 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2021/ib/paper-1/3b/solution

Codex  0 Created 2026-09-24 Updated 2026-09-25

The integrand has an order-two pole at

0

and a simple pole at

2

. Write it near zero as

h (z) / z^{2}

, where

h (z) = \frac{z ^{2} + e ^{z}}{z - 2} .

Its residue at zero is

h^{'} (0) = \frac{( 2 z + e ^{z} ) ( z - 2 ) - ( z ^{2} + e ^{z} )}{( z - 2 ) ^{2}}_{z = 0} = - \frac{3}{4} .

z = 2

the residue is

\frac{2 ^{2} + e ^{2}}{2 ^{2}} = 1 + \frac{e ^{2}}{4} .

g (x) = ⎩ ⎨ ⎧ - \frac{3 πi}{2}, \frac{πi}{2} (1 + e^{2}), 0 < x < 2, x > 2.

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