Q.Let $A = \{x \in W : x < 2\}$, where W is whole number; $B = \{x \in N : 1 < x \le 4\}$, where N is natural number; $C = \{3,5\}$. Verify that $A \times (B \cup C) = (A \times B) \cup (A \times C)$ OR Find the domain and range of the function $f(x) = \sqrt{9 - x^2}$.

Showing the 12 most recent of 27 on this concept.

1. CBSE 2026Set ANNUAL1 markMCQ
   Q.If $A = \{1, 2\}$ and $B = \{3, 4, 5\}$ then number of relations from $A$ to $B$ is

   (a) 6

   (b) 36

   (c) 32

   (d) 64

   ›Reveal solutionSolution

   A relation from $A$ to $B$ is any subset of $A \times B$; with $|A \times B| = 6$, there are $2^6 = 64$ subsets, hence 64 relations.

   A relation from $A$ to $B$ is defined as any subset of the Cartesian product $A \times B$. Here $|A| = 2$ and $|B| = 3$, so $|A \times B| = 2 \times 3 = 6$. The total number of subsets of a set with 6 elements is $2^6 = 64$. Since every subset of …

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2. CBSE 2026Set ANNUAL1 markMCQ
   Q.Let $A = \{1, 2\}$ and $B = \{3, 4\}$, then the number of relations from set A to set B will be:

   (a) 4

   (b) $2^4$

   (c) 2

   (d) 1

   ›Reveal solutionSolution

   The number of relations from a set A to a set B is $2^{|A|\cdot|B|}$, since every relation is a subset of $A \times B$.

   Given $A=\{1,2\}$, $B=\{3,4\}$, so $|A|=2$, $|B|=2$.

   $A \times B$ has $|A|\times|B| = 2 \times 2 = 4$ ordered pairs: $(1,3),(1,4),(2,3),(2,4)$.

   …

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3. CBSE 2026Set ANNUAL1 mark
   Q.Write True/False: If Cartesian product of two sets A and B is $A \times B = \{(p, q), (p, r)\}$, then $A = \{p, q, r\}$.

   ›Reveal solutionSolution

   In $A\times B=\{(p,q),(p,r)\}$, A is the set of first coordinates and B is the set of second coordinates; here $A=\{p\}$, not $\{p,q,r\}$.

   Every ordered pair $(x,y)\in A\times B$ has $x\in A$ and $y\in B$.

   Here both ordered pairs $(p,q)$ and $(p,r)$ have first component $p$, so the set of first components — which is exactly A — is $A=\{p\}$.

   …

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4. CBSE 2025Set ANNUAL1 markMCQ
   Q.$A = \{1, 2\}, B = \{3\} \Rightarrow A \times B =$

   (a) $\{1, 2, 3\}$

   (b) $\{(1,3), (2,3), (1,2)\}$

   (c) $\{(1,3), (2,3)\}$

   (d) $\{(1,2), (3,1)\}$

   ›Reveal solutionSolution

   $A \times B = \{(1,3), (2,3)\}$.

   For sets $A$ and $B$, $A \times B = \{(a, b) : a \in A, b \in B\}$ — every element of $A$ is paired, in order, with every element of $B$.

   …

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5. CBSE 2025Set ANNUAL1 markMCQ
   Q.If $A = \{a, b\}, B = \{c, d\}$ then the number of relations from $A$ to $B =$

   (a) 8

   (b) 16

   (c) 32

   (d) 64

   ›Reveal solutionSolution

   The number of relations from $A$ to $B$ is $2^{|A \times B|} = 2^4 = 16$.

   A relation from $A$ to $B$ is defined as any subset of the Cartesian product $A \times B$. If $A$ has $m$ elements and $B$ has $n$ elements, $A \times B$ has $mn$ elements, and a set with $mn$ elements has $2^{mn}$ subsets.

   …

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6. CBSE 2025Set ANNUAL1 markMCQ
   Q.If A = {1, 2} and B = {3, 4}, then A × B is:

   (a) {3, 4, 6, 8}

   (b) {3, 8}

   (c) {(1, 3), (1, 4), (2, 3), (2, 4)}

   (d) None of these

   ›Reveal solutionSolution

   $A \times B$ is the set of all ordered pairs $(a,b)$ with $a \in A$, $b \in B$.

