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 which of the following is/are true

• A. $a^2+b^2=9$
• B. $a^2+b^2=10$
• C. $a-b=4$
• D. $a-b=2$

Answer 1

Answer: (B), (C)

$a-b=4$

Since the curve passes through $(1,2),$
$a+b=2\cdots \left(i\right)$
By observation the curve also passes through $(0,0)$
Therefore the area enclosed by the curve, $x$ -axis and $x=4$ is given by
$A=\underset{0}{\overset{4}{\int }}(a{x}^{\frac{1}{2}}+bx)dx={[\frac{2a}{3}{x}^{\frac{3}{2}}+\frac{b{x}^2}{2}]}_0^4=\frac{16a}{3}+\frac{16b}{2}=8\Rightarrow \frac{2a}{3}+b=1\cdots \left(ii\right)$
Solving $\left(i\right)$ and $\left(ii\right)$, we get
$a=3,b=-1$
$\Rightarrow a^2+b^2=10,$
$a-b=4$