JEE PYQ: Area Under Curves Question 20
Question 20 - 2020 (09 Jan Shift 2)
Given: $f(x) = \begin{cases} x, & 0 \leq x < \frac{1}{2} \ \frac{1}{2}, & x = \frac{1}{2} \ 1-x, & \frac{1}{2} < x \leq 1 \end{cases}$ and $g(x) = \left(x - \frac{1}{2}\right)^2$, $x \in \mathbb{R}$. Then the area (in sq. units) of the region bounded by the curves $y = f(x)$ and $y = g(x)$ between the lines $2x = 1$ and $2x = \sqrt{3}$, is:
(1) $\frac{1}{3} + \frac{\sqrt{3}}{4}$
(2) $\frac{\sqrt{3}}{4} - \frac{1}{3}$
(3) $\frac{1}{2} - \frac{\sqrt{3}}{4}$
(4) $\frac{1}{2} + \frac{\sqrt{3}}{4}$
Show Answer
Answer: (2) $\frac{\sqrt{3}}{4} - \frac{1}{3}$
Solution
Between $x = 1/2$ and $x = \sqrt{3}/2$: Area $= \int_{1/2}^{\sqrt{3}/2}[(1-x) - (x-1/2)^2],dx = \frac{\sqrt{3}}{4} - \frac{1}{3}$.
BETA
