Let A=left{n ∈ N mid n^{2} ≤ n+10,000right}, B={3 k+1 mid k ∈ N} and C={2 k mid k ∈ N} , the

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1.Set Theory

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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 $.....$

$832$

$412$

$963$

$123$

(JEE MAIN-2021)

$\mathrm{B}-\mathrm{C} \equiv\{7,13,19, \ldots 97, \ldots\}$

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

hard

hard

(JEE MAIN-2018)

easy

hard

(JEE MAIN-2023)

hard

(JEE MAIN-2020)