Challenging +1.2 This question requires students to apply error analysis to a Taylor series approximation, comparing an approximate sum to the true value and calculating percentage error. While it involves multiple steps (computing the approximation, finding the true value, calculating error, and verifying the bound), the techniques are standard for Further Maths Taylor series work. The calculation itself is straightforward once the setup is understood, making it moderately above average difficulty but not requiring novel insight.
15 The expression given in line 34 is used to calculate \(\sum _ { r = 1 } ^ { 6 } \frac { 1 } { r }\).
Show that the error in the result is less than \(1.5 \%\) of the true value.
AO 1.1. 2.45 (need not have working). May be done BC
\(\ln 6 + \frac{13}{24} + \frac{36+5}{12\times6\times7} = 2.41477\ldots\) % error \(= 100\times\frac{0.0352\ldots}{2.45} = 1.438\%\), this is less than \(1.5\%\). OR \(1.5\%\) of actual and compare with \(0.03523\)
B1 [2]
AO 2.2a. Convincing completion
The appendix pages shown contain only the headers/titles for exemplar response tables for Q9(b) and Q9(d)(iii), but the actual response content and mark allocations within those tables are blank/empty in the images provided.
Answer
Marks
There is no extractable mark scheme content from these particular pages — the tables show column headers ("Response"
"Mark") but all data cells are empty.
To access the full mark scheme content for H640/03 June 2023, I would recommend checking the OCR website directly at ocr.org.uk/qualifications/resource-finder, where the complete mark scheme should be available.
## Question 15:
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\sum_{r=1}^{6}\frac{1}{r} = \frac{49}{20}$ | B1 | AO 1.1. 2.45 (need not have working). May be done BC |
| $\ln 6 + \frac{13}{24} + \frac{36+5}{12\times6\times7} = 2.41477\ldots$ % error $= 100\times\frac{0.0352\ldots}{2.45} = 1.438\%$, this is less than $1.5\%$. OR $1.5\%$ of actual and compare with $0.03523$ | B1 [2] | AO 2.2a. Convincing completion |
The appendix pages shown contain only the **headers/titles** for exemplar response tables for Q9(b) and Q9(d)(iii), but the actual response content and mark allocations within those tables are **blank/empty** in the images provided.
There is no extractable mark scheme content from these particular pages — the tables show column headers ("Response" | "Mark") but all data cells are empty.
To access the full mark scheme content for H640/03 June 2023, I would recommend checking the **OCR website** directly at ocr.org.uk/qualifications/resource-finder, where the complete mark scheme should be available.
15 The expression given in line 34 is used to calculate $\sum _ { r = 1 } ^ { 6 } \frac { 1 } { r }$.\\
Show that the error in the result is less than $1.5 \%$ of the true value.
\hfill \mbox{\textit{OCR MEI Paper 3 2023 Q15 [2]}}