CAIE P1 2019 November — Question 2 3 marks

Exam BoardCAIE
ModuleP1 (Pure Mathematics 1)
Year2019
SessionNovember
Marks3
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicChain Rule
TypeDetermine if function is increasing/decreasing
DifficultyStandard +0.3 This question requires students to find when f'(x) > 0 for an increasing function by factorizing a quadratic and solving an inequality. It's a straightforward application of the relationship between derivatives and increasing functions, requiring only factorization of x² - 6x + 8 = (x-2)(x-4) and identifying that f'(x) > 0 when x > 4, making n = 4. Slightly above average difficulty due to the need to connect multiple concepts (derivatives, increasing functions, inequalities) but still a standard textbook-style question.
Spec1.07i Differentiate x^n: for rational n and sums1.07o Increasing/decreasing: functions using sign of dy/dx

2 An increasing function, f , is defined for \(x > n\), where \(n\) is an integer. It is given that \(\mathrm { f } ^ { \prime } ( x ) = x ^ { 2 } - 6 x + 8\). Find the least possible value of \(n\).

Question 2:
AnswerMarks Guidance
AnswerMarks Guidance
Attempt to solve \(f'(x) = 0\) or \(f'(x) > 0\) or \(f'(x) \geqslant 0\)M1 SOI
\((x-2)(x-4)\)A1 2 and 4 seen
Least possible value of \(n\) is 4A1 Accept \(n = 4\) or \(n \geqslant 4\)
Total: 3
**Question 2:**

| Answer | Marks | Guidance |
|--------|-------|----------|
| Attempt to solve $f'(x) = 0$ or $f'(x) > 0$ or $f'(x) \geqslant 0$ | M1 | SOI |
| $(x-2)(x-4)$ | A1 | 2 and 4 seen |
| Least possible value of $n$ is 4 | A1 | Accept $n = 4$ or $n \geqslant 4$ |
| **Total: 3** | | |

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2 An increasing function, f , is defined for $x > n$, where $n$ is an integer. It is given that $\mathrm { f } ^ { \prime } ( x ) = x ^ { 2 } - 6 x + 8$. Find the least possible value of $n$.\\

\hfill \mbox{\textit{CAIE P1 2019 Q2 [3]}}