Review Exercise 3 — Class 9 Mathematics Solutions

Solution

Solution

1. Analyze the pattern of the given elements.

   Given set:

   $$\{1,\frac{1}{3},\frac{1}{5},\frac{1}{7},\dots \}=\{\frac{1}{1},\frac{1}{3},\frac{1}{5},\frac{1}{7},\dots \}$$

   The numerators are always $1$, and the denominators are the sequence of odd numbers: $1,3,5,7,\dots$.

2. Express the general term using whole numbers $W=\{0,1,2,3,\dots \}$.

   $$\begin{array}{cc}n=0& ⟹\frac{1}{2(0)+1}=\frac{1}{1}=1\\ n=1& ⟹\frac{1}{2(1)+1}=\frac{1}{3}\\ n=2& ⟹\frac{1}{2(2)+1}=\frac{1}{5}\\ n=3& ⟹\frac{1}{2(3)+1}=\frac{1}{7}\end{array}$$

   Thus, the elements follow the algebraic rule $x=\frac{1}{2n+1}$ for $n\in W$.

3. Identify the correct option.

   This matches option (b).

Solution

1. Recall the definition and formula for power set.

   The power set $P(A)$ is the set containing all possible subsets of set $A$. The total number of subsets is given by $2^{n(A)}$.

2. Evaluate for the empty set $A=\{\}=\varnothing$.

   Since $n(A)=0$:

   $$\begin{array}{cc}n(P(A))& =2^0\&=1\end{array}$$

   The only subset of the empty set is $\varnothing$ (or $\{\}$).

3. Write the power set.

   $$P(A)=\{\{\}\}(\text{or }\{\varnothing \})$$

4. Identify the correct option.

   This matches option (c).

Solution

1. Find the intersection $A\cap B$.

   $$\begin{array}{cc}A\cap B& =\{1,2,3\}\cap \{3,4,5\}\&=\{3\}\end{array}$$

2. Compute the set difference $U-(A\cap B)$.

   $$\begin{array}{cc}U-(A\cap B)& =\{1,2,3,4,5\}-\{3\}\&=\{1,2,4,5\}\end{array}$$

3. Identify the correct option.

   This corresponds to option (a).

Solution

1. Understand the definition of set difference for overlapping sets.

   For any two overlapping sets $A$ and $B$, the set $A-B$ consists of elements that belong to $A$ but do not belong to $B$.

2. Express cardinality in terms of intersections.

   $$\begin{array}{cc}n(A-B)& =n(A)-n(A\cap B)\end{array}$$

3. Identify the correct option.

   This corresponds to option (d).

Solution

1. Utilize the subset property $A\subseteq B$.

   Since $A$ is a proper subset of $B$:

   $$\begin{array}{ccc}A\cap B& =A\text{n}(A\cap B)& =n(A)\end{array}$$

2. Apply the formula for difference of sets.

   $$\begin{array}{cc}n(B-A)& =n(B)-n(A\cap B)\&=n(B)-n(A)\end{array}$$

3. Identify the correct option.

   This matches option (d).

Solution

1. State the principle of inclusion and exclusion.

   $$\begin{array}{cc}n(A\cup B)& =n(A)+n(B)-n(A\cap B)\end{array}$$

2. Substitute the given values.

   $$\begin{array}{ccc}50& =30+35-n(A\cap B)\text{5}0& =65-n(A\cap B)\end{array}$$

3. Solve for $n(A\cap B)$.

   $$\begin{array}{cc}n(A\cap B)& =65-50\&=15\end{array}$$

4. Identify the correct option.

   This matches option (b).

Solution

1. Count the number of elements in each set.

   • For set $A=\{1,2,3,4\}$: $n(A)=4$
   • For set $B=\{x,y,z\}$: $n(B)=3$

2. Apply the Cartesian product cardinality formula.

   $$\begin{array}{cc}n(A\times B)& =n(A)\times n(B)\&=4\times 3\&=12\end{array}$$

3. Identify the correct option.

   This matches option (b).

