Question

If the straight line$x=b$ divides the area enclosed by $y=(1-x{)}^2,y=0,\text{and}x=0$ into two parts ${\mathrm{R}}_1(0\le \mathrm{x}\le \mathrm{b})\text{and}{\mathrm{R}}_2(\mathrm{b}\le \mathrm{x}\le 1)$ such that ${\mathrm{R}}_1-{\mathrm{R}}_2=\frac{1}{4}$. Then $b$ equals:

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

Answer 1

The correct option is B

$\frac{1}{2}$

Explanation of correct option.

Step 1. Find $R_1$ and $R_2$

We can write,

$$\begin{gathered} {\mathrm{R}}_1={\int }_0^{\mathrm{b}}(\mathrm{x}-1{)}^2\mathrm{dx} \\ ={\left[\frac{(\mathrm{x}-1{)}^3}{3}\right]}_0^{\mathrm{b}} \\ =\frac{(\mathrm{b}-1{)}^3+1}{3} \\ {\mathrm{R}}_2={\int }_{\mathrm{b}}^1(\mathrm{x}-1{)}^2\mathrm{dx} \\ ={\left[\frac{(\mathrm{x}-1{)}^3}{3}\right]}_{\mathrm{b}}^1 \\ =-\frac{(\mathrm{b}-1{)}^3}{3} \end{gathered}$$

Step 2. Find the value of $b$

Now, it is given that,

${\mathrm{R}}_1-{\mathrm{R}}_2=\frac{1}{4}$

Therefore, $\frac{(\mathrm{b}-1{)}^3+1}{3}+\frac{(\mathrm{b}-1{)}^3}{3}=\frac{1}{4}$

$$\begin{gathered} \Rightarrow \frac{2(b-1{)}^3+1}{3}=\frac{1}{4} \\ \Rightarrow 2(b-1{)}^3=\frac{3}{4}-1=-\frac{1}{4} \\ \Rightarrow (\mathrm{b}-1{)}^3=-\frac{1}{8} \\ \Rightarrow \left(b-1\right)=-\frac{1}{2} \\ \Rightarrow b=\frac{1}{2} \end{gathered}$$

Hence the correct option is (B).