| Exam Board | Edexcel |
|---|---|
| Module | M2 (Mechanics 2) |
| Year | 2014 |
| Session | January |
| 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 and curve sketching question requiring standard techniques: expand and integrate a quadratic, use a boundary condition to find the constant, factorise, and sketch using standard features. While it has multiple parts (12 marks total), each step follows routine A-level procedures with no novel problem-solving required, making it slightly easier than average. |
| Spec | 1.02j Manipulate polynomials: expanding, factorising, division, factor theorem1.08b Integrate x^n: where n != -1 and sums |
| Answer | Marks | Guidance |
|---|---|---|
| (a) \(f'(x) = (x-2)(3x+4)\) | B1 | Writes \((x-2)(3x+4)\) as \(3x^2 - 2x - 8\) |
| \(= 3x^2 - 2x - 8\) | M1A1 | \(x^n \to x^{n+1}\) 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 |
| \(y = \int 3x^2 - 2x - 8dx = 3x \cdot \frac{x^3}{3} - 2x \cdot \frac{x^2}{2} - 8x + c\) | M1 | Substitutes \(x=3\) and \(y=6\) into their \(f(x)\) containing a constant, \(c\) and proceed to find its value. |
| \(x = 3, y = 6 \Rightarrow 6 = 27 - 9 - 24 + c\) | A1 | Cso \(f(x) = x^3 - x^2 - 8x + 12\). Allow \(y = ...\) |
| \(c = ...\) | ||
| \(f(x) = x^3 - x^2 - 8x + 12\) cso | Do not accept an answer produced from part (b) | |
| (5 marks) | ||
| (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 |
| \(f(x) = x^3 - 4x^2 + 3x^2 + 4x - 12x + 12\) | M1A1 | |
| \(f(x) = x^3 - x^2 - 8x + 12\) cso | ||
| (3 marks) | ||
| (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) B1ft | 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. | |
| (4 marks) | ||
| (12 marks) |
(a) $f'(x) = (x-2)(3x+4)$ | B1 | Writes $(x-2)(3x+4)$ as $3x^2 - 2x - 8$
$= 3x^2 - 2x - 8$ | M1A1 | $x^n \to x^{n+1}$ 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
$y = \int 3x^2 - 2x - 8dx = 3x \cdot \frac{x^3}{3} - 2x \cdot \frac{x^2}{2} - 8x + c$ | M1 | Substitutes $x=3$ and $y=6$ into their $f(x)$ containing a constant, $c$ and proceed to find its value.
$x = 3, y = 6 \Rightarrow 6 = 27 - 9 - 24 + c$ | A1 | Cso $f(x) = x^3 - x^2 - 8x + 12$. Allow $y = ...$
$c = ...$ | |
$f(x) = x^3 - x^2 - 8x + 12$ cso | | Do not accept an answer produced from part (b)
| | **(5 marks)**
(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
$f(x) = x^3 - 4x^2 + 3x^2 + 4x - 12x + 12$ | M1A1 |
$f(x) = x^3 - x^2 - 8x + 12$ cso | |
| | **(3 marks)**
(c) | | **Shape** B1 | $+x^3$ 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: [shows examples of curves that don't satisfy shape]
| | **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)** B1ft | 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.
| | **(4 marks)**
| | **(12 marks)**
---A curve with equation $y = f(x)$ passes through the point $(3, 6)$. Given that
$$f'(x) = (x - 2)(3x + 4)$$
\begin{enumerate}[label=(\alph*)]
\item use integration to find $f(x)$. Give your answer as a polynomial in its simplest form. [5]
\item Show that $f(x) = (x - 2)^2(x + p)$, where $p$ is a positive constant. State the value of $p$. [3]
\item Sketch the graph of $y = f(x)$, showing the coordinates of any points where the curve touches or crosses the coordinate axes. [4]
\end{enumerate}
\hfill \mbox{\textit{Edexcel M2 2014 Q9 [12]}}