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(D) Given,$\sqrt{r} = a e^{	heta \cot \alpha}$  Squaring both sides,we get $r = a^{2} e^{2 	heta \cot \alpha}$.  Differentiating with respect to $

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(D) Given,$\sqrt{r} = a e^{	heta \cot \alpha}$  Squaring both sides,we get $r = a^{2} e^{2 	heta \cot \alpha}$.  Differentiating with respect to $	heta$:  $\frac{dr}{d	heta} = a^{2} \cdot e^{2 	heta \cot \alpha} \cdot (2 \cot \alpha) = 2 a^{2} \cot \alpha \cdot e^{2 	heta \cot \alpha}$.  Differentiating again with respect to $	heta$:  $\frac{d^{2}r}{d	heta^{2}} = 2 a^{2} \cot \alpha \cdot e^{2 	heta \cot \alpha} \cdot (2 \cot \alpha) = 4 a^{2} \cot^{2} \alpha \cdot e^{2 	heta \cot \alpha}$.  Since $r = a^{2} e^{2 	heta \cot \alpha}$,we can substitute this into the expression:  $\frac{d^{2}r}{d	heta^{2}} = 4 \cot^{2} \alpha \cdot (a^{2} e^{2 	heta \cot \alpha}) = 4 r \cot^{2} \alpha$.  Therefore,$\frac{d^{2}r}{d	heta^{2}} - 4 r \cot^{2} \alpha = 0$.

If $\sqrt{r} = a e^{\theta \cot \alpha}$ where $a$ and $\alpha$ are real numbers,then $\frac{d^{2} r}{d \theta^{2}} - 4 r \cot^{2} \alpha$ is

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