JEE Main 2022 — Mathematics Calculus

Given, the area of the region under this curve, above the $x$x-axis and between the abscissae $3$3 and $x(>3)$x(>3) be ${(\frac{y}{x})}^{3}$(xy​)3

So, $\displaystyle {\int }_{3}^{x}ydx={(\frac{y}{x})}^{3}$∫3x​ydx=(xy​)3

$$\Rightarrow {x}^{3}{\int }_{3}^{x}ydx={y}^{3}$$⇒x3∫3x​ydx=y3

Now differentiating both side w.r.t $x$x, we get

$\Rightarrow {x}^{3}y+3{x}^{2}\frac{{y}^{3}}{{x}^{3}}=3{y}^{2}\frac{dy}{dx}$⇒x3y+3x2x3y3​=3y2dxdy​ $\displaystyle {\text{as given }{\int }_{3}^{x}ydx={(\frac{y}{x})}^{3}}$as given ∫3x​ydx=(xy​)3

$$\Rightarrow {x}^{4}+3{y}^{2}=3yx\frac{dy}{dx}$$⇒x4+3y2=3yxdxdy​

$$\Rightarrow 3xy\frac{dy}{dx}=3{y}^{2}+{x}^{4}$$⇒3xydxdy​=3y2+x4

Now put ${y}^{2}=t,y\frac{dy}{dx}=\frac{1}{2}\frac{dt}{dx}$y2=t,ydxdy​=21​dxdt​

$$\Rightarrow \frac{dt}{dx}-\frac{2}{x}t=\frac{2}{3}{x}^{3}$$⇒dxdt​−x2​t=32​x3

Now $\displaystyle IF={e}^{\int -\frac{2}{x}dx}=\frac{1}{{x}^{2}}$IF=e∫−x2​dx=x21​

So, solution is given by

$$t\times IF=\int \frac{2}{3}{x}^{3}\times IF$$t×IF=∫32​x3×IF

$$\Rightarrow \frac{t}{{x}^{2}}=\frac{{x}^{2}}{3}+C$$⇒x2t​=3x2​+C

Now given curve passes through $(3,3)$(3,3) so $C=-2$C=−2

$$\Rightarrow \frac{{y}^{2}}{{x}^{2}}=\frac{{x}^{2}}{3}-2$$⇒x2y2​=3x2​−2

So the equation of curve is $3{y}^{2}={x}^{4}-6{x}^{2}$3y2=x4−6x2

Now curve also passes through $(\alpha ,6\sqrt{10})$(α,610​)

So, ${\alpha }^{4}-6{\alpha }^{2}=1080$α4−6α2=1080

$$\Rightarrow \alpha =6$$⇒α=6