Solution

1. Substitute $x=a+1$ into the given function.

   $$\begin{array}{ccc}f(x)& =x^2-3x+2\text{f}(a+1)& =(a+1{)}^2-3(a+1)+2\end{array}$$

2. Expand the algebraic expressions.

   $$\begin{array}{ccc}(a+1{)}^2& =a^2+2a+1\text{-}3(a+1)& =-3a-3\end{array}$$

   So:

   $$\begin{array}{cc}f(a+1)& =(a^2+2a+1)-3a-3+2\end{array}$$

3. Combine like terms.

   $$\begin{array}{cc}f(a+1)& =a^2+(2a-3a)+(1-3+2)\&=a^2-a+0\&=a^2-a\end{array}$$

4. Identify the correct option.

   This corresponds to option (d).

Solution

1. Set up the linear equation.

   $$\begin{array}{ccc}f(x)& =28\text{3}x+1& =28\end{array}$$

2. Isolate the variable $x$.

   Subtract $1$ from both sides:

   $$\begin{array}{ccc}3x& =28-1\text{3}x& =27\end{array}$$

   Divide both sides by $3$:

   $$\begin{array}{ccc}x& =\frac{27}{3}\text{x}& =9\end{array}$$

3. Identify the correct option.

   This corresponds to option (a).

Solution

1. Determine the range of the function.

   From the set of ordered pairs $f=\{(1,a),(2,b),(3,b)\}$:

   $$\begin{array}{cc}\text{Domain}(f)& =\{1,2,3\}=A\\ textRange(f)& =\{a,b\}=B\end{array}$$

2. Test for surjectivity (onto function).

   Since $\text{Range}(f)=B$ (the co-domain is completely covered by the image set), the function is surjective (onto).

3. Test for injectivity (one-to-one).

   Observe that $f(2)=b$ and $f(3)=b$. Since two distinct elements $2\ne 3$ map to the same image $b$, $f$ is not injective.

4. Conclude the correct statement.

   Since $f$ is surjective but not injective, it cannot be bijective. Therefore, statement (b) is true.

Solution

Solution

1. Identify the allowed values of $n$.

   The variable $n$ belongs to whole numbers $W=\{0,1,2,3,\dots \}$ and satisfies $n<11$:

   $$\begin{array}{c}n\in \{0,1,2,3,4,5,6,7,8,9,10\}\end{array}$$

2. Compute each element $x=11n$.

   $$\begin{array}{cc}n=0& ⟹x=11(0)=0\\ n=1& ⟹x=11(1)=11\\ n=2& ⟹x=11(2)=22\\ n=3& ⟹x=11(3)=33\\ n=4& ⟹x=11(4)=44\\ n=5& ⟹x=11(5)=55\\ n=6& ⟹x=11(6)=66\\ n=7& ⟹x=11(7)=77\\ n=8& ⟹x=11(8)=88\\ n=9& ⟹x=11(9)=99\\ n=10& ⟹x=11(10)=110\end{array}$$

3. Write the set in tabular form.

   $$\begin{array}{cc}\{x|x=11n,n\in W\wedge n<11\}& =\{0,11,22,33,44,55,66,77,88,99,110\}\end{array}$$

Solution

1. Examine the range between 4 and 6.

   The only integer strictly between $4$ and $6$ is $5$.

2. Check if this integer is even.

   Since $5$ is an odd integer, $5\notin E$.

3. Conclude the tabular form.

   There is no even number strictly between $4$ and $6$, so the set has no elements.

   $$\begin{array}{cc}\{x|x\in E\wedge 4<x<6\}& =\varnothing (\text{or }\{\})\end{array}$$

Solution

1. Examine the range between 5 and 7.

   The only integer strictly between $5$ and $7$ is $6$.

2. Check if this integer is odd.

   Since $6$ is an even integer, $6\notin O$.

