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(D) Given the curve $y = a\sqrt{x} + bx$ passes through $(1, 2)$,we have $2 = a(1) + b(1)$,so $a + b = 2$. The area bounded by the curve,$x=4$,and the

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$4$

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(D) Given the curve $y = a\sqrt{x} + bx$ passes through $(1, 2)$,we have $2 = a(1) + b(1)$,so $a + b = 2$. The area bounded by the curve,$x=4$,and the $X$-axis is given by $\int_{0}^{4} (a\sqrt{x} + bx) dx = 8$. Evaluating the integral: $[a \cdot \frac{2}{3}x^{3/2} + b \cdot \frac{x^2}{2}]_{0}^{4} = 8$. Substituting $x=4$: $a \cdot \frac{2}{3}(8) + b \cdot \frac{16}{2} = 8$,which simplifies to $\frac{16}{3}a + 8b = 8$. Dividing by $8$: $\frac{2}{3}a + b = 1$. We have the system: $1) a + b = 2$ $2) \frac{2}{3}a + b = 1$ Subtracting $(2)$ from $(1)$: $(a - \frac{2}{3}a) = 2 - 1$,so $\frac{1}{3}a = 1$,which gives $a = 3$. Substituting $a=3$ into $a+b=2$: $3 + b = 2$,so $b = -1$. Therefore,$a - b = 3 - (-1) = 4$.

If a curve $y=a \sqrt{x}+b x$ passes through the point $(1,2)$ and the area bounded by this curve,line $x=4$ and the $X$-axis is $8$ sq.units,then the value of $a-b$ is

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