Calculus – Area under Curves – JEE Main 28 Jan 2026 Shift 2

Solution:

Area bounded by $P_1$ and $P_2$:

First, find the points of intersection of $y=4x^2$ and $y=x^2+27$.
$$4x^2 = x^2 + 27 \Rightarrow 3x^2 = 27 \Rightarrow x^2 = 9 \Rightarrow x = \pm 3$$
The area $A_1$ is given by:
$$A_1 = \int_{-3}^{3} (y_{upper} – y_{lower}) dx = \int_{-3}^{3} [(x^2+27) – 4x^2] dx$$
$$A_1 = \int_{-3}^{3} (27 – 3x^2) dx = 2\int_{0}^{3} (27 – 3x^2) dx$$
$$A_1 = 2 [27x – x^3]_{0}^{3} = 2 [27(3) – 27] = 2(54) = 108$$

Area bounded by $P_1$ and line $y=\alpha x$:
Find intersection points of $y=4x^2$ and $y=\alpha x$ ($\alpha > 0$).
$$4x^2 = \alpha x \Rightarrow x(4x – \alpha) = 0 \Rightarrow x=0, x=\frac{\alpha}{4}$$
The area $A_2$ is:
$$A_2 = \int_{0}^{\frac{\alpha}{4}} (\alpha x – 4x^2) dx$$
$$A_2 = [\frac{\alpha x^2}{2} – \frac{4x^3}{3}]_{0}^{\frac{\alpha}{4}}$$
$$A_2 = \frac{\alpha}{2}(\frac{\alpha^2}{16}) – \frac{4}{3}(\frac{\alpha^3}{64}) = \frac{\alpha^3}{32} – \frac{\alpha^3}{48}$$
$$A_2 = \frac{3\alpha^3 – 2\alpha^3}{96} = \frac{\alpha^3}{96}$$

Using the given condition $A_1 = 6 A_2$:
$$108 = 6 \left( \frac{\alpha^3}{96} \right)$$
$$108 = \frac{\alpha^3}{16}$$
$$\alpha^3 = 108 \times 16 = 27 \times 4 \times 16 = 3^3 \times 4^3$$
$$\alpha = 3 \times 4 = 12$$

Ans. (3)