The figure illustrates a right circular cone shaped mountain. If a shortest distance road is built for sight-seeing around the mountain, in which road starts at point A and ends at B (A, B and Q are collinear). Let $f (x)$ be the total length of road. The road will go up-hill for some distance $L (x)$ and down-hill for the distance of $g (x)$ .
Given $P Q = 6; O P = 2; A B = x$ , then

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$f (x) = \sqrt{x^{2} - 18 x + 108} \forall x \in [0, 6]$

$g (x) = \frac{f (x)}{\sqrt{x^{2} - 15 x + 54}} \forall x \in [0, 6]$

Maximum value of $g (x)$ is $3 \sqrt{3}$

If at $x = x_{0}, g (x)$ take a maximum value, then $f (x_{0}) = 6 \sqrt{3}$

answer is A, B, D.

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Detailed Solution

$I n △ PQ B cos \frac{2 π}{3} = \frac{6^{2} + {(6 - x)}^{2} - {(f (x))}^{2}}{12 (6 - x)} = - \frac{1}{2} \Rightarrow f (x) = \sqrt{x^{2} - 18 x + 108} \forall x \in [0, 6]$

$I n △ P Q B cos α = \frac{g (x)}{6 - x} = \frac{{(6 - x)}^{2} + {(f (x))}^{2} - 6^{2}}{2 f (x) \cdot (6 - x)} \Rightarrow g (x) = \frac{x^{2} - 15 x + 54}{\sqrt{x^{2} - 18 x + 108}}$