JEE Main 12 Jan 2019 Shift 2 Solved Paper

Section: Mathematics

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Question : 45 of 90

Marks: +1, -0

lim

n→∞

(

n

n2+12

+

n

n2+22

+

n

n2+32

+ ... +

1

5n

)
is equal to :

[12 Jan 2019 Shift 2]

π
4
tan−1 (2)
tan−1 (3)
π
2

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