JEE Main 2021 — Differential Equations Question with Solution

Given, $$\frac{dy}{dx}$$ = 2(x + 1)

Integrating both sides, we get

$$y=x^2+2x+c$$

Let the two roots of the quadratic equation $$\alpha$$ and $$\beta$$



As parabola intercept the x axis so D > 0

From figure, AB = |$$\alpha$$ - $$\beta$$| = $$\frac{\sqrt{D}}{\left|a\right|}$$ = $$\sqrt{D}$$

and BC = $$-\frac{D}{4a}$$ = $$-\frac{D}{4}$$

$$\therefore$$ Area of rectangle (ABCD) = AB $$\times$$ BC = $$\sqrt{D}\times \frac{D}{4}$$

From property we know,

Area of parabola with the x axis = $$\frac{2}{3}$$(Area of rectangle)

$$\Rightarrow$$ $$\frac{4\sqrt{8}}{3}$$ = $$\frac{2}{3}\times \sqrt{D}\times \frac{D}{4}$$

$$\Rightarrow$$ $$D\sqrt{D}$$ = $$8\sqrt{8}$$

$$\Rightarrow$$ D = 8

$$\therefore$$ b² - 4ac = 8

$$\Rightarrow$$ 4 - 4c = 8

$$\Rightarrow$$ 1 $$-$$ c = 2 $$\Rightarrow$$ c = $$-$$ 1

Equation of f(x) = x² + 2x $$-$$ 1

$$\therefore$$ f(1) = 1 + 2 $$-$$ 1 = 2