Exercise 3.3 — Class 9 Mathematics Solutions

Solution

1. Understand the Cartesian product $A \times A$.

   For $A = \{1, 2, 3, 4\}$:

   $$\begin{aligned} A \times A = \{ &(1,1), (1,2), (1,3), (1,4), \\ &(2,1), (2,2), (2,3), (2,4), \\ &(3,1), (3,2), (3,3), (3,4), \\ &(4,1), (4,2), (4,3), (4,4) \} \end{aligned}$$

2. Find ordered pairs satisfying $y = x$.

   The condition $y = x$ requires the second element to equal the first element:

   $$R_1 = \{(1, 1), (2, 2), (3, 3), (4, 4)\}$$

3. Determine the Domain and Range.

   • Domain: Set of all first elements:

   $$\text{Dom}(R_1) = \{1, 2, 3, 4\}$$

   • Range: Set of all second elements:

   $$\text{Range}(R_1) = \{1, 2, 3, 4\}$$

4. Cartesian Graph Representation.

   Cartesian Graph of Relation R₁ (y = x)

   Figure: Points (1,1), (2,2), (3,3), (4,4) plotted on coordinate axes.

Solution

1. Identify pairs in $A \times A$ where $x + y = 5$.

   Testing elements from $A = \{1, 2, 3, 4\}$:

   $$\begin{aligned} \text{If } x = 1 &\implies y = 5 - 1 = 4 \in A \implies (1, 4) \\ \text{If } x = 2 &\implies y = 5 - 2 = 3 \in A \implies (2, 3) \\ \text{If } x = 3 &\implies y = 5 - 3 = 2 \in A \implies (3, 2) \\ \text{If } x = 4 &\implies y = 5 - 4 = 1 \in A \implies (4, 1) \end{aligned}$$

   Thus, the relation is:

   $$R_2 = \{(1, 4), (2, 3), (3, 2), (4, 1)\}$$

2. Determine the Domain and Range.

   • Domain:

   $$\text{Dom}(R_2) = \{1, 2, 3, 4\}$$

   • Range:

   $$\text{Range}(R_2) = \{1, 2, 3, 4\}$$

3. Cartesian Graph Representation.

   Cartesian Graph of Relation R₂ (y + x = 5)

   Figure: Points (1,4), (2,3), (3,2), (4,1) plotted on coordinate axes.

Solution

1. Identify pairs in $A \times A$ where sum $x + y < 5$.

   Checking combinations from $A = \{1, 2, 3, 4\}$:

   $$\begin{aligned} x = 1 &\implies y = 1 \; (1+1=2 < 5), \; y = 2 \; (1+2=3 < 5), \; y = 3 \; (1+3=4 < 5) \\ x = 2 &\implies y = 1 \; (2+1=3 < 5), \; y = 2 \; (2+2=4 < 5) \\ x = 3 &\implies y = 1 \; (3+1=4 < 5) \\ x = 4 &\implies 4 + 1 = 5 \not< 5 \quad (\text{no valid } y) \end{aligned}$$

   Thus, the relation is:

   $$R_3 = \{(1, 1), (1, 2), (1, 3), (2, 1), (2, 2), (3, 1)\}$$

2. Determine the Domain and Range.

   • Domain:

   $$\text{Dom}(R_3) = \{1, 2, 3\}$$

   • Range:

   $$\text{Range}(R_3) = \{1, 2, 3\}$$

3. Cartesian Graph Representation.

   Cartesian Graph of Relation R₃ (x + y < 5)

   Figure: Points (1,1), (1,2), (1,3), (2,1), (2,2), (3,1) plotted on coordinate axes.

Solution

1. Identify pairs in $A \times A$ where sum $x + y > 5$.

   Checking combinations from $A = \{1, 2, 3, 4\}$:

   $$\begin{aligned} x = 1 &\implies 1 + 4 = 5 \not> 5 \quad (\text{no valid } y) \\ x = 2 &\implies 2 + 4 = 6 > 5 \implies (2, 4) \\ x = 3 &\implies 3 + 3 = 6 > 5 \implies (3, 3), \quad 3 + 4 = 7 > 5 \implies (3, 4) \\ x = 4 &\implies 4 + 2 = 6 > 5 \implies (4, 2), \quad 4 + 3 = 7 > 5 \implies (4, 3), \quad 4 + 4 = 8 > 5 \implies (4, 4) \end{aligned}$$

   Thus, the relation is:

   $$R_4 = \{(2, 4), (3, 3), (3, 4), (4, 2), (4, 3), (4, 4)\}$$

2. Determine the Domain and Range.

   • Domain:

   $$\text{Dom}(R_4) = \{2, 3, 4\}$$

   • Range:

   $$\text{Range}(R_4) = \{2, 3, 4\}$$

3. Cartesian Graph Representation.

   Cartesian Graph of Relation R₄ (x + y > 5)

   Figure: Points (2,4), (3,3), (3,4), (4,2), (4,3), (4,4) plotted on coordinate axes.

