Q25 / 69 · JEE Advanced 2011 (Paper 1) · Mathematics
Let the straight line $x=b$ divide the area enclosed by $y=(1-x)^2, y=0$ and $x=0$ into two parts $R_1(0 \leq x \leq b)$ and $R_2(b \leq x \leq 1)$ such that $R_1-R_2=\frac{1}{4}$. Then, $b$ equals to
- A$\frac{3}{4}$
- Correct answerB$\frac{1}{2}$
- C$\frac{1}{3}$
- D$\frac{1}{4}$
Solution
Here, area between 0 to $b$ is $R_1$ and $b$ to
$$
\begin{aligned}
& 1 \text { is } R_2 \\
& \therefore \int_0^b(1-x)^2 d x-\int_b^1(1-x)^2 d x=\frac{1}{4} \\
& \Rightarrow \quad\left(\frac{(1-x)^3}{-3}\right)_0^b-\left(\frac{(1-x)^3}{-3}\right)_b^1=\frac{1}{4} \\
& \Rightarrow-\frac{1}{3}\left\{(1-b)^3-1\right\}+\frac{1}{3}\left\{0-(1-b)^3\right\} \\
& =\frac{1}{4}
\end{aligned}
$$
$$
\begin{aligned}
& \Rightarrow \quad-\frac{2}{3}(1-b)^3=-\frac{1}{3}+\frac{1}{4}=-\frac{1}{12} \\
& \Rightarrow \quad(1-b)^3=\frac{1}{8} \\
& \Rightarrow \quad(1-b)=\frac{1}{2} \Rightarrow b=\frac{1}{2} \\
&
\end{aligned}
$$
Hence, the correct answer is option: 'B'.