10. The data in the adjacent table depicts the length of a person's forehand and their corresponding height. Relationship: \( y = ax + b \).

(i) Check if this relation is a function.

Each value of forehand length \( x \) is uniquely paired with a height \( y \). Hence, it is a function.

(ii) Find \( a \) and \( b \).

Using given points \( (35, 56) \) and \( (45, 65) \):

1) \( 35a + b = 56 \)

2) \( 45a + b = 65 \)

Subtracting equation (1) from (2):

\( 10a = 9 \implies a = \mathbf{0.9} \)

Substitute \( a = 0.9 \) in (1):

\( 35(0.9) + b = 56 \implies 31.5 + b = 56 \implies b = \mathbf{24.5} \)

Equation: \( y = 0.9x + 24.5 \)

(iii) Find the height of a person whose forehand length is 40 cm.

Given \( x = 40 \):

\( y = 0.9(40) + 24.5 = 36 + 24.5 = \mathbf{60.5} \text{ inches} \)

(iv) Find the length of forehand of a person if the height is 53.5 inches.

Given \( y = 53.5 \):

\( 53.5 = 0.9x + 24.5 \)

\( 0.9x = 53.5 - 24.5 = 29 \)

\( x = \frac{29}{0.9} = \mathbf{32.22} \text{ cm} \)