Question

The line $y=mx$ bisects the area enclosed by the curve $y=1+4x-x^2$ and the lines $x=0,x=\frac{3}{2}$ and $y=0$. Then the value of $m$ is

• A. $\frac{13}{6}$
• B. $\frac{6}{13}$
• C. $\frac{3}{2}$
• D. $4$

Answer 1

Answer: (A)

$\frac{13}{6}$

first plot all the given functions

now the the line in the black is $y=mx$

and the graph in the green is $y=1+4x-x^2$

and the one in the yellow is $x=\frac{3}{2}$

now look at the pic with labelled points

intersection point of $y=mxandx=\frac{3}{2}$ is

$(\frac{3}{2},m\frac{3}{2})$

the area of the triangle needs to be half of the total area

the area of the triangle will be $\frac{1}{2}\left(1.5\right)\left(1.5m\right)=a$

$\frac{9}{8}m=a$

the total area will be ${\int }_0^{1.5}1+4x-x^2=A$

$A=\frac{39}{8}$

now $A=2a$

hence $a=\frac{39}{16}$

so the value of m is $\frac{9}{8}m=\frac{39}{16}$

$m=\frac{13}{6}$