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 ≤ x ≤ b) and R_2(b ≤ x ≤ 1) such that R_1-R_2=(1)/(4) . Then, b equals to

Correct answer: B. (1)/(2) Here, area between 0 to b is R_1 and b to aligned & 1 is R_2 & ∫_0^b(1-x)^2 d x-∫_b^1(1-x)^2 d x=(1)/(4) & (((1-x)^3)/(-3))_0^b-(((1-x)^3)/(-3))_b^1=(1)/(4) & -(1)/(3)\ (1-b)^3-1\ +(1)/(3)\ 0-(1-b)^3\ & =(1)/(4) aligned aligned & -(2)/(3)(1-b)^3=-(1)/(3)+(1)/(4)=-(1)/(12) & (1-b)^3=(1)/(8) & (1-b)=(1)/(2) =(1)/(2) & aligned

JEE Advanced 2011Paper 1

Exam: JEE Advanced
Year: 2011
Paper: Paper 1
Subject: Mathematics
Chapter: Area Under Curves

JEE Advanced 2011 · Paper 1 · Mathematics · Area Under Curves

Let the straight line x=b divide the area enclosed by y=(1-x)^2,...

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

Options

A $\frac{3}{4}$
B $\frac{1}{2}$
C $\frac{1}{3}$
D $\frac{1}{4}$

Correct answer

B. $\frac{1}{2}$

Step-by-step 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} \Rightarrowb=\frac{1}{2}\\ & \end{aligned}

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