Question

For $b>0$, let $A_1$ be the area bounded by $x=0,x+y=1,y=b{x}^2$ and $A_2$ be the area bounded by $y=0,x+y=1,y=b{x}^2$ such that $A_1:A_2=11:16$, then the value of $b$ is

• A. 2
• B. 3
• C. 4
• D. 6

Answer 1

Answer: (D)

$A_2=\underset{0}{\overset{x_1}{\int }}\left(b{x}^2\right)dx+\frac{1}{2}\left(1-x_1\right)\cdot b{x}_1^2$
$=\frac{b{x}_1^2}{6}\cdot \left(3-x_1\right)=\frac{\left(1-x_1\right)\left(3-x_1\right)}{6}\left(As,b{x}_1^2=1-x_1\right)$
Now,$\frac{A_1}{A_2}=\frac{11}{16}\Rightarrow \frac{A_1+A_2}{A_2}=\frac{27}{16}\left(\text{ As, }{A}_1+A_2=\frac{1}{2}(1)(1)=\frac{1}{2}\right)$
$\Rightarrow A_2=\frac{8}{27}.$
$\therefore \frac{\left(1-x_1\right)\left(3-x_1\right)}{6}=\frac{8}{27}\Rightarrow x_1=\frac{1}{3}\Rightarrow b=6$