| Exam Board | Edexcel |
|---|---|
| Module | C34 (Core Mathematics 3 & 4) |
| Year | 2015 |
| Session | June |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Curve Sketching |
| Type | Sketch |f(x)| only |
| Difficulty | Standard +0.3 This is a straightforward C3/C4 question requiring product rule differentiation to find a stationary point, then basic analysis of the function's behavior. Part (a) is standard calculus, part (b) requires understanding of function behavior from the sketch, and part (c) is routine modulus transformation. Slightly easier than average due to the guided structure and standard techniques. |
| Spec | 1.02s Modulus graphs: sketch graph of |ax+b|1.07j Differentiate exponentials: e^(kx) and a^(kx)1.07n Stationary points: find maxima, minima using derivatives |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(f'(x) = e^x \times 2 + (2x-5)e^x\) | M1A1 | Product rule giving form \(Ae^x + (2x-5)e^x\) |
| \(f'(x) = 0 \Rightarrow (2x-3)e^x = 0 \Rightarrow x = \frac{3}{2}\) | M1A1 | Sets \(f'(x)=0\) and solves |
| Coordinates of \(A = \left(\frac{3}{2}, -2e^{\frac{3}{2}}\right)\), obtains \(y = -2e^{\frac{3}{2}}\) | A1ft | Follow through their \(x\); allow even if positive; may not be given as coordinates |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(-2e^{\frac{3}{2}} < k < 0\) | M1A1 | M1 for using their minimum \(y\) value as lower limit; both inequalities must be strictly less than; need \(k\) not \(y\); need exact lower limit |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Correct shape including cusp | B1 | Curve in first two quadrants only, maximum in first quadrant, tends to \(x\)-axis as \(x \to -\infty\), touches \(x\)-axis with discontinuous gradient (cusp) |
| \(\left(\frac{5}{2}, 0\right)\) only | B1 | Must be only crossing point on \(x\)-axis; 2.5 sufficient on sketch |
| \((0, 5)\) only | B1 | Must be only crossing point on \(y\)-axis; 5 sufficient on sketch |
## Question 3:
### Part (a):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $f'(x) = e^x \times 2 + (2x-5)e^x$ | M1A1 | Product rule giving form $Ae^x + (2x-5)e^x$ |
| $f'(x) = 0 \Rightarrow (2x-3)e^x = 0 \Rightarrow x = \frac{3}{2}$ | M1A1 | Sets $f'(x)=0$ and solves |
| Coordinates of $A = \left(\frac{3}{2}, -2e^{\frac{3}{2}}\right)$, obtains $y = -2e^{\frac{3}{2}}$ | A1ft | Follow through their $x$; allow even if positive; may not be given as coordinates |
**(5 marks)**
### Part (b):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $-2e^{\frac{3}{2}} < k < 0$ | M1A1 | M1 for using their minimum $y$ value as lower limit; both inequalities must be strictly less than; need $k$ not $y$; need exact lower limit |
**(2 marks)**
### Part (c):
| Answer/Working | Marks | Guidance |
|---|---|---|
| Correct shape including cusp | B1 | Curve in first two quadrants only, maximum in first quadrant, tends to $x$-axis as $x \to -\infty$, touches $x$-axis with discontinuous gradient (cusp) |
| $\left(\frac{5}{2}, 0\right)$ only | B1 | Must be only crossing point on $x$-axis; 2.5 sufficient on sketch |
| $(0, 5)$ only | B1 | Must be only crossing point on $y$-axis; 5 sufficient on sketch |
**(3 marks)**
---3.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{4c08fbab-283e-4c92-89a4-10f68f37e133-05_799_885_118_534}
\captionsetup{labelformat=empty}
\caption{Figure 1}
\end{center}
\end{figure}
Figure 1 shows a sketch of part of the curve with equation $y = \mathrm { f } ( x )$, where
$$\mathrm { f } ( x ) = ( 2 x - 5 ) \mathrm { e } ^ { x } , \quad x \in \mathbb { R }$$
The curve has a minimum turning point at $A$.
\begin{enumerate}[label=(\alph*)]
\item Use calculus to find the exact coordinates of $A$.
Given that the equation $\mathrm { f } ( x ) = k$, where $k$ is a constant, has exactly two roots,
\item state the range of possible values of $k$.
\item Sketch the curve with equation $y = | \mathrm { f } ( x ) |$.
Indicate clearly on your sketch the coordinates of the points at which the curve crosses or meets the axes.
\end{enumerate}
\hfill \mbox{\textit{Edexcel C34 2015 Q3 [10]}}