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Integrals · Question 32

JEE Main 2026 (Online) 21st January Evening ShiftNumericalMedium📈 Topic in 3% papers+4
1

Let [ ] denote the greatest integer function and f(x) = n 1n3 k=1n [ k23x ]. Then 12 j=1 f(j) is equal to .

Answer

2

Solution

[ k23x ] = k23x - \ k23x \, where 0 \ \ < 1

0 1n3 k=1n \ k23x \ < nn3 0

f(x) = n 13x 1n3 k=1n k2

= 13x n n(n + 1)(2n + 1)6n3 = 13x 13 = 13x+1

j=1 f(j) = 19 + 127 + = 191 - 13 = 16

12 j=1 f(j) = 12 16 = 2

Answer: 2

JEE Main 2026 (Online) 21st January Evening Shift

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