KVPY SX SB Exam 19-Nov-2017 Question Paper

for a cone
r will be cone slant height
$\mathit{V}=2\sqrt{3}$
Let x radius of cone
h be height then
$\frac{1}{3}\mathit{\pi }{\mathit{r}}^2\mathit{h}=2\sqrt{3}\mathit{\pi }$
${\mathit{x}}^2\mathit{h}=6\sqrt{3}\Rightarrow {\mathit{x}}^2=\frac{6\sqrt{3}}{\mathit{h}}$
least Diameter ⇒ least slant height of cone
${\mathit{l}}^2={\mathit{x}}^2+{\mathit{h}}^2$
${\mathit{r}}^2={\mathit{x}}^2+{\mathit{h}}^2$
${\mathit{r}}^2=\frac{6\sqrt{3}}{\mathit{h}}+{\mathit{h}}^2$
Diff. w. to x
$2\mathit{r}\frac{\mathit{d}\mathit{r}}{\mathit{d}\mathit{h}}=\frac{-6\sqrt{3}}{{\mathit{h}}^2}+2\mathit{h}$
for maximum & minimum $\frac{\mathit{d}\mathit{r}}{\mathit{d}\mathit{h}}=0$
$\mathit{h}=3\Rightarrow {\mathit{h}}^2=3$
${\mathit{x}}^2=\frac{6\sqrt{3}}{\sqrt{3}}$
${\mathit{x}}^2=6$
${\mathit{r}}^{62}=6+3$
${\mathit{r}}^2=9$
r = 3
d = 2r
d = 6