JEE Main 2022 · 28 Jul · Shift 2 · Mathematics Q68

← JEE Main 2022 · 28 Jul · Shift 2

MathematicsSingle correct+4 / -1Question 68 of 90

Let $x(t)=2\sqrt2\cos t\sqrt{\sin2t}$ and $y(t)=2\sqrt2\sin t\sqrt{\sin2t}$ , $t\in\left(0,\dfrac\pi2\right)$ . Then $\dfrac{1+\left(\frac{dy}{dx}\right)^2}{\frac{d^2y}{dx^2}}$ at $t=\dfrac\pi4$ is equal to :

\dfrac{-2\sqrt2}3

\dfrac23

\dfrac13

\dfrac{-2}3

Answer and worked solution

Official answer: D

Curve. $r^2=8\sin2t$ in polar form; $t=\frac\pi4$ is the point (2, 2).

Derivatives there. Differentiating parametrically gives $\frac{dy}{dx}=-1$ and $\frac{d^2y}{dx^2}=-3$ .

Value $\frac{1+1}{-3}=-\frac23$ . (D)

Topics:Parametric differentiation Higher-order derivatives

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JEE Main 2022 · 28 Jul · Shift 2, Mathematics question 68