JEE Main 11 Jan 2019 Shift 2 Solved Paper

Section: Mathematics

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Question : 59 of 90

Marks: +1, -0

Let x, y be positive real numbers and m, n positiveintegers. The maximum value of the expression

xmym

(1+x2m)(1+y2n)

is :

[11 Jan 2019 Shift 2]

1
2
1
4
m+n
6mn
1

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