   Given $A=\{1,2\}$ and $B=\{3,4\}$.

   Pair each element of $A$ with every element of $B$:

   $1 \to (1,3),(1,4)$

   $2 \to (2,3),(2,4)$

   …

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7. CBSE 2025Set ANNUAL1 mark
   Q.If $A = \{x, y, z\}$ and $B = \{1, 2\}$, write the number of relations from $A$ to $B$.

   ›Reveal solutionSolution

   A relation from $A$ to $B$ is any subset of $A \times B$; with $|A|=3$ and $|B|=2$, $|A \times B| = 6$, so there are $2^6 = 64$ relations.

   $A = \{x, y, z\}$ has 3 elements, $B = \{1, 2\}$ has 2 elements.

   …

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8. CBSE 2024Set ANNUAL1 markMCQ
   Q.If set $A$ has 3 elements and set $B = \{3, 4, 5\}$, then number of elements in $(A \times B)$ will be —

   (a) 8

   (b) 9

   (c) 10

   (d) 6

   ›Reveal solutionSolution

   If $A$ has $m$ elements and $B$ has $n$ elements, then $A\times B$ has $mn$ elements.

   The Cartesian product $A\times B$ consists of all ordered pairs $(a,b)$ with $a\in A$ and $b\in B$. Since $A$ has 3 elements and $B=\{3,4,5\}$ has 3 elements, each of the 3 choices for the first coordinate can be paire …

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9. CBSE 2024Set ANNUAL1 markMCQ
   Q.Let A = {x, y, z} and B = {1, 2}, then number of relations from A into B will be:

   (a) 6

   (b) 9

   (c) 24

   (d) 64

   ›Reveal solutionSolution

   A relation from A to B is any subset of $A\times B$; with $|A|=3,|B|=2$, there are $2^{3\times2}=64$ subsets.

   A relation from set $A$ to set $B$ is defined as any subset of the Cartesian product $A \times B$.

   $|A| = 3$ (elements $x,y,z$), $|B| = 2$ (elements $1,2$).

   So $|A \times B| = 3 \times 2 = 6$.

   …

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10. CBSE 2024Set ANNUAL1 markMCQ
    Q.If $n((A\times B)\cap(A\times C))=8$ and $n(B\cap C)=2$ then $n(A)$ is:

    (a) $8$

    (b) $6$

    (c) $16$

    (d) $4$

    ›Reveal solutionSolution

    Using $(A\times B)\cap(A\times C)=A\times(B\cap C)$, we get $n(A)=4$.

    For sets, $(A\times B)\cap(A\times C)=A\times(B\cap C)$, so

    $n\big((A\times B)\cap(A\times C)\big)=n(A)\times n(B\cap C).$ …

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11. CBSE 2024Set ANNUAL1 markMCQ
    Q.If R is a relation on a finite set A having $n$ elements, then the number of relations on A is:

    (a) $2^n$

    (b) $2^{n^2}$

    (c) $n^2$

    (d) $n^n$.

    ›Reveal solutionSolution

    The number of relations on a set of $n$ elements is $2^{n^2}$.

    A relation $R$ on set $A$ is defined as any subset of the Cartesian product $A\times A$. Since $A$ has $n$ elements, $A\times A$ has $n\times n = n^2$ ordered pairs. The number of subsets of a set with $n^2$ elements is $2^{n^2}$ (by the subset-counting rule). Each such subset is a valid relation on $A$ (including the …

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12. CBSE 2023Set ANNUAL1 markMCQ
    Q.If $A = \{a, b, c\}$ and $B = \{p, q, r\}$, then $n(A \times B) =$

    (a) 8

    (b) 5

    (c) $2^9$

    (d) 9

    ›Reveal solutionSolution

    The Cartesian product's size is the product of the two sets' sizes: $3 \times 3 = 9$.

    The Cartesian product $A \times B$ consists of all ordered pairs $(a, b)$ where $a \in A$ and $b \in B$. For each of the $n(A)$ choices of $a$, there are $n(B)$ choices of $b$, giving:

    $n(A \times B) = n(A) \cdot n(B)$ …

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