The curve y=ax^2+bx passes through the point | vedclass

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(B) Given,the curve $y=ax^2+bx$ passes through the point $(1,2)$.$	herefore 2 = a(1)^2 + b(1) \Rightarrow a+b=2$ ... $(i)$Given that the curve lies a

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(B) Given,the curve $y=ax^2+bx$ passes through the point $(1,2)$.$	herefore 2 = a(1)^2 + b(1) \Rightarrow a+b=2$ ... $(i)$Given that the curve lies above the $X$-axis for $0 \leq x \leq 8$,the area enclosed by the curve,the $X$-axis,and the line $x=6$ is given by:$\int_0^6 (ax^2+bx) dx = 108$$\Rightarrow \left[ \frac{ax^3}{3} + \frac{bx^2}{2} ight]_0^6 = 108$$\Rightarrow \frac{a(216)}{3} + \frac{b(36)}{2} = 108$$\Rightarrow 72a + 18b = 108$Dividing by $18$,we get $4a + b = 6$ ... $(ii)$Subtracting $(i)$ from $(ii)$:$(4a+b) - (a+b) = 6 - 2$$3a = 4 \Rightarrow a = \frac{4}{3}$Substituting $a = \frac{4}{3}$ in $(i)$:$\frac{4}{3} + b = 2 \Rightarrow b = 2 - \frac{4}{3} = \frac{2}{3}$Now,calculate $2b-a$:$2b-a = 2\left(\frac{2}{3}ight) - \frac{4}{3} = \frac{4}{3} - \frac{4}{3} = 0$

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

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