Area Ques 37
Question
- If the line $x=\alpha$ divides the area of region $R=\left\{(x, y) \in R^{2}: x^{3} \leq y \leq x, 0 \leq x \leq 1\right\}$ into two equal parts, then
(2017 Adv.)
(a) $2 \alpha^{4}-4 \alpha^{2}+1=0$
(b) $\alpha^{4}+4 \alpha^{2}-1=0$
(c) $\frac{1}{2}<\alpha<1$
(d) $0<\alpha \leq \frac{1}{2}$
Show Answer
Answer:
Correct Answer: 37.(a,c)
Solution:
Formula:
- $\int _{0}^{1}\left(x-x^{3}\right) d x=2 \int _{0}^{\alpha}\left(x-x^{3}\right) d x$
$ \begin{aligned} \frac{1}{4} & =2 (\frac{\alpha^{2}}{2}-\frac{\alpha^{4}}{4}) \\ \Rightarrow \quad 2 \alpha^{4}-4 \alpha^{2}+1 & =0 \\ \alpha^{2} & =\frac{4-\sqrt{16-8}}{4} \\ \alpha^{2} & =1-\frac{1}{\sqrt{2}} \end{aligned} $
BETA
