JEE Main 20208 Jan, Evening shiftMathematicsMediumIntegral Calculus369× since 2002If I=∫12dx2x3−9x2+12x+4I = \int\limits_1^2 {{{dx} \over {\sqrt {2{x^3} - 9{x^2} + 12x + 4} }}}I=1∫22x3−9x2+12x+4dx, then :AOption A. 116<I2<19{1 \over 16} < {I^2} < {1 \over 9}161<I2<91BOption B. 18<I2<14{1 \over 8} < {I^2} < {1 \over 4}81<I2<41COption C. 19<I2<18{1 \over 9} < {I^2} < {1 \over 8}91<I2<81DOption D. 16<I2<12{1 \over 6} < {I^2} < {1 \over 2}61<I2<21CheckSkip · show solution
