Area Under Curves Question 11
Question 11 - 2024 (31 Jan Shift 1)
The area of the region
$\left{(x, y): y^{2} \leq 4 x, x<4, \frac{x y(x-1)(x-2)}{(x-3)(x-4)}>0, x \neq 3\right}$ is
(1) $\frac{16}{3}$
(2) $\frac{64}{3}$
(3) $\frac{8}{3}$
(4) $\frac{32}{3}$
Show Answer
Answer (4)
Solution
$y^{2} \leq 4 x, x<4$
$\frac{x y(x-1)(x-2)}{(x-3)(x-4)}>0$
Case-I : $y>0$
$\frac{x(x-1)(x-2)}{(x-3)(x-4)}>0$
$x \in(0,1) \cup(2,3)$
Case -II:y $<0$
$\frac{x(x-1)(x-2)}{(x-3)(x-4)}<0, x \in(1,2) \cup(3,4)$
$$ \begin{aligned} & \text { Area }=2 \int_{0}^{4} \sqrt{x} d x \ & =2 \cdot \frac{2}{3}\left[x^{3 / 2}\right]_{0}^{4}=\frac{32}{3} \end{aligned} $$
