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=\int\limits_0^{x_1}\left(b x^2\right) d x+\frac{1}{2}\left(1-x_1\right) \cdot b x_1^2$
$=\frac{ bx _1^2}{6} \cdot\left(3- x _1\right)=\frac{\left(1- x _1\right)\left(3- x _1\right)}{6}\left( As , bx _1^2=1- x _1\right)$
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} . $
$\therefore \frac{\left(1- x _1\right)\left(3- x _1\right)}{6}=\frac{8}{27} \Rightarrow x _1=\frac{1}{3} \Rightarrow b =6$