JEE Advanced · Mathematics · 28. Area Under Curves
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
\(\frac{3}{4}\)
- B
\(\frac{1}{2}\)
- C
\(\frac{1}{3}\)
- D
\(\frac{1}{4}\)
Answer & Solution
Correct Answer
(B)
\(\frac{1}{2}\)
Step-by-step Solution
Detailed explanation
Here, area between 0 to \(b\) is \(R_1\) and \(b\) to
\[
\begin{aligned}
& 1 \text { is } R_2 \\
& \therefore \int_0^b(1-x)^2 d x-\int_b^1(1-x)^2 d x=\frac{1}{4} \\
& \Rightarrow \quad\left(\frac{(1-x)^3}{-3}\right)_0^b-\left(\frac{(1-x)^3}{-3}\right)_b^1=\frac{1}{4} \\
& \Rightarrow-\frac{1}{3}\left\{(1-b)^3-1\right\}+\frac{1}{3}\left\{0-(1-b)^3\right\} \\
& =\frac{1}{4}
\end{aligned}
\]
\[
\begin{aligned}
& \Rightarrow \quad-\frac{2}{3}(1-b)^3=-\frac{1}{3}+\frac{1}{4}=-\frac{1}{12} \\
& \Rightarrow \quad(1-b)^3=\frac{1}{8} \\
& \Rightarrow \quad(1-b)=\frac{1}{2} \Rightarrow b=\frac{1}{2} \\
&
\end{aligned}
\]
\[
\begin{aligned}
& 1 \text { is } R_2 \\
& \therefore \int_0^b(1-x)^2 d x-\int_b^1(1-x)^2 d x=\frac{1}{4} \\
& \Rightarrow \quad\left(\frac{(1-x)^3}{-3}\right)_0^b-\left(\frac{(1-x)^3}{-3}\right)_b^1=\frac{1}{4} \\
& \Rightarrow-\frac{1}{3}\left\{(1-b)^3-1\right\}+\frac{1}{3}\left\{0-(1-b)^3\right\} \\
& =\frac{1}{4}
\end{aligned}
\]
\[
\begin{aligned}
& \Rightarrow \quad-\frac{2}{3}(1-b)^3=-\frac{1}{3}+\frac{1}{4}=-\frac{1}{12} \\
& \Rightarrow \quad(1-b)^3=\frac{1}{8} \\
& \Rightarrow \quad(1-b)=\frac{1}{2} \Rightarrow b=\frac{1}{2} \\
&
\end{aligned}
\]
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