Area Ques 31
Question
- If 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
(a) $\frac{3}{4}$
(b) $\frac{1}{2}$
(c) $\frac{1}{3}$
(d) $\frac{1}{4}$
(2011)
Show Answer
Answer:
Correct Answer: 31.(b)
Solution:
Formula:
- Here, area between 0 to $b$ is $R _{1}$ and $b$ to 1 is $R _{2}$.
$\therefore \quad \int _{0}^{b}(1-x)^{2} d x-\int _{b}^{1}(1-x)^{2} d x=\frac{1}{4}$
$ \Rightarrow \quad {[\frac{(1-x)^{3}}{-3}]}{ } _{0}^{b}-{[\frac{(1-x)^{3}}{-3}]} _{b}^{1}=\frac{1}{4} $
$\Rightarrow-\frac{1}{3}\left[(1-b)^{3}-1\right]+\frac{1}{3}\left[0-(1-b)^{3}\right]=\frac{1}{4}$
$\Rightarrow \quad-\frac{2}{3}(1-b)^{3}=-\frac{1}{3}+\frac{1}{4}=-\frac{1}{12} \Rightarrow(1-b)^{3}=\frac{1}{8}$
$\Rightarrow \quad(1-b)=\frac{1}{2} \quad \Rightarrow \quad b=\frac{1}{2}$
BETA
