BitsatArea Under The CurvesBITSAT 2024MCQ+3 / -12024The line $ y=m x $ bisects the area unclosed by lines $ x=0, y=0 $ and $ x=\frac{3}{2} $ and the curve $ y=1+4 x-x^{2} $. Then, the value of $ m $ isA$ \frac{13}{6} $B$ \frac{13}{2} $C$ \frac{13}{5} $D$ \frac{13}{7} $Check AnswerClear SelectionReveal AnswerShow Explanation