3. Conclude the tabular form.

   There is no odd number strictly between $5$ and $7$, so the set is empty.

   $$\begin{array}{cc}\{x|x\in O\wedge 5<x<7\}& =\varnothing (\text{or }\{\})\end{array}$$

Solution

1. Solve the equation $x^2=2$.

   Taking the square root on both sides:

   $$\begin{array}{cc}{x}^2& =2\\ x& =\pm \sqrt{2}\end{array}$$

2. Check membership in rational numbers $Q$.

   Both $\sqrt{2}$ and $-\sqrt{2}$ are irrational numbers, so:

   $$\begin{array}{cc}\sqrt{2}& \notin Q\\ -\sqrt{2}& \notin Q\end{array}$$

3. Conclude the tabular form.

   No rational number satisfies $x^2=2$. Hence, the set contains no elements.

   $$\begin{array}{cc}\{x|x\in Q\wedge x^2=2\}& =\varnothing (\text{or }\{\})\end{array}$$

Solution

1. Solve the equation $x=-x$.

   Add $x$ to both sides:

   $$\begin{array}{cc}x+x& =0\\ 2x& =0\\ x& =0\end{array}$$

2. Verify whether $0\in Q$.

   Since $0=\frac{0}{1}$, $0$ is a rational number ($0\in Q$).

3. Write the set in tabular form.

   $$\begin{array}{cc}\{x|x\in Q\wedge x=-x\}& =\{0\}\end{array}$$

Solution

1. Recall the classification of real numbers.

   The set of real numbers $R$ is the union of rational numbers $Q$ and irrational numbers $Q^{\prime }$:

   $$\begin{array}{cc}R& =Q\cup Q^{\prime }(\text{where }Q\cap Q^{\prime }=\varnothing )\end{array}$$

2. Apply the set difference definition.

   The elements of $R$ that do not belong to $Q^{\prime }$ represent the set difference $R-Q^{\prime }$:

   $$\begin{array}{cc}\{x|x\in R\wedge x\notin Q^{\prime }\}& =R-Q^{\prime }\\ & =Q\end{array}$$

3. Conclude the result.

   The set is the set of all rational numbers $Q$.

Solution

Solution

1. State the definition of complement.

   $$\begin{array}{cc}{A}^{\prime }& =U-A\end{array}$$

2. Substitute the given sets.

   $$\begin{array}{cc}{A}^{\prime }& =\{1,2,3,4,5,6,7,8,9,10\}-\{2,4,6,8,10\}\end{array}$$

3. Remove all even numbers from $U$.

   $$\begin{array}{cc}{A}^{\prime }& =\{1,3,5,7,9\}\end{array}$$

Solution

1. State the definition of complement.

   $$\begin{array}{cc}{B}^{\prime }& =U-B\end{array}$$

2. Substitute the given sets.

   $$\begin{array}{cc}{B}^{\prime }& =\{1,2,3,4,5,6,7,8,9,10\}-\{1,2,3,4,5\}\end{array}$$

3. Remove the first 5 natural numbers from $U$.

   $$\begin{array}{cc}{B}^{\prime }& =\{6,7,8,9,10\}\end{array}$$

Solution

1. Set up the union of $A$ and $B$.

   $$\begin{array}{cc}A\cup B& =\{2,4,6,8,10\}\cup \{1,2,3,4,5\}\end{array}$$

2. Collect all unique elements in ascending order.

   $$\begin{array}{cc}A\cup B& =\{1,2,3,4,5,6,8,10\}\end{array}$$

Solution

1. Set up the difference $A-B$.

   $$\begin{array}{cc}A-B& =\{2,4,6,8,10\}-\{1,2,3,4,5\}\end{array}$$

2. Eliminate elements of $B$ that appear in $A$.

   The common elements are $2$ and $4$. Removing them from $A$:

   $$\begin{array}{cc}A-B& =\{6,8,10\}\end{array}$$

Solution

1. Set up the intersection $A\cap C$.

   $$\begin{array}{cc}A\cap C& =\{2,4,6,8,10\}\cap \{1,3,5,7,9\}\end{array}$$

2. Identify common elements.

   Since set $A$ consists exclusively of even numbers and set $C$ consists exclusively of odd numbers, they share no elements in common.

   $$\begin{array}{cc}A\cap C& =\varnothing (\text{or }\{\})\end{array}$$

Solution

1. Compute $A^{\prime }$.

   $$\begin{array}{cc}{A}^{\prime }& =U-A\\ & =\{1,2,3,4,5,6,7,8,9,10\}-\{2,4,6,8,10\}\\ & =\{1,3,5,7,9\}\end{array}$$

