| Exam Board | Edexcel |
|---|---|
| Module | C1 (Core Mathematics 1) |
| Year | 2014 |
| Session | January |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Indefinite & Definite Integrals |
| Type | Find curve from gradient |
| Difficulty | Moderate -0.8 This is a straightforward C1 integration question requiring expansion of brackets, term-by-term integration, finding the constant using a given point, and basic curve sketching. All steps are routine procedures with no problem-solving insight needed, making it easier than average but not trivial due to the algebraic manipulation required. |
| Spec | 1.02j Manipulate polynomials: expanding, factorising, division, factor theorem1.08a Fundamental theorem of calculus: integration as reverse of differentiation1.08b Integrate x^n: where n != -1 and sums |
| Answer | Marks | Guidance |
|---|---|---|
| (a) | \(f'(x) = (x-2)(3x+4)\) | B1 |
| \(= 3x^2 - 2x - 8\) | ||
| \(y = \int 3x^2 - 2x - 8dx = 3x \cdot \frac{x^3}{3} - 2x \cdot \frac{x^2}{2} - 8x + c\) | M1 A1 | \(x^n \to x^{n+1}\) in any one term. For this M to be scored there must have been an attempt to expand the brackets and obtain a quadratic expression. A completely correct (unsimplified) expression for \(f(x)\), no need for \(+c\). Accept \(3 \frac{x^3}{3} - 2 \frac{x^2}{2} - 8x\). |
| \(x = 3, y = 6 \Rightarrow 6 = 27 - 9 - 24 + c\) | M1 | Substitutes \(x=3\) and \(y=6\) into their \(f(x)\) containing a constant ,\(c\)'' and proceed to find its value. |
| \(c = ...\) | A1 | Cso \(f(x) = x^3 - x^2 - 8x + 12\). Allow \(y = ...\). |
| \(f(x) = x^3 - x^2 - 8x + 12\) cso | Do not accept an answer produced from part (b). | |
| (b) | \(f(x) = (x-2)^2(x+p) \quad p = 3\) | B1 |
| \(f(x) = (x^2 - 4x + 4)(x + 3)\) | M1 | Multiplies out a pair of brackets first, usually \((x-2)^2\) and then attempts to multiply by the third. The minimum criteria should be the first multiplication is a 3T quadratic with correct first and last terms and the second is a 4T cubic with correct first and last terms. Accept an expression involving \(p\) for M1. |
| \(f(x) = x^3 - 4x^2 + 3x^2 + 4x - 12x + 12\) | ||
| \(f(x) = x^3 - x^2 - 8x + 12\) cso | M1 A1 | cso \(f(x) = x^3 - x^2 - 8x + 12\), which must be the same as their answer for part (a). |
| (c) | Shape | B1 |
| Min at \((2,0)\) | B1 | There is a turning point at \((2, 0)\). Accept 2 marked as a maximum or minimum on the \(x\)-axis. |
| Crosses \(x\)-axis at \((-3,0)\) | B1 ft | Graph crosses the \(x\)-axis at \((-3, 0)\). Accept \(-3\) marked at the point where the curve crosses the \(x\)-axis. You may follow through on their values of '– \(p\)' as long as \(p < 2\). |
| Crosses \(y\)-axis at \((0, 12)\) | B1 | Graph crosses the \(y\)-axis at \((0,12)\). Accept 12 marked on the \(y\)-axis. |
(a) | $f'(x) = (x-2)(3x+4)$ | B1 | Writes $(x-2)(3x+4)$ as $3x^2 - 2x - 8$.
| $= 3x^2 - 2x - 8$ | | |
| $y = \int 3x^2 - 2x - 8dx = 3x \cdot \frac{x^3}{3} - 2x \cdot \frac{x^2}{2} - 8x + c$ | M1 A1 | $x^n \to x^{n+1}$ in any one term. For this M to be scored there must have been an attempt to expand the brackets and obtain a quadratic expression. A completely correct (unsimplified) expression for $f(x)$, no need for $+c$. Accept $3 \frac{x^3}{3} - 2 \frac{x^2}{2} - 8x$.
| $x = 3, y = 6 \Rightarrow 6 = 27 - 9 - 24 + c$ | M1 | Substitutes $x=3$ and $y=6$ into their $f(x)$ containing a constant ,$c$'' and proceed to find its value.
| $c = ...$ | A1 | Cso $f(x) = x^3 - x^2 - 8x + 12$. Allow $y = ...$.
| $f(x) = x^3 - x^2 - 8x + 12$ cso | | Do not accept an answer produced from part (b).
(b) | $f(x) = (x-2)^2(x+p) \quad p = 3$ | B1 | States $p = 3$. This may be obtained from subbing $(3,6)$ into $f(x) = (x-2)^2(x + p)$.
| $f(x) = (x^2 - 4x + 4)(x + 3)$ | M1 | Multiplies out a pair of brackets first, usually $(x-2)^2$ and then attempts to multiply by the third. The minimum criteria should be the first multiplication is a 3T quadratic with correct first and last terms and the second is a 4T cubic with correct first and last terms. Accept an expression involving $p$ for M1.
| $f(x) = x^3 - 4x^2 + 3x^2 + 4x - 12x + 12$ | | |
| $f(x) = x^3 - x^2 - 8x + 12$ cso | M1 A1 | cso $f(x) = x^3 - x^2 - 8x + 12$, which must be the same as their answer for part (a).
(c) | Shape | B1 | Shape $+x^3$ graph with one maximum and one minimum. Its position is not important. It must appear to tend to $+ \infty$ at the rhs and $-\infty$ at the lhs. The curve must extend beyond its "maximum" point and minimum points. Eg. These are NOT acceptable.
| Min at $(2,0)$ | B1 | There is a turning point at $(2, 0)$. Accept 2 marked as a maximum or minimum on the $x$-axis.
| Crosses $x$-axis at $(-3,0)$ | B1 ft | Graph crosses the $x$-axis at $(-3, 0)$. Accept $-3$ marked at the point where the curve crosses the $x$-axis. You may follow through on their values of '– $p$' as long as $p < 2$.
| Crosses $y$-axis at $(0, 12)$ | B1 | Graph crosses the $y$-axis at $(0,12)$. Accept 12 marked on the $y$-axis.
---9. A curve with equation $y = \mathrm { f } ( x )$ passes through the point ( 3,6 ). Given that
$$f ^ { \prime } ( x ) = ( x - 2 ) ( 3 x + 4 )$$
\begin{enumerate}[label=(\alph*)]
\item use integration to find $\mathrm { f } ( x )$. Give your answer as a polynomial in its simplest form.
\item Show that $\mathrm { f } ( x ) \equiv ( x - 2 ) ^ { 2 } ( x + p )$, where $p$ is a positive constant. State the value of $p$.
\item Sketch the graph of $y = \mathrm { f } ( x )$, showing the coordinates of any points where the curve touches or crosses the coordinate axes.
\end{enumerate}
\hfill \mbox{\textit{Edexcel C1 2014 Q9 [12]}}