MHT CET · Maths · Definite Integration
If the straight-line \(x=b\) divides the area enclosed by \(y=(1-x)_{-}^2\) \(y=0\) and \(x=0\) in 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 \(\mathrm{b}\) equals
- A \(\frac{1}{2}\)
- B \(\frac{1}{4}\)
- C \(\frac{3}{4}\)
- D \(\frac{1}{3}\)
Answer & Solution
Correct Answer
(A) \(\frac{1}{2}\)
Step-by-step Solution
Detailed explanation
\(\begin{aligned} & \int_0^b(1-x)^2 \mathrm{~d} x-\int_b^1(1-x)^2 \mathrm{~d} x=\frac{1}{4} \\ & \Rightarrow\left[\frac{(1-x)^3}{-3}\right]_0^b-\left[\frac{(1-x)^3}{-3}\right]_b^1=\frac{1}{4}\end{aligned}\)
\(\begin{aligned} & \Rightarrow \frac{(1-b)^3-(1-0)^3-(1-1)^3+(1-b)^3}{-3}=\frac{1}{4} \\ & 2(1-b)^3-1=\frac{-3}{4} \\ & \Rightarrow 2(1-b)^3-1=\frac{-3}{4} \\ & \Rightarrow(1-b)^3=\frac{1}{8} \\ & \Rightarrow b=\frac{1}{2}\end{aligned}\)
\(\begin{aligned} & \Rightarrow \frac{(1-b)^3-(1-0)^3-(1-1)^3+(1-b)^3}{-3}=\frac{1}{4} \\ & 2(1-b)^3-1=\frac{-3}{4} \\ & \Rightarrow 2(1-b)^3-1=\frac{-3}{4} \\ & \Rightarrow(1-b)^3=\frac{1}{8} \\ & \Rightarrow b=\frac{1}{2}\end{aligned}\)
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