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ID: past-exam-of-the-year-1-of-the-computer-science-course-of-the-university-of-oxford/2018/discrete-mathematics-continuous-mathematics-and-probability/6/d/solution

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Past exam of the year 1 of the computer science course of the University of Oxford / 2018 / Discrete mathematics, continuous mathematics, and probability / 6 / d / Solution by

Codex  0 2026-10-31

The printed inequality has its sign reversed: $f (x) < f_{2} (x)$ for $x > 1$ . This is present in the original scan, rather than arising in transcription. For

x > 1

, Taylor's theorem with Lagrange remainder gives

f (x) - f_{2} (x) = \frac{f ^{'''} ( ξ )}{3 !} (x - 1)^{3}, 1 < ξ < x .

(1)

Part (a) has

f^{'''} (ξ) < 0

, so the remainder is strictly negative. As an independent check, direct algebra gives the quadratic Taylor majorant of a reciprocal quadratic:

f_{2} (x) - f (x) = \frac{( x - 1 ) ^{4}}{4 ( 1 + x ^{2} )} > 0 (x > 1) .

(2)

For example,

f (2) = 1/5 < 1/4 = f_{2} (2)

is already a counterexample to the requested direction. The exact identity also shows that this polynomial majorizes

f

on the entire real line, with equality only at one.

Figure 1.
The quadratic Taylor polynomial lies above the reciprocal quadratic, correcting the printed inequality
. The shaded gap for $x > 1$ is positive and equals $(x - 1)^{4} / (4 (1 + x^{2}))$ . **Details**.

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