 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2024/ii/paper-1/2g/e/solution

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

x = cos θ

and taking the limit as

θ \to 0

gives

T_{n}^{'} (1) = lim_{θ \to 0} \frac{n s i n ( n θ )}{s i n θ} = n^{2} .

Part (d), with

r = 1

, shows that

T_{n}^{'}

is increasing on

[1, \infty)

, so

T_{n} (x) - T_{n} (1) = \int_{1}^{x} T_{n}^{'} (t) d t \geq n^{2} (x - 1) .

Since

T_{n} (1) = 1

and

n^{2} \geq n

T_{n} (x) \geq 1 + n (x - 1) (x \geq 1) .

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