f(x)=6t-t^2

differentiate

f'(x)=6-2t

Differentiate again

f''(x)=-2. Hence maxima will occur at f'(x)=0

0=6-2t

t=3

Hence at t=3 f(x)=>f(3)=9

Now f(x) represents maximum displacement. Maximum displacement in a circle is diameter of circle.

hence diameter of circle D=9 radius R=4.5

Now full distance will be circumference of circle=2piR= 2pi 4.5=9pi

1/6 of that distance will be 1.5pi

Now look at the figure, 1.5pi is the arc of the circle.

Central angle will be 60

Hence cord will be 4.5 i.e the displacement is 4.5 when the distance is 1/6th.

Now, f(x)=6t-t^2

t^2-6t+4.5=0

Solve for t

t=0.88 or t=5.12.

hence OA B.

Main concept here is to understand difference between displacement and distance. Please let me know if any step is not clear. I have omitted obvious calculations.

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