Question

The curve $y=a{x}^2+bx$ passes through the point $(1,2)$ and lies above the $X$-axis for $0\le x\le 8.$ If the area enclosed by this curve, the X-axis and the line $x=6$ is $108$ square units, then $2b-a=$

• A. 2
• B. 0
• C. 1
• D. -1

Answer 1

Answer: (B)

Given, curve $y=a{x}^2+bx$ passes through point $(1,2)$.
$\therefore 2=a(1{)}^2+b(1)$
$\Rightarrow a+b=2\dots$(i)
Given, area under curve $=\underset{0}{\overset{6}{\int }}\left(a{x}^2+bx\right)dx=108$
$\Rightarrow {\left[\frac{a{x}^3}{3}+\frac{b{x}^2}{2}\right]}_0^6=108$
$\Rightarrow 72a+18b=108$
$\Rightarrow 4a+b=6\dots$(ii)
By solving Eqn. (i) and (ii), we get
$a=\frac{4}{3}$ and $b=\frac{2}{3}$
$\therefore 2b-a=2\times \frac{2}{3}-\frac{4}{3}$
$=\frac{4}{3}-\frac{4}{3}=0$