| Exam Board | Edexcel |
|---|---|
| Module | Paper 2 (Paper 2) |
| Year | 2020 |
| Session | October |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Chain Rule |
| Type | Find curve equation from derivative |
| Difficulty | Standard +0.3 This is a straightforward integration problem requiring students to find f(x) from f'(x), then use two conditions to determine constants. It involves routine integration, applying the factor theorem, and solving simultaneous equations—all standard A-level techniques with no novel insight required. Slightly easier than average due to clear structure and familiar methods. |
| Spec | 1.02j Manipulate polynomials: expanding, factorising, division, factor theorem1.08a Fundamental theorem of calculus: integration as reverse of differentiation |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(f'(x) = 6x^2 + ax - 23 \Rightarrow f(x) = 2x^3 + \frac{1}{2}ax^2 - 23x + c\) | M1 | Integrates \(f'(x)\) with two correct indices |
| Fully correct integration (unsimplified acceptable) | A1 | \(+c\) must be seen or implied by the \(-12\) |
| \(``c" = -12\) | B1 | Deduces constant term is \(-12\) |
| \(f(-4) = 0 \Rightarrow 2(-4)^3 + \frac{1}{2}a(-4)^2 - 23(-4) - 12 = 0\) | dM1 | Sets up linear equation in \(a\) using \(f(-4)=0\); depends on some integration |
| \(a = \ldots \,(6)\) | dM1 | Solves linear equation in \(a\) |
| \(f(x) = 2x^3 + 3x^2 - 23x - 12\) or \((x+4)(2x^2-5x-3)\) or \((x+4)(2x+1)(x-3)\) | A1cso |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(f(x) = (x+4)(Ax^2+Bx+C)\) | M1 A1 | Uses \((x+4)\) as factor of cubic |
| \(f(x) = Ax^3+(4A+B)x^2+(4B+C)x+4C \Rightarrow C=-3\) | B1 | Deduces \(C=-3\) |
| \(f'(x) = 3Ax^2+2(4A+B)x+(4B+C)\) and \(f'(x)=6x^2+ax-23 \Rightarrow A=\ldots\) | dM1 | |
| Full method to get \(A\), \(B\) and \(C\) | dM1 | |
| \(f(x) = (x+4)(2x^2-5x-3)\) | A1cso |
# Question 8:
| Answer/Working | Mark | Guidance |
|---|---|---|
| $f'(x) = 6x^2 + ax - 23 \Rightarrow f(x) = 2x^3 + \frac{1}{2}ax^2 - 23x + c$ | M1 | Integrates $f'(x)$ with two correct indices |
| Fully correct integration (unsimplified acceptable) | A1 | $+c$ must be seen or implied by the $-12$ |
| $``c" = -12$ | B1 | Deduces constant term is $-12$ |
| $f(-4) = 0 \Rightarrow 2(-4)^3 + \frac{1}{2}a(-4)^2 - 23(-4) - 12 = 0$ | dM1 | Sets up linear equation in $a$ using $f(-4)=0$; depends on some integration |
| $a = \ldots \,(6)$ | dM1 | Solves linear equation in $a$ |
| $f(x) = 2x^3 + 3x^2 - 23x - 12$ or $(x+4)(2x^2-5x-3)$ or $(x+4)(2x+1)(x-3)$ | A1cso | |
**Alternative (via factor):**
| Answer/Working | Mark | Guidance |
|---|---|---|
| $f(x) = (x+4)(Ax^2+Bx+C)$ | M1 A1 | Uses $(x+4)$ as factor of cubic |
| $f(x) = Ax^3+(4A+B)x^2+(4B+C)x+4C \Rightarrow C=-3$ | B1 | Deduces $C=-3$ |
| $f'(x) = 3Ax^2+2(4A+B)x+(4B+C)$ and $f'(x)=6x^2+ax-23 \Rightarrow A=\ldots$ | dM1 | |
| Full method to get $A$, $B$ and $C$ | dM1 | |
| $f(x) = (x+4)(2x^2-5x-3)$ | A1cso | |\begin{enumerate}
\item A curve $C$ has equation $y = \mathrm { f } ( x )$
\end{enumerate}
Given that
\begin{itemize}
\item $\mathrm { f } ^ { \prime } ( x ) = 6 x ^ { 2 } + a x - 23$ where $a$ is a constant
\item the $y$ intercept of $C$ is - 12
\item ( $x + 4$ ) is a factor of $\mathrm { f } ( x )$\\
find, in simplest form, $\mathrm { f } ( x )$
\end{itemize}
\hfill \mbox{\textit{Edexcel Paper 2 2020 Q8 [6]}}