Question

If the curve $y=a{x}^{\frac{1}{2}}+bx$ passes through the point $(1,2)$ and $y\ge 0$ 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

• A. $a=1,b=1$
• B. $a=3,b=-1$
• C. $a=3,b=1$
• D. $a=1,b=-1$

Answer 1

The correct option is B

$a=3,b=-1$

Since the curve passes through $(1,2),$
$a+b=2\cdots \left(1\right)$
By observation the curve also passes through $(0,0)$
Therefore the area enclosed by the curve, $x$ -axis and $x=4$ is given by
$A={\int }_0^4(a{x}^{\frac{1}{2}}+bx)dxA={[\frac{2a}{3}{x}^{\frac{3}{2}}+\frac{b{x}^2}{2}]}_0^{4}A=\frac{16a}{3}+\frac{16b}{2}=8\frac{2a}{3}+b=1\cdots \left(2\right)$
Solving $\left(1\right)$ and $\left(2\right)$, we get
$a=3,b=-1$