JEE Mains · Maths · STD 11 - 1. set theory
Let \(A=\left\{n \in N \mid n^{2} \leq n+10,000\right\}, B=\{3 k+1 \mid k \in N\}\) and \(C=\{2 k \mid k \in N\}\), then the sum of all the elements of the set \(A \cap(B-C)\) is equal to \(.....\)
- A \(832\)
- B \(412\)
- C \(963\)
- D \(123\)
Answer & Solution
Correct Answer
(A) \(832\)
Step-by-step Solution
Detailed explanation
\(\mathrm{B}-\mathrm{C} \equiv\{7,13,19, \ldots 97, \ldots\}\) 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…
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