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1.Set Theory
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જો $A=\left\{n \in N \mid n^{2} \leq n+10,000\right\}, B=\{3 k+1 \mid k \in N\}$ અને $C=\{2 k \mid k \in N\}$ હોય તો ગણ $A \cap(B-C)$ ના બધાજ ઘટકોનો સરવાળો મેળવો.
A
$832$
B
$412$
C
$963$
D
$123$
(JEE MAIN-2021)
Solution
$\mathrm{B}-\mathrm{C} \equiv\{7,13,19, \ldots 97, \ldots\}$
Now, $n^{2}-n \leq 100 \times 100$
$\Rightarrow \mathrm{n}(\mathrm{n}-1) \leq 100 \times 100$
$\Rightarrow \mathrm{A}=\{1,2, \ldots, 100\}$
So, $A \cap(B-C)=\{7,13,19, \ldots, 97\}$
Hence, sum $=\frac{16}{2}(7+97)=832$
Standard 11
Mathematics