SOLUTION —
Let $u=a \sin ^3 t$ and $v=a \cos ^3 t$
$\therefore \quad \frac{d u}{d t}=3 a \sin ^2 t \cos t$
and $\frac{d v}{d t}=3 a \cos ^2 t(-\sin t)$
$\Rightarrow \quad \frac{d^2 u}{d t^2}=3 a\left[2 \sin t \cos ^2 t-\sin ^3 t\right]$
and $\frac{d^2 v}{d t^2}=-3 a\left[-2 \cos t \sin ^2 t+\cos ^3 t\right]$
$\therefore \quad \frac{d^2 u}{d^2 v}=\frac{\left[2 \sin t \cos ^2 t-\sin ^3 t\right]}{\left[2 \cos t \sin ^2 t-\cos ^3 t\right]}$
At $t=\pi / 4$,
$\frac{d^2 u}{d^2 v}=\frac{\left[2 \times \frac{1}{\sqrt{2}} \times \frac{1}{2}-\frac{1}{2 \sqrt{2}}\right]}{\left[2 \times \frac{1}{\sqrt{2}} \times \frac{1}{2}-\frac{1}{2 \sqrt{2}}\right]}=1$
So, The correct option will be (D).