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The line $y=m x$ bisects the area enclosed by lines $x$ $=0, \mathrm{y}=0$ and $\mathrm{x}=3 / 2$ and the curve $\mathrm{y}=1+4 \mathrm{x}-\mathrm{x}^{2}$
Then the value of $\mathrm{m}$ is
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Then the value of $\mathrm{m}$ is
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Verified Answer
The correct answer is:
$\frac{13}{6}$
$y=1+4 x-x^{2}=5-(x-2)^{2}$
We have $\int_{0}^{3 / 2}\left(1+4 x-x^{2}\right) \mathrm{dx}=2 \int_{0}^{3 / 2} \operatorname{mx} \mathrm{dx}$
$=\frac{3}{2}+2\left(\frac{9}{4}\right)-\frac{1}{3}\left(\frac{27}{8}\right)=\mathrm{m} \cdot \frac{9}{4}$
On solving we get $\mathrm{m}=\frac{13}{6}$
We have $\int_{0}^{3 / 2}\left(1+4 x-x^{2}\right) \mathrm{dx}=2 \int_{0}^{3 / 2} \operatorname{mx} \mathrm{dx}$
$=\frac{3}{2}+2\left(\frac{9}{4}\right)-\frac{1}{3}\left(\frac{27}{8}\right)=\mathrm{m} \cdot \frac{9}{4}$
On solving we get $\mathrm{m}=\frac{13}{6}$
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