| Exam Board | Edexcel |
|---|---|
| Module | PMT Mocks (PMT Mocks) |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Curve Sketching |
| Type | Parameter values from curve properties |
| Difficulty | Standard +0.3 This is a straightforward curve sketching question requiring students to find constants using given conditions (substituting coordinates and using stationary point conditions), then interpret the derivative and sketch. All techniques are standard A-level procedures with no novel problem-solving required, making it slightly easier than average. |
| Spec | 1.02n Sketch curves: simple equations including polynomials1.02p Interpret algebraic solutions: graphically1.07n Stationary points: find maxima, minima using derivatives1.07o Increasing/decreasing: functions using sign of dy/dx |
| Answer | Marks | Guidance |
|---|---|---|
| (a) | \(y = -x(x-3)^2\) | A1 |
| (b) | \(1 < x < 3\) or equivalent such as \(x > 1, x < 3\) or \(x \in (1,3)\), \(\{x: x > 1\} \cap \{x: x < 3\}\) | B1 |
| (c) | \(\{k: k > 0\} \cup \{k: k < -4\}\) | A1 |
**(a)** | $y = -x(x-3)^2$ | A1 | M1 Realises that the equation of $C$ is of the form $y = ax(x-3)^2$. Condone with $a = 1$ for this mark. dM1 Substitutes $(1,-4)$ into the form $y = ax(x-3)^2$ and attempts to find the value for $a$. E.g. $-4 = a(1)((1-3))^2 \Rightarrow -4 = 4a \Rightarrow a = \ldots$. A1 Uses all of the information to form a correct equation for $C$. E.g. $a = -1 \Rightarrow y = -x(x-3)^2$. OR M1 Realises that the equation of $C$ is of the form $y = ax^3 + bx^2 + cx$ and forms two equations in $a, b$ and $c$. Condone with $a = 1$ for this mark. There are four equations that could be formed, only two are necessary for this mark. Using $(1,-4)$: $-4 = a + b + c$. Using $(3,0)$: $0 = 27a + 9b + 3c \Rightarrow 0 = 9a + 3b + c$. Using $\frac{dy}{dx} = 0$ at $x = 1$: $3ax^2 + 2bx + c = 0 \Rightarrow 3a + 2b + c = 0$. Using $\frac{dy}{dx} = 0$ at $x = 3$: $3ax^2 + 2bx + c = 0 \Rightarrow 27a + 6b + c = 0$. dM1 Forms and solves three different equations, one of which must be using $(1,-4)$ to find values for $a, b$ and $c$. E.g. $a + b + c = -4$ (1), $9a + 3b + c = 0$ (2) subtract (1) from (2) $\Rightarrow 8a + 2b = 4$, $9a + 3b + c = 0$ (2), $3a + 2b + c = 0$ (3) subtract (3) from (2) $\Rightarrow 6a + b = 0$. Solves $8a + 2b = 4$ and $6a + b = 0$ simultaneously $\Rightarrow a = -1, b = 6, c = -9$. A1 Uses all of the information to form a correct equation for $C$. $y = -x^3 + 6x^2 - 9x = -x(x^2 - 6x + 9) = -x(x-3)^2$ |
**(b)** | $1 < x < 3$ or equivalent such as $x > 1, x < 3$ or $x \in (1,3)$, $\{x: x > 1\} \cap \{x: x < 3\}$ | B1 | Deduces $1 < x < 3$ or equivalent such as $x > 1, x < 3$ |
**(c)** | $\{k: k > 0\} \cup \{k: k < -4\}$ | A1 | M1 States either $k > 0$ or $k < -4$ |
---6.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{48f9a252-61a2-491d-94d0-8470aee96942-07_864_995_299_495}
\captionsetup{labelformat=empty}
\caption{Figure 1}
\end{center}
\end{figure}
The figure 1 shows sketch of the curve $C$ with equation $y = \mathrm { f } ( x )$.
$$f ( x ) = a x ( x - b ) ^ { 2 } , x \in R$$
where $a$ and $b$ are constants.\\
The curve passes through the origin and touches the $x$-axis at the point $( 3,0 )$.\\
There is a minimum point at $( 1 , - 4 )$ and a maximum point at $( 3,0 )$.\\
a. Find the equation of $C$.\\
b. Deduce the values of $x$ for which
$$\mathrm { f } ^ { \prime } ( x ) > 0$$
Given that the line with equation $y = k$, where $k$ is a constant, intersects $C$ at exactly one point,\\
c. State the possible values for $k$.\\
\hfill \mbox{\textit{Edexcel PMT Mocks Q6 [6]}}