Question

The curve $y= ax^2 + bx$ 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 $y=a x^{2}+b x$ passes through point $(1,2)$.
$ \therefore 2=a(1)^{2}+b(1) $
$ \Rightarrow a + b = 2 \dots $(i)
Given, area under curve $=\int\limits_{0}^{6}\left(a x^{2}+b x\right) d x=108$
$\Rightarrow \left[\frac{a x^{3}}{3}+\frac{b x^{2}}{2}\right]_{0}^{6}=108$
$\Rightarrow 72 a+18 b=108 $
$ \Rightarrow 4 a+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$