If the line y=mx bisects the area enclosed by | vedclass

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(D) The total area $A$ enclosed by the lines $x=0, y=0, x=\frac{3}{2}$ and the curve $y=1+4x-x^2$ is given by:$A = \int_{0}^{3/2} (1+4x-x^2) \, dx$$A

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$26$

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(D) The total area $A$ enclosed by the lines $x=0, y=0, x=\frac{3}{2}$ and the curve $y=1+4x-x^2$ is given by:$A = \int_{0}^{3/2} (1+4x-x^2) \, dx$$A = [x + 2x^2 - \frac{x^3}{3}]_{0}^{3/2}$$A = (\frac{3}{2} + 2(\frac{9}{4}) - \frac{27}{24}) - 0$$A = \frac{3}{2} + \frac{9}{2} - \frac{9}{8} = 6 - \frac{9}{8} = \frac{48-9}{8} = \frac{39}{8}$Since the line $y=mx$ bisects this area,the area of the triangle formed by the lines $x=0, x=\frac{3}{2}$ and $y=mx$ must be half of the total area.The area of the triangle with vertices $(0,0), (3/2, 0)$ and $(3/2, 3m/2)$ is:$A_{triangle} = \frac{1}{2} 	imes 	ext{base} 	imes 	ext{height} = \frac{1}{2} 	imes \frac{3}{2} 	imes \frac{3m}{2} = \frac{9m}{8}$Equating the two areas:$\frac{9m}{8} = \frac{1}{2} 	imes \frac{39}{8}$$9m = \frac{39}{2}$$m = \frac{39}{18} = \frac{13}{6}$Therefore,$12m = 12 	imes \frac{13}{6} = 2 	imes 13 = 26$.

If the line $y=mx$ bisects the area enclosed by the lines $x=0, y=0, x=\frac{3}{2}$ and the curve $y=1+4x-x^2$,then $12m$ is equal to ..... .

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