JEE Main 2022 — Mathematics Calculus

Given

$$x=2\sqrt{2}\mathrm{cos}t\sqrt{\mathrm{sin}2t}$$x=22​costsin2t​

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

$$\frac{dx}{dt}=\frac{2\sqrt{2}\mathrm{cos}3t}{\sqrt{\mathrm{sin}2t}}......(1)$$dtdx​=sin2t​22​cos3t​......(1)

Also given $y(t)=2\sqrt{2}\mathrm{sin}t\sqrt{\mathrm{sin}2t}$y(t)=22​sintsin2t​

Again differentiating w.r.t $t$t both side we get,

$$\frac{dy}{dt}=\frac{2\sqrt{2}\mathrm{sin}3t}{\sqrt{\mathrm{sin}2t}}......(2)$$dtdy​=sin2t​22​sin3t​......(2)

Now dividing equation $(2)$(2) from $(1)$(1) we get,

$$\frac{dy}{dx}=\mathrm{tan}3t$$dxdy​=tan3t

Now finding the value of $\frac{dy}{dx}$dxdy​ at $t=\frac{\pi }{4}$t=4π​ we get,

$$\frac{dy}{dx}=-1$$dxdy​=−1

Now finding $\frac{{d}^{2}y}{d{x}^{2}}$dx2d2y​ we get,

$$\frac{{d}^{2}y}{d{x}^{2}}=\frac{3}{2\sqrt{2}}\frac{{\mathrm{sec}}^{2}3t\cdot \sqrt{\mathrm{sin}2t}}{\mathrm{cos}3t}$$dx2d2y​=22​3​cos3tsec23t⋅sin2t​​

Now finding value of $\frac{{d}^{2}y}{d{x}^{2}}$dx2d2y​ at $t=\frac{\pi }{4}$t=4π​ we get,

$$\frac{{d}^{2}y}{d{x}^{2}}=-3$$dx2d2y​=−3

Now putting the value of \frac{dy}{dx}&\frac{{d}^{2}y}{d{x}^{2}} in$\frac{1+{(\frac{dy}{dx})}^{2}}{\frac{{d}^{2}y}{d{x}^{2}}}$dx2d2y​1+(dxdy​)2​ we get,

$$\frac{1+{(\frac{dy}{dx})}^{2}}{\frac{{d}^{2}y}{d{x}^{2}}}=\frac{1+1}{-3}=-\frac{2}{3}$$dx2d2y​1+(dxdy​)2​=−31+1​=−32​