Solution

1. Write the given sets and relation.

   $$\begin{aligned} A &= \{1, 2, 3\} \\ B &= \{a, b, c, d\} \\ R &= \{(1, a), (1, b), (2, c), (3, d)\} \end{aligned}$$

2. Apply the definition of a function.

   A relation from $A$ to $B$ is a function if and only if:

   1. Every element in domain $A$ has an image in codomain $B$.
   2. Each element in domain $A$ maps to exactly one unique image in $B$.

3. Conclusion.

   In this relation, the first element $1 \in A$ is repeated in two distinct ordered pairs $(1, a)$ and $(1, b)$ (i.e. element $1$ has two images $a$ and $b$). Therefore, $R$ is NOT a function.

Solution

1. Write the given sets and relation.

   $$\begin{aligned} A &= \{a, b, c\} \\ B &= \{1, 3, 5\} \\ R &= \{(a, 1), (b, 3), (c, 5)\} \end{aligned}$$

2. Test for function validity and type.

   1. Function check: Every element in $A$ maps to exactly one image in $B$ $\implies$ It is a function.
   2. One-to-One (Injective): Distinct elements of $A$ map to distinct elements of $B$ ($a \to 1, b \to 3, c \to 5$).
   3. Onto (Surjective):

   $$\text{Range}(R) = \{1, 3, 5\} = B \quad (\text{Codomain})$$

3. Conclusion.

   Since the function is both one-to-one (injective) and onto (surjective), it is a bijective function.

Solution

1. Write the given sets and relation.

   $$\begin{aligned} A &= \{1, 2, 3\} \\ B &= \{a, b, c\} \\ R &= \{(1, a), (2, b), (3, c)\} \end{aligned}$$

2. Test for function validity and type.

   1. Function check: Every element in $A$ has a unique image in $B$ $\implies$ It is a function.
   2. One-to-One (Injective): Distinct elements of $A$ have distinct images in $B$ ($1 \to a, 2 \to b, 3 \to c$).
   3. Onto (Surjective):

   $$\text{Range}(R) = \{a, b, c\} = B \quad (\text{Codomain})$$

3. Conclusion.

   Since it is both one-to-one and onto, it is a bijective function.

Solution

1. Write the given sets and relation.

   $$\begin{aligned} A &= \{l, m, n\} \\ B &= \{x, y, z\} \\ R &= \{(l, x), (m, x), (n, z)\} \end{aligned}$$

2. Test for function validity and type.

   1. Function check: Every element of $A$ has a unique image in $B$ $\implies$ It is a function.
   2. Many-to-One: Distinct elements $l$ and $m$ share the same image $x$.
   3. Into function:

   $$\text{Range}(R) = \{x, z\} \subsetneq B = \{x, y, z\}$$

   Since element $y \in B$ has no pre-image in $A$, $\text{Range}(R) \neq \text{Codomain}(B)$.

3. Conclusion.

   Therefore, it is an into function (specifically a many-to-one into function).

Solution

1. Substitute $x = 0$ into $g(x) = 3x + 2$.

   $$\begin{aligned} g(0) &= 3(0) + 2 \\ &= 0 + 2 \\ &= 2 \end{aligned}$$

Solution

1. Substitute $x = -3$ into $g(x) = 3x + 2$.

   $$\begin{aligned} g(-3) &= 3(-3) + 2 \\ &= -9 + 2 \\ &= -7 \end{aligned}$$

Solution

1. Substitute $x = \frac{2}{3}$ into $g(x) = 3x + 2$.

   $$\begin{aligned} g\left(\frac{2}{3}\right) &= 3\left(\frac{2}{3}\right) + 2 \\ &= 2 + 2 \\ &= 4 \end{aligned}$$

