The figure illustrates a right circular cone shaped mountain. The 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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$f (x) = \sqrt{x^{2} - 18 x + 108} \forall x \in [0, 6]$

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

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

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

answer is A, B, D.

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Detailed Solution

$I n Δ P Q B \cos \frac{2 π}{3} = \frac{6^{2} + {(6 - x)}^{2} - {(f (x))}^{2}}{12 (6 - x)} = - \frac{1}{2}$

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

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

R is a point from where down-hill road will begin
$\Rightarrow g (x)$ will be maximum where x=0 i.e. when QR is maximum
$\Rightarrow g (0) = \frac{54}{\sqrt{2.54}} = 3 \sqrt{3} a n d f (0) = 6 \sqrt{3}$