2. Compute $C^{\prime }$.

   $$\begin{array}{cc}{C}^{\prime }& =U-C\\ & =\{1,2,3,4,5,6,7,8,9,10\}-\{1,3,5,7,9\}\\ & =\{2,4,6,8,10\}\end{array}$$

3. Take the union $A^{\prime }\cup C^{\prime }$.

   $$\begin{array}{cc}{A}^{\prime }\cup C^{\prime }& =\{1,3,5,7,9\}\cup \{2,4,6,8,10\}\\ & =\{1,2,3,4,5,6,7,8,9,10\}\\ & =U\end{array}$$

Solution

1. Compute $A^{\prime }$.

   $$\begin{array}{cc}{A}^{\prime }& =U-A\\ & =\{1,2,3,4,5,6,7,8,9,10\}-\{2,4,6,8,10\}\\ & =\{1,3,5,7,9\}\end{array}$$

2. Take the union $A^{\prime }\cup C$.

   Given $C=\{1,3,5,7,9\}$:

   $$\begin{array}{cc}{A}^{\prime }\cup C& =\{1,3,5,7,9\}\cup \{1,3,5,7,9\}\\ & =\{1,3,5,7,9\}\\ & =C\end{array}$$

Solution

1. State the definition of the universal complement.

   $$\begin{array}{cc}{U}^{\prime }& =U-U\end{array}$$

2. Subtract $U$ from itself.

   $$\begin{array}{cc}{U}^{\prime }& =\{1,2,3,4,5,6,7,8,9,10\}-\{1,2,3,4,5,6,7,8,9,10\}\\ & =\varnothing (\text{or }\{\})\end{array}$$

Solution

Solution

1. Recall the definition of the complement of a set.

   The complement of a set $A$ with respect to the universal set $U$, denoted by $A^{\prime }$ (or $U\setminus A$), is defined as the set of all elements in $U$ that do not belong to $A$:

   $$\begin{array}{cc}{A}^{\prime }& =\{x\in U|x\notin A\}\end{array}$$

2. Express as a single set difference.

   Subtracting set $A$ from the universal set $U$:

   $$\begin{array}{cc}{A}^{\prime }& =U-A\end{array}$$

3. Venn Diagram Representation.

   Venn Diagram of A' = U - Avenn diagram

   Figure: The region inside rectangle U but outside circle A represents A' = U - A.

   Therefore, the single set equal to $A^{\prime }$ is $U-A$.

Solution

1. Understand the definition of intersection with the universal set.

   The intersection $A\cap U$ contains all elements that belong simultaneously to set $A$ and the universal set $U$:

   $$\begin{array}{cc}A\cap U& =\{x|x\in A\wedge x\in U\}\end{array}$$

2. Apply the subset property and Identity Law.

   Since set $A$ is a subset of the universal set $U$ ($A\subseteq U$), every element of $A$ is already contained in $U$. Hence, taking the common elements yields:

   $$\begin{array}{cc}A\cap U& =A\end{array}$$

3. Venn Diagram Interpretation.

   Venn Diagram of A ∩ U = Avenn diagram

   Figure: The region common to circle A and rectangle U is the entire circle A.

   Therefore, the single set equal to $A\cap U$ is $A$.

Solution

1. Recall the definition of union with the universal set.

   The union $A\cup U$ contains all elements belonging to set $A$, universal set $U$, or both:

   $$\begin{array}{cc}A\cup U& =\{x|x\in A\vee x\in U\}\end{array}$$

2. Apply the Dominance Law (Universal Bound Law).

   Because $A$ is entirely contained within $U$ ($A\subseteq U$), the collection of all elements in either $A$ or $U$ is simply the universal set $U$:

   $$\begin{array}{cc}A\cup U& =U\end{array}$$

3. Venn Diagram Interpretation.

   Venn Diagram of A ∪ U = Uvenn diagram

   Figure: Combining circle A with the universal rectangle U covers the entire rectangle U.

   Therefore, the single set equal to $A\cup U$ is $U$.