Solution

1. Substitute $x = 1$ into $h(x) = x^2 + 1$.

   $$\begin{aligned} h(1) &= (1)^2 + 1 \\ &= 1 + 1 \\ &= 2 \end{aligned}$$

Solution

1. Substitute $x = -4$ into $h(x) = x^2 + 1$.

   $$\begin{aligned} h(-4) &= (-4)^2 + 1 \\ &= 16 + 1 \\ &= 17 \end{aligned}$$

Solution

1. Substitute $x = -\frac{1}{2}$ into $h(x) = x^2 + 1$.

   $$\begin{aligned} h\left(-\frac{1}{2}\right) &= \left(-\frac{1}{2}\right)^2 + 1 \\ &= \frac{1}{4} + 1 \\ &= \frac{1 + 4}{4} \\ &= \frac{5}{4} \end{aligned}$$

Solution

1. Use the given condition $f(3) = 8$.

   Substitute $x = 3$ into $f(x) = ax + b + 1$:

   $$\begin{aligned} f(3) &= a(3) + b + 1 \\ 8 &= 3a + b + 1 \\ 3a + b &= 8 - 1 \\ 3a + b &= 7 \quad \text{--- (1)} \end{aligned}$$

2. Use the given condition $f(6) = 14$.

   Substitute $x = 6$ into $f(x) = ax + b + 1$:

   $$\begin{aligned} f(6) &= a(6) + b + 1 \\ 14 &= 6a + b + 1 \\ 6a + b &= 14 - 1 \\ 6a + b &= 13 \quad \text{--- (2)} \end{aligned}$$

3. Solve the simultaneous equations for $a$.

   Subtract equation (1) from equation (2):

   $$\begin{aligned} (6a + b) - (3a + b) &= 13 - 7 \\ 3a &= 6 \\ a &= \frac{6}{3} \\ a &= 2 \end{aligned}$$

4. Substitute $a = 2$ into equation (1) to find $b$.

   $$\begin{aligned} 3(2) + b &= 7 \\ 6 + b &= 7 \\ b &= 7 - 6 \\ b &= 1 \end{aligned}$$

Solution

1. Use the given condition $g(-1) = 0$.

   Substitute $x = -1$ into $g(x) = ax + b + 5$:

   $$\begin{aligned} g(-1) &= a(-1) + b + 5 \\ 0 &= -a + b + 5 \\ a - b &= 5 \quad \text{--- (1)} \end{aligned}$$

2. Use the given condition $g(2) = 10$.

   Substitute $x = 2$ into $g(x) = ax + b + 5$:

   $$\begin{aligned} g(2) &= a(2) + b + 5 \\ 10 &= 2a + b + 5 \\ 2a + b &= 10 - 5 \\ 2a + b &= 5 \quad \text{--- (2)} \end{aligned}$$

3. Solve the simultaneous equations for $a$.

   Add equation (1) and equation (2):

   $$\begin{aligned} (a - b) + (2a + b) &= 5 + 5 \\ 3a &= 10 \\ a &= \frac{10}{3} \end{aligned}$$

4. Substitute $a = \frac{10}{3}$ into equation (1) to find $b$.

   $$\begin{aligned} \frac{10}{3} - b &= 5 \\ b &= \frac{10}{3} - 5 \\ b &= \frac{10 - 15}{3} \\ b &= -\frac{5}{3} \end{aligned}$$

Solution

1. Set up the linear equation.

   Given that $f(x) = 5x + 1$ and $f(x) = 32$:

   $$5x + 1 = 32$$

2. Solve for $x$.

   $$\begin{aligned} 5x &= 32 - 1 \\ 5x &= 31 \\ x &= \frac{31}{5} \end{aligned}$$

Solution

1. Use the given condition $f(1) = 6$.

   Substitute $x = 1$ into $f(x) = cx^2 + d$:

   $$\begin{aligned} f(1) &= c(1)^2 + d \\ 6 &= c(1) + d \\ c + d &= 6 \quad \text{--- (1)} \end{aligned}$$

2. Use the given condition $f(-2) = 10$.

   Substitute $x = -2$ into $f(x) = cx^2 + d$:

   $$\begin{aligned} f(-2) &= c(-2)^2 + d \\ 10 &= c(4) + d \\ 4c + d &= 10 \quad \text{--- (2)} \end{aligned}$$

3. Solve the simultaneous equations for $c$.

   Subtract equation (1) from equation (2):

   $$\begin{aligned} (4c + d) - (c + d) &= 10 - 6 \\ 3c &= 4 \\ c &= \frac{4}{3} \end{aligned}$$

4. Substitute $c = \frac{4}{3}$ into equation (1) to find $d$.

   $$\begin{aligned} \frac{4}{3} + d &= 6 \\ d &= 6 - \frac{4}{3} \\ d &= \frac{18 - 4}{3} \\ d &= \frac{14}{3} \end{aligned}$$