JEE Main 30 June 2022 Shift 1 Solved Paper

Section: Mathematics

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Question : 6 of 90

Marks: +1, -0

For two positive real numbers a and b such that ‌

1

a2

+‌

1

b3

=4, then minimum value of the constant term in the expansion of (ax‌

1

8

+bx−‌

1

12

)10 is :

[30-Jun-2022-Shift-1]

‌
105
2
‌
105
4
‌
105
8
‌
105
16

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