 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2022/ib/paper-1/17c/b/solution

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

L

annihilates polynomials of degree below

m

, then

L (f) = \int_{a}^{b} K (t) f^{(m)} (t) d t, K (t) = L (\frac{( x - t ) _{+}^{m - 1}}{( m - 1 )!}) .

Here

m = 3

and

L (f) = f^{''} (0) - f (- 1) + 2 f (0) - f (1) .

Applying

L

(x - t)_{+}^{2} /2

gives

K (t) = {\frac{1}{2} (1 + t)^{2}, - \frac{1}{2} (1 - t)^{2}, - 1 \leq t < 0, 0 \leq t \leq 1.

Hence

∣ L (f) ∣ \leq ∥ f^{'''} ∥_{\infty} \int_{- 1}^{1} ∣ K (t) ∣ d t = \frac{1}{3} ∥ f^{'''} ∥_{\infty} .

The smallest candidate is therefore

c = \frac{1}{3} .

 New to topics? Read the docs here!