| Exam Board | Edexcel |
|---|---|
| Module | C1 (Core Mathematics 1) |
| Year | 2018 |
| Session | June |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Curve Sketching |
| Type | Curve from derivative information |
| Difficulty | Moderate -0.3 This is a straightforward integration question requiring students to integrate a quadratic, use a point to find the constant, factorise the result, and sketch using standard techniques. While it involves multiple parts and skills (integration, factorisation, curve sketching), each step follows routine C1 procedures with no novel problem-solving required, making it slightly easier than average. |
| Spec | 1.02j Manipulate polynomials: expanding, factorising, division, factor theorem1.02n Sketch curves: simple equations including polynomials1.07i Differentiate x^n: for rational n and sums1.08b Integrate x^n: where n != -1 and sums |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \((x-3)(3x+5) = 3x^2 - 4x - 15\); allow \(3x^2 + 5x - 9x - 15\) | B1 | Correct expansion, simplified or unsimplified |
| \(f(x) = x^3 - 2x^2 - 15x + c\) | M1A1 | M1: \(x^n \to x^{n+1}\) for any term, follow through on incorrect indices but not for \("+c"\); A1: all terms correct, no need for \(+c\) |
| \(x=1, y=20 \Rightarrow 20 = 1-2-15+c \Rightarrow c = 36\) | dM1 | Substitutes \(x=1\) and \(y=20\) into their \(f(x)\) to find \(c\). Must have \(+c\). Dependent on first M1 |
| \(f(x) = x^3 - 2x^2 - 15x + 36\) | A1 | cao, all together on one line |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(A = 4\) | B1 | Correct value (may be implied) |
| Way 1: \(f(x) = (x-3)^2(x+A) = (x^2-6x+9)(x+A)\); \(f(x) = x^3+(A-6)x^2+(9-6A)x+9A\); \(A-6=-2 \Rightarrow A=4\); \(9-6A=-15 \Rightarrow A=4\); \(9A=36 \Rightarrow A=4\) | M1A1 | M1: Expands \((x-3)^2(x+A)\) and compares coefficients with \(f(x)\) from (a) to form 3 equations and attempts to solve at least two; A1: fully correct proof using all 3 coefficients |
| Way 2: \(f(x)=(x-3)^2(x+4) = (x^2-6x+9)(x+4) = x^3-6x^2+4x^2+9x-24x+36 = x^3-2x^2-15x+36\) | M1A1 | M1: Expands \((x-3)^2(x+\text{"4"})\) fully; A1: fully correct proof (condone invisible brackets around \(x+4\) if sufficient working shown) |
| Way 3: \((x^3-2x^2-15x+36) \div (x-3) = x^2+x-12\); \((x^2+x-12)\div(x-3) = x+4\) or \((x^2+x-12)=(x+4)(x-3)\) | M1A1 | M1: Divides \(f(x)\) by \((x-3)\) then divides quotient by \((x-3)\); A1: fully correct proof |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Positive cubic shape (curve, ends not turning back) | B1 | Must be a curve, not straight lines; ends must not clearly turn back |
| Touches \(x\)-axis at \((3, 0)\) (could be a maximum) | B1 | Accept 3 marked on \(x\)-axis; allow \((0,3)\) if in correct place; sketch takes precedence if ambiguous |
| Crosses \(x\)-axis at \((-4, 0)\) | B1ft | FT on their \(-A\) from part (b); allow \(-A\) and "made up" \(A\); sketch takes precedence |
| Crosses \(y\)-axis at \((0, 36)\) with a maximum in the second quadrant | B1ft | FT on their numerical \(c\) from part (a) only; sketch takes precedence |
## Question 9(a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $(x-3)(3x+5) = 3x^2 - 4x - 15$; allow $3x^2 + 5x - 9x - 15$ | B1 | Correct expansion, simplified or unsimplified |
| $f(x) = x^3 - 2x^2 - 15x + c$ | M1A1 | M1: $x^n \to x^{n+1}$ for any term, follow through on incorrect indices but not for $"+c"$; A1: all terms correct, no need for $+c$ |
| $x=1, y=20 \Rightarrow 20 = 1-2-15+c \Rightarrow c = 36$ | dM1 | Substitutes $x=1$ and $y=20$ into their $f(x)$ to find $c$. Must have $+c$. **Dependent on first M1** |
| $f(x) = x^3 - 2x^2 - 15x + 36$ | A1 | cao, all together on one line |
---
## Question 9(b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $A = 4$ | B1 | Correct value (may be implied) |
| **Way 1:** $f(x) = (x-3)^2(x+A) = (x^2-6x+9)(x+A)$; $f(x) = x^3+(A-6)x^2+(9-6A)x+9A$; $A-6=-2 \Rightarrow A=4$; $9-6A=-15 \Rightarrow A=4$; $9A=36 \Rightarrow A=4$ | M1A1 | M1: Expands $(x-3)^2(x+A)$ and compares coefficients with $f(x)$ from (a) to form 3 equations and attempts to solve at least two; A1: fully correct proof using all 3 coefficients |
| **Way 2:** $f(x)=(x-3)^2(x+4) = (x^2-6x+9)(x+4) = x^3-6x^2+4x^2+9x-24x+36 = x^3-2x^2-15x+36$ | M1A1 | M1: Expands $(x-3)^2(x+\text{"4"})$ fully; A1: fully correct proof (condone invisible brackets around $x+4$ if sufficient working shown) |
| **Way 3:** $(x^3-2x^2-15x+36) \div (x-3) = x^2+x-12$; $(x^2+x-12)\div(x-3) = x+4$ or $(x^2+x-12)=(x+4)(x-3)$ | M1A1 | M1: Divides $f(x)$ by $(x-3)$ then divides quotient by $(x-3)$; A1: fully correct proof |
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## Question 9(c):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Positive cubic shape (curve, ends not turning back) | B1 | Must be a curve, not straight lines; ends must not clearly turn back |
| Touches $x$-axis at $(3, 0)$ (could be a maximum) | B1 | Accept 3 marked on $x$-axis; allow $(0,3)$ if in correct place; sketch takes precedence if ambiguous |
| Crosses $x$-axis at $(-4, 0)$ | B1ft | FT on their $-A$ from part (b); allow $-A$ and "made up" $A$; sketch takes precedence |
| Crosses $y$-axis at $(0, 36)$ **with a maximum in the second quadrant** | B1ft | FT on their numerical $c$ from part (a) only; sketch takes precedence |
---\begin{enumerate}
\item The curve $C$ has equation $y = \mathrm { f } ( x )$, where
\end{enumerate}
$$f ^ { \prime } ( x ) = ( x - 3 ) ( 3 x + 5 )$$
Given that the point $P ( 1,20 )$ lies on $C$,\\
(a) find $\mathrm { f } ( x )$, simplifying each term.\\
(b) Show that
$$f ( x ) = ( x - 3 ) ^ { 2 } ( x + A )$$
where $A$ is a constant to be found.\\
(c) Sketch the graph of $C$. Show clearly the coordinates of the points where $C$ cuts or meets the $x$-axis and where $C$ cuts the $y$-axis.\\
\hfill \mbox{\textit{Edexcel C1 2018 Q9 [12]}}