If the curve $\mathrm{y}={\mathrm{ax}}^{1/2}+\mathrm{bx}$ passes through the point (1, 2) and lies above the x-axis for $0\le x\le 9$ and the area enclosed
by the curve, the x-axis, and the line x = 4 is 8 sq. units, then

Since the curve $\mathrm{y}={\mathrm{ax}}^{1/2}+\mathrm{bx}$ passes through the point (1, 2)
.'. 2=a + b  ----------(1)
By observation the curve also passes through (0, 0).
Therefore, the area enclosed by the curve, x-axis and x=4 is given by

$\mathrm{A}={\int }_0^4 \left({\mathrm{ax}}^{1/2}+\mathrm{bx}\right)\mathrm{dx}=8$

$\begin{array}{l}\mathrm{or} \frac{2\mathrm{a}}{3}\times 8+\frac{\mathrm{b}}{2}\times 16=8\\ \mathrm{or} \frac{2\mathrm{a}}{3}+\mathrm{b}=1----\left(2\right)\end{array}$

Solving (1) and (2), we get a=3, b = -1