Solution

1. Recall the definition of union with the empty set.

   The union $A\cup \varnothing$ contains all elements that belong to set $A$ or to the empty set $\varnothing$:

   $$\begin{array}{cc}A\cup \varnothing & =\{x|x\in A\vee x\in \varnothing \}\end{array}$$

2. Apply the Identity Law for union.

   Since the empty set contains no elements ($n(\varnothing )=0$), uniting $A$ with $\varnothing$ does not introduce any additional elements:

   $$\begin{array}{cc}A\cup \varnothing & =A\end{array}$$

3. Venn Diagram Interpretation.

   In a Venn diagram, the empty set $\varnothing$ corresponds to a region containing no points. Uniting circle $A$ with no additional region leaves circle $A$ unchanged.

   Therefore, the single set equal to $A\cup \varnothing$ is $A$.

Solution

1. Recall the definition of intersection of sets.

   The intersection $\varnothing \cap \varnothing$ is the set containing all elements common to both empty sets:

   $$\begin{array}{cc}\varnothing \cap \varnothing & =\{x|x\in \varnothing \wedge x\in \varnothing \}\end{array}$$

2. Apply the Idempotent Law for intersection.

   By the idempotent law ($X\cap X=X$), intersecting any set with itself returns the original set:

   $$\begin{array}{cc}\varnothing \cap \varnothing & =\varnothing \end{array}$$

3. Conclusion.

   Since neither operand contains any elements, their intersection is also completely empty.

   Therefore, the single set equal to $\varnothing \cap \varnothing$ is $\varnothing$.

Solution

Solution

Solution

1. Partition the Universal Set $U$ into Disjoint Regions.

   For two overlapping sets $A$ and $B$ in universal set $U$:

   • Region I: Elements in $A$ only ($A-B$)
   • Region II: Elements in both $A$ and $B$ ($A\cap B$)
   • Region III: Elements in $B$ only ($B-A$)
   • Region IV: Elements outside both sets ($(A\cup B{)}^{\prime }$)

   Thus:

   $$\begin{array}{cc}U& =\text{Region I}\cup \text{Region II}\cup \text{Region III}\cup \text{Region IV}\\ A& =\text{Region I}\cup \text{Region II}\\ B& =\text{Region II}\cup \text{Region III}\end{array}$$

   ---

2. Evaluate the Left-Hand Side (L.H.S).

   First, determine the set difference $A-B$:

   $$\begin{array}{cc}A-B& =\text{Region I}\end{array}$$

   Next, find its complement $(A-B{)}^{\prime }=U-(A-B)$:

   $$\begin{array}{cc}(A-B{)}^{\prime }& =U-\text{Region I}\\ & =\text{Region II}\cup \text{Region III}\cup \text{Region IV}\end{array}$$

   Now, intersect with set $B$:

   $$\begin{array}{cc}\text{L.H.S}& =(A-B{)}^{\prime }\cap B\\ & =(\text{Region II}\cup \text{Region III}\cup \text{Region IV})\cap (\text{Region II}\cup \text{Region III})\\ & =\text{Region II}\cup \text{Region III}\text{— (1)}\end{array}$$

   ---

3. Evaluate the Right-Hand Side (R.H.S).

   $$\begin{array}{cc}\text{R.H.S}& =B\\ & =\text{Region II}\cup \text{Region III}\text{— (2)}\end{array}$$

   From (1) and (2), $\text{L.H.S}=\text{R.H.S}=\text{Region II}\cup \text{Region III}$.

   ---

4. Venn Diagram Representation

   Venn Diagram Verification: (A - B)' ∩ B = Bvenn diagram

   Figure: (A - B)' contains Regions II, III, and IV. Its intersection with B (Regions II and III) yields Set B exactly.

   ---

5. Conclusion

   The Venn diagram regional analysis confirms:

   $$\begin{array}{cc}\text{L.H.S}& =\text{Region II}\cup \text{Region III}\\ \text{R.H.S}& =\text{Region II}\cup \text{Region III}\\ \therefore (A-B{)}^{\prime }\cap B& =B\end{array}$$

   Hence, the identity is verified using Venn diagrams.