 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2018/ia/paper-1/10f/b/solution

Codex  0 Created 2026-09-24 Updated 2026-10-03

f

is continuous on

[a, b]

and differentiable on

(a, b)

, then

f (b) - f (a) = f^{'} (c) (b - a)

for some

c \in (a, b)

For

F (t) = cos (e^{- t})

and

t \geq 0

∣ F^{'} (t) ∣ = e^{- t} ∣ sin (e^{- t}) ∣ \leq 1.

The mean value theorem gives $∣ cos (e^{- x}) - cos (e^{- y}) ∣ \leq ∣ x - y ∣$ .

For the second inequality put

t = 1 + x \geq 1

. It becomes

2 lo g t \leq t - t^{- 1}

G (t) = t - t^{- 1} - 2 lo g t

satisfies

G (1) = 0

and

G^{'} (t) = 1 + t^{- 2} - 2 t^{- 1} = \frac{( t - 1 ) ^{2}}{t ^{2}} \geq 0.

Thus $lo g (1 + x) \leq x / 1 + x$ .

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