Question

If the curve $y=a{x}^{\frac{1}{2}}+bx$ 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

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

Answer 1

Answer: (C), (D)

$a=3$
$b=-1$

Since the curve $y=a{x}^{\frac{1}{2}}+bx$ passes through the point $(1,2)$
$\therefore 2=a+b\left(1\right)$
By observation the curve also passes through $(0,0)$.
Therefore, the area enclosed by the curve, $x-axis$ and $x=4$ is given by
$A={\int }_0^4(a{x}^{\frac{1}{2}}+bx)dx=8\Rightarrow \frac{2a}{3}\times 8+\frac{b}{12}\times 16=8$
$\Rightarrow \frac{2a}{3}+b=1\left(2\right)$
Solving $\left(1\right),\left(2\right)$ we get $a=3,b=-1$.