Question

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 – x². The value of m is -

• A. $\frac { 13 } { 6 }$
• B. $\frac { 13 } { 8 }$
• C. $\frac { 8 } { 13 }$
• D. $\frac { 6 } { 13 }$

Answer 1

Answer: (A)

$\frac { 13 } { 6 }$

Answer 2

y = 1 + 4x – x² .. (1) gives (x – 2)² = – (y – 5), which is parabola with vertex at (2, 5). Also it cuts x-axis, where (x – 2)² = – (0 – 5)

̃ x = 2 ± $\sqrt { 5 }$

̃ x = 2 + $\sqrt { 5 }$, x =2 – $\sqrt { 5 }$ < 0

The line y = mx divides the shaded area ODEBCO bounded by x = 0, y = 0, x = 3/2 and parabola into two equal parts ODEO and OEBCO.

\ Area ODEBCO = 2. Area ODE.

$\int _ { 0 } ^ { 3 / 2 } \left( 1 + 4 x - x ^ { 2 } \right) d x = 2$ $\int _ { 0 } ^ { 3 / 2 } m x d x$

̃ $\frac { 3 } { 2 }$+ 2$\left( \frac { 3 } { 2 } \right) ^ { 2 }$– $\frac { 1 } { 3 }$ $\left( \frac { 3 } { 2 } \right) ^ { 3 }$ = m$\left( \frac { 3 } { 2 } \right) ^ { 2 }$

̃ m =$\frac { 13 } { 6 }$