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ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2022/ia/paper-1/11d/b/solution

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Past exam of the mathematics course of the University of Cambridge / 2022 / ia / Paper 1 / 11D / b / Solution by

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

Let

m = \frac{f ( b ) - f ( a )}{b - a}, h (x) = f (x) - f (a) - m (x - a) .

Then

h (a) = h (b) = 0

. Since

f

is not linear,

h (c) \neq = 0

for some

c \in (a, b)

. If

h (c) > 0

, the mean value theorem on

[a, c]

gives a point

ξ

with

h^{'} (ξ) = \frac{h ( c ) - h ( a )}{c - a} > 0.

If

h (c) < 0

, apply it on

[c, b]

to obtain

h^{'} (ξ) = \frac{h ( b ) - h ( c )}{b - c} > 0.

In either case

h^{'} = f^{'} - m

, and therefore

f^{'} (ξ) > \frac{f ( b ) - f ( a )}{b - a} .

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