Edexcel P1 2020 October — Question 9 6 marks

Exam BoardEdexcel
ModuleP1 (Pure Mathematics 1)
Year2020
SessionOctober
Marks6
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicStandard Integrals and Reverse Chain Rule
TypeFind curve equation from derivative (straightforward integration + point)
DifficultyModerate -0.3 This is a straightforward integration question requiring simplification of the derivative into standard power form, then term-by-term integration and finding the constant using the given point. While it involves multiple steps and algebraic manipulation, it's a routine P1 exercise with no conceptual challenges beyond standard techniques, making it slightly easier than average.
Spec1.08a Fundamental theorem of calculus: integration as reverse of differentiation1.08b Integrate x^n: where n != -1 and sums

9. A curve with equation \(y = \mathrm { f } ( x )\) passes through the point \(( 9,10 )\). Given that $$f ^ { \prime } ( x ) = 27 x ^ { 2 } - \frac { 21 x ^ { 3 } - 5 x } { 2 \sqrt { x } } \quad x > 0$$ find \(\mathrm { f } ( x )\), fully simplifying each term.

Question 9:
AnswerMarks Guidance
Answer/WorkingMark Guidance
\(\dfrac{21x^3 - 5x}{2\sqrt{x}} = \alpha x^{\frac{5}{2}} + ...\) or \(... + \beta x^{\frac{1}{2}}\)M1 Uses correct index laws to obtain at least one correct index from splitting the fraction. Award for \(\alpha x^{\frac{5}{2}} + ...\) or \(... + \beta x^{\frac{1}{2}}\)
\(f(x) = \dfrac{27}{3}x^3 - \dfrac{21}{2} \times \dfrac{2}{7}x^{\frac{7}{2}} + \dfrac{5}{2} \times \dfrac{2}{3}x^{\frac{3}{2}} (+c)\) \(= 9x^3 - 3x^{\frac{7}{2}} + \dfrac{5}{3}x^{\frac{3}{2}} (+c)\)M1A1A1 Second M1: \(x^n \to x^{n+1}\) correctly seen on one term (usually \(27x^2 \to ...x^3\)). Two correct terms (A1); all correct simplified or unsimplified, \(+c\) not required (A1)
\(f(9) = 10 \Rightarrow 9(9)^3 - 3(9)^{\frac{7}{2}} + \dfrac{5}{3}(9)^{\frac{3}{2}} + c = 10 \Rightarrow c = ...\)dM1 Uses \(f(9) = 10\) and attempts to find \(c\); dependent on previous M1
\(f(x) = 9x^3 - 3x^{\frac{7}{2}} + \dfrac{5}{3}x^{\frac{3}{2}} - 35\)A1 All correct and simplified; accept equivalent forms e.g. \(\frac{1}{3}(27x^3 - 9x^{\frac{7}{2}} + 5x^{\frac{3}{2}} - 105)\)
Total: 6 marks
## Question 9:

| Answer/Working | Mark | Guidance |
|---|---|---|
| $\dfrac{21x^3 - 5x}{2\sqrt{x}} = \alpha x^{\frac{5}{2}} + ...$ or $... + \beta x^{\frac{1}{2}}$ | M1 | Uses correct index laws to obtain at least one correct index from splitting the fraction. Award for $\alpha x^{\frac{5}{2}} + ...$ or $... + \beta x^{\frac{1}{2}}$ |
| $f(x) = \dfrac{27}{3}x^3 - \dfrac{21}{2} \times \dfrac{2}{7}x^{\frac{7}{2}} + \dfrac{5}{2} \times \dfrac{2}{3}x^{\frac{3}{2}} (+c)$ $= 9x^3 - 3x^{\frac{7}{2}} + \dfrac{5}{3}x^{\frac{3}{2}} (+c)$ | M1A1A1 | Second M1: $x^n \to x^{n+1}$ correctly seen on one term (usually $27x^2 \to ...x^3$). Two correct terms (A1); all correct simplified or unsimplified, $+c$ not required (A1) |
| $f(9) = 10 \Rightarrow 9(9)^3 - 3(9)^{\frac{7}{2}} + \dfrac{5}{3}(9)^{\frac{3}{2}} + c = 10 \Rightarrow c = ...$ | dM1 | Uses $f(9) = 10$ and attempts to find $c$; dependent on previous M1 |
| $f(x) = 9x^3 - 3x^{\frac{7}{2}} + \dfrac{5}{3}x^{\frac{3}{2}} - 35$ | A1 | All correct and simplified; accept equivalent forms e.g. $\frac{1}{3}(27x^3 - 9x^{\frac{7}{2}} + 5x^{\frac{3}{2}} - 105)$ |

**Total: 6 marks**
9. A curve with equation $y = \mathrm { f } ( x )$ passes through the point $( 9,10 )$.

Given that

$$f ^ { \prime } ( x ) = 27 x ^ { 2 } - \frac { 21 x ^ { 3 } - 5 x } { 2 \sqrt { x } } \quad x > 0$$

find $\mathrm { f } ( x )$, fully simplifying each term.\\

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\hfill \mbox{\textit{Edexcel P1 2020 Q9 [6]}}