QJEEM21Ma69c5: JEE Main 2021 Mathematics - Class 12 JEEM21Ma69c5 Solution: Let the curve y = y(x) be the solution of the differential…

Given, ${{dy} \over {dx}}$ = 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 = | $\alpha$ - $\beta$ | = ${{\sqrt D } \over {\left| a \right|}}$ = $\sqrt D$

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

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

From property we know,

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

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

$\Rightarrow$ $D\sqrt D$ = $8\sqrt 8$

$\Rightarrow$ 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