 Solution

ID: past-exam-of-the-mathematics-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2018/paper-1/16/f/solution

The new angular interval has width

π /2

2/ π

there. Its endpoints stay away from the poles of the tangent function, making

Y

E [Y] = \frac{2 x _{0}}{π} \int_{- π /6}^{π /3} tan θ d θ = \frac{2 x _{0}}{π} [- ln (cos θ)]_{- π /6}^{π /3} .

The two cosines are

3 /2

and

1/2

. Hence

E [Y] = \frac{2 x _{0}}{π} ln 3 = \frac{x _{0} ln 3}{π} .

It has units of length and is positive because the permitted positive angles extend farther than the negative ones.

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