JEE MAIN - Mathematics Hindi (2020 - 8th January Evening Slot - No. 9)
यदि $$I = \int\limits_1^2 {{{dx} \over {\sqrt {2{x^3} - 9{x^2} + 12x + 4} }}} $$, तो :
$${1 \over 16} < {I^2} < {1 \over 9}$$
$${1 \over 8} < {I^2} < {1 \over 4}$$
$${1 \over 9} < {I^2} < {1 \over 8}$$
$${1 \over 6} < {I^2} < {1 \over 2}$$
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