Let $x(t)=2 \sqrt{2} \cos t \sqrt{\sin 2 t}$ and $y(t)=2 \sqrt{2} \sin t \sqrt{\sin 2 t}, t \in\left(0, \frac{\pi}{2}\right)$. Then $\frac{1+\left(\frac{d y}{d x}\right)^{2}}{\frac{d^{2} y}{d x^{2}}}$ at $t=\frac{\pi}{4}$ is equal to

JEE Main · 2022 · 28 Jul 2022 · Shift 2 · Differentiation

Let

x (t) = 22 cos t sin 2 t

and

y (t) = 22 sin t sin 2 t, t \in (0, \frac{π}{2})

. Then

\frac{1 + ( \frac{d y}{d x} ) ^{2}}{\frac{d ^{2} y}{d x ^{2}}}

t = \frac{π}{4}

is equal to

Answer: (D)

Compute the derivatives:

\frac{d x}{d t} = \frac{2 2 c o s 3 t}{s i n 2 t}

\frac{d y}{d t} = \frac{2 2 s i n 3 t}{s i n 2 t}

Thus

\frac{d y}{d x} = tan 3 t

t = \frac{π}{4}

\frac{d y}{d x} = tan \frac{3 π}{4} = - 1

For the second derivative:

\frac{d ^{2} y}{d x ^{2}} = \frac{d}{d t} (\frac{d y}{d x}) \cdot \frac{d t}{d x} = 3 sec^{2} (3 t) \cdot \frac{s i n 2 t}{2 2 c o s 3 t}

t = \frac{π}{4}

\frac{d ^{2} y}{d x ^{2}} = - 3

Therefore:

\frac{1 + ( \frac{d y}{d x} ) ^{2}}{\frac{d ^{2} y}{d x ^{2}}} = \frac{1 + 1}{- 3} = \frac{- 2}{3}

JEE Main · 2022 · 28 Jul 2022 · Shift 2 · Single correct option