JEE MAIN - Mathematics (2024 - 31st January Morning Shift - No. 15)
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
$$\frac{32}{3}$$
$$\frac{16}{3}$$
$$\frac{8}{3}$$
$$\frac{64}{3}$$
Explanation
$$\begin{aligned} & y^2 \leq 4 x, x<4 \\ & \frac{x y(x-1)(x-2)}{(x-3)(x-4)}>0 \\ & \text { Case - I}: y>0 \\ & \frac{x(x-1)(x-2)}{(x-3)(x-4)}>0 \\ & x \in(0,1) \cup(2,3) \\ & \text { Case }- \text { II : y<0 } \\ & \frac{x(x-1)(x-2)}{(x-3)(x-4)}<0, x \in(1,2) \cup(3,4) \end{aligned}$$
_31st_January_Morning_Shift_en_15_1.png)
$$\begin{aligned} & \text { Area }=2 \int_\limits0^4 \sqrt{x} d x \\ & =2 \cdot \frac{2}{3}\left[x^{3 / 2}\right]_0^4=\frac{32}{3} \end{aligned}$$
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