(A) Given,$n(A)=p, n(B)=q$ and $n(A \times B)=7$. Since,$n(A \times B)=n(A) \times n(B)$,we have $p \times q = 7$. Since $p$ and $q$ are the number of

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(A) Given,$n(A)=p, n(B)=q$ and $n(A \times B)=7$. Since,$n(A \times B)=n(A) \times n(B)$,we have $p \times q = 7$. Since $p$ and $q$ are the number of elements in sets,they must be positive integers. The factors of $7$ are $1$ and $7$. Thus,the possible values for $(p, q)$ are $(7, 1)$ or $(1, 7)$. In both cases,$p^2+q^2 = 7^2 + 1^2 = 49 + 1 = 50$.

Class 11 Mathematics Relations and Functions Cartesian Products of Sets

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Suppose that the number of elements in set $A$ is $p$,the number of elements in set $B$ is $q$ and the number of elements in $A \times B$ is $7$,then $p^2+q^2=$

KCET 2022 All KCET

A
$50$
B
$51$
C
$42$
D
$49$

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Suppose that the number of elements in set $A$ is $p$,the number of elements in set $B$ is $q$ and the number of elements in $A \times B$ is $7$,then $p^2+q^2=$

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