Exercise 3.3 — Class 9 Mathematics Solutions

Solution

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

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

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

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)cartesian plane

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

   Diagram Key & Elementsmobile-optimized

   (1, 1)

   (2, 2)

   (3, 3)

   (4, 4)

Solution

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

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

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

   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)cartesian plane

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

   Diagram Key & Elementsmobile-optimized

   (1, 4)

   (2, 3)

   (3, 2)

   (4, 1)

Solution

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

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

   $$\begin{array}{cc}x=1& ⟹y=1(1+1=2<5),y=2(1+2=3<5),y=3(1+3=4<5)\\ x=2& ⟹y=1(2+1=3<5),y=2(2+2=4<5)\\ x=3& ⟹y=1(3+1=4<5)\\ x=4& ⟹4+1=5≮5(\text{no valid }y)\end{array}$$

   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)cartesian plane

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

   Diagram Key & Elementsmobile-optimized

   (1, 1)

   (1, 2)

   (1, 3)

   (2, 1)

   (2, 2)

   (3, 1)

Solution

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

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

   $$\begin{array}{cc}x=1& ⟹1+4=5≯5(\text{no valid }y)\\ x=2& ⟹2+4=6>5⟹(2,4)\\ x=3& ⟹3+3=6>5⟹(3,3),3+4=7>5⟹(3,4)\\ x=4& ⟹4+2=6>5⟹(4,2),4+3=7>5⟹(4,3),4+4=8>5⟹(4,4)\end{array}$$

   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)cartesian plane

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

   Diagram Key & Elementsmobile-optimized

   (2, 4)

   (3, 3)

   (3, 4)

   (4, 2)

   (4, 3)

   (4, 4)

Solution

1. Write the given sets and relation.

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

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.

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

2. Test for function validity and type.

   1. Function check: Every element in $A$ maps to exactly one image in $B$ $⟹$ 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):

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

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.

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

2. Test for function validity and type.

   1. Function check: Every element in $A$ has a unique image in $B$ $⟹$ 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):

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

3. Conclusion.

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

Solution

1. Write the given sets and relation.

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

2. Test for function validity and type.

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

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

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

3. Conclusion.

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

Solution

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Solution

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

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

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

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

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

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

3. Solve the simultaneous equations for $a$.

   Subtract equation (1) from equation (2):

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

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

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

Solution

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

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

   $$\begin{array}{cc}g(-1)& =a(-1)+b+5\\ 0& =-a+b+5\\ a-b& =5\text{— (1)}\end{array}$$

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

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

   $$\begin{array}{cc}g(2)& =a(2)+b+5\\ 10& =2a+b+5\\ 2a+b& =10-5\\ 2a+b& =5\text{— (2)}\end{array}$$

3. Solve the simultaneous equations for $a$.

   Add equation (1) and equation (2):

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

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

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

Solution

1. Set up the linear equation.

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

   $$5x+1=32$$

2. Solve for $x$.

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

Solution

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

   Substitute $x=1$ into $f(x)=c{x}^2+d$:

   $$\begin{array}{cc}f(1)& =c(1{)}^2+d\\ 6& =c(1)+d\\ c+d& =6\text{— (1)}\end{array}$$

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

   Substitute $x=-2$ into $f(x)=c{x}^2+d$:

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

3. Solve the simultaneous equations for $c$.

   Subtract equation (1) from equation (2):

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

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

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