 Solution

ID: past-exam-of-the-mathematics-course-of-the-year-2-of-the-natural-sciences-course-of-the-university-of-cambridge/2018/paper-1/9/b/iii/solution

The initial and final slopes from part (i) are

(1 + α)^{- 1/3}

and

(1 + α)^{- 1/2}

. The prescribed ratio gives

(1 + α)^{- 1/2} = \frac{1}{2} (1 + α)^{- 1/3} ⟺ (1 + α)^{1/6} = 2.

Consequently

α = 63 .

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