$\begin{array}{1 1}x=2 \;\tan \;\theta \\ x= \large\frac{2}{\tan \theta}\\x= \sqrt 2 \cot \theta \\x= \large\frac{\sqrt 2}{\cot \theta}\end{array} $

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$x=2 \;\tan \;\theta$

The acceleration of the body down the plane is $g \sin \theta - \mu g \cos \theta -g (\sin \theta -\mu \cos \theta )= g(\sin \theta -0.5 x \cos \theta ).$

Therefore , the body will first accelerate up to $ x <2 \tan \theta $ .

The velocity will be maximum at $ x= 2 \tan \theta$, because for $ x > 2\;\tan \theta $, the body starts decelerating .

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