| Exam Board | Edexcel |
|---|---|
| Module | C4 (Core Mathematics 4) |
| Year | 2013 |
| Session | January |
| Marks | 15 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Areas by integration |
| Type | Area under polynomial curve |
| Difficulty | Moderate -0.3 This is a standard C4 area under curve question with routine steps: finding coordinates, finding a normal line equation, and integrating to find area. While it requires multiple techniques (differentiation for the normal, integration for area), these are all textbook applications with no novel problem-solving required. Slightly easier than average due to straightforward setup and standard methods. |
| Spec | 1.03g Parametric equations: of curves and conversion to cartesian1.07m Tangents and normals: gradient and equations1.07s Parametric and implicit differentiation1.08e Area between curve and x-axis: using definite integrals |
| Answer | Marks | Guidance |
|---|---|---|
| \(\{x=0\Rightarrow\} 0=1-\frac{1}{2}t \Rightarrow t=2\) | M1 | Applies \(x=0\) to obtain a value for \(t\) |
| When \(t=2\), \(y=2^2-1=3\) | A1 | Correct value for \(y\) |
| Answer | Marks | Guidance |
|---|---|---|
| \(\{y=0\Rightarrow\} 0=2^t-1\Rightarrow t=0\) | M1 | Applies \(y=0\) to obtain a value for \(t\) (must be seen in part (b)) |
| When \(t=0\), \(x=1-\frac{1}{2}(0)=1\) | A1 | \(x=1\) |
| Answer | Marks | Guidance |
|---|---|---|
| \(\frac{dx}{dt}=-\frac{1}{2}\) and \(\frac{dy}{dt}=2^t\ln 2\) or \(\frac{dy}{dt}=e^{t\ln 2}\ln 2\) | B1 | Both \(\frac{dx}{dt}\) and \(\frac{dy}{dt}\) correct |
| \(\frac{dy}{dx} = \frac{2^t\ln 2}{-\frac{1}{2}}\) | M1 | Attempts \(\frac{dy}{dt}\) divided by \(\frac{dx}{dt}\) |
| At \(A\), \(t=2\), so \(m(\text{T})=-8\ln 2 \Rightarrow m(\text{N})=\frac{1}{8\ln 2}\) | M1 | Applies \(t=2\) and \(m(\text{N})=\frac{-1}{m(\text{T})}\) |
| \(y-3=\frac{1}{8\ln 2}(x-0)\) or \(y=3+\frac{1}{8\ln 2}x\) | M1 A1 cso | Correct normal equation |
| Answer | Marks | Guidance |
|---|---|---|
| \(\text{Area}(R)=\int(2^t-1)\cdot\left(-\frac{1}{2}\right)dt\) | M1 | Complete substitution for both \(y\) and \(dx\) |
| \(x=-1\to t=4\) and \(x=1\to t=0\) | B1 | Correct changed limits |
| \(=\left\{-\frac{1}{2}\right\}\left(\frac{2^t}{\ln 2}-t\right)\) | M1* A1 | Integrates \(2^t\) correctly to give \(\frac{2^t}{\ln 2}\); correct integration of \((2^t-1)\) |
| \(\left\{-\frac{1}{2}\left[\frac{2^t}{\ln 2}-t\right]_4^0\right\} = -\frac{1}{2}\left(\left(\frac{1}{\ln 2}\right)-\left(\frac{16}{\ln 2}-4\right)\right)\) | dM1* | Substitutes changed limits and subtracts either way round |
| \(=\frac{15}{2\ln 2}-2\) | A1 | \(\frac{15}{2\ln 2}-2\) or equivalent |
| Answer | Marks | Guidance |
|---|---|---|
| \(\{x=0 \Rightarrow\}\ y = 2^2 - 1 = 3\) | M1, A1 | M1: Applies \(x=0\) in their Cartesian equation. A1: Correct answer of 3. [2] |
| Answer | Marks | Guidance |
|---|---|---|
| \(\{y=0 \Rightarrow\}\ 0 = 2^{2-2x} - 1 \Rightarrow 0 = 2-2x \Rightarrow x=1\) | M1, A1 | M1: Applies \(y=0\) to obtain a value for \(x\). A1: \(x=1\). Must be seen in part (b). [2] |
| Answer | Marks | Guidance |
|---|---|---|
| \(\frac{dy}{dx} = -2(2^{2-2x})\ln 2\) | M1, A1 | M1: \(\pm\lambda 2^{2-2x},\ \lambda \neq 1\). A1: \(-2(2^{2-2x})\ln 2\) or equivalent. |
| At \(A\), \(x=0\), so \(m(\mathbf{T}) = -8\ln 2 \Rightarrow m(\mathbf{N}) = \frac{1}{8\ln 2}\) | M1 | Applies \(x=0\) and \(m(\mathbf{N}) = \frac{-1}{m(\mathbf{T})}\) |
| \(y - 3 = \frac{1}{8\ln 2}(x-0)\) or \(y = 3 + \frac{1}{8\ln 2}x\) or equivalent | M1 A1 oe | As in the original scheme. [5] |
| Answer | Marks | Guidance |
|---|---|---|
| \(\text{Area}(R) = \int(2^{2-2x}-1)\,dx\) | M1 | Form the integral of their Cartesian equation of \(C\). |
| \(= \int_{-1}^{1}(2^{2-2x}-1)\,dx\) | B1 | For \(2^{2-2x}-1\) with limits \(x=-1\) and \(x=1\). |
| \(= \left(\frac{2^{2-2x}}{-2\ln 2} - x\right)\) | M1* | Either \(2^{2-2x} \to \frac{2^{2-2x}}{-2\ln 2}\), or \((2^{2-2x}-1) \to \frac{2^{2-2x}}{\pm\alpha(\ln 2)} - x\), or \((2^{2-2x}-1) \to \pm\alpha(\ln 2)(2^{2-2x}) - x\) |
| \(\left(\frac{2^{2-2x}}{-2\ln 2} - x\right)\) | A1 | \((2^{2-2x}-1) \to \frac{2^{2-2x}}{-2\ln 2} - x\) |
| \(= \left(\frac{1}{-2\ln 2}-1\right) - \left(\frac{16}{-2\ln 2}+1\right) = \frac{15}{2\ln 2} - 2\) | dM1*, A1 | dM1*: Depends on previous method mark. Substitutes limits of \(-1\) and their \(x_B\) and subtracts either way round. A1: \(\frac{15}{2\ln 2} - 2\) or equivalent. [6] |
# Question 5:
## Part (a):
| $\{x=0\Rightarrow\} 0=1-\frac{1}{2}t \Rightarrow t=2$ | M1 | Applies $x=0$ to obtain a value for $t$ |
| When $t=2$, $y=2^2-1=3$ | A1 | Correct value for $y$ |
## Part (b):
| $\{y=0\Rightarrow\} 0=2^t-1\Rightarrow t=0$ | M1 | Applies $y=0$ to obtain a value for $t$ (must be seen in part (b)) |
| When $t=0$, $x=1-\frac{1}{2}(0)=1$ | A1 | $x=1$ |
## Part (c):
| $\frac{dx}{dt}=-\frac{1}{2}$ and $\frac{dy}{dt}=2^t\ln 2$ or $\frac{dy}{dt}=e^{t\ln 2}\ln 2$ | B1 | Both $\frac{dx}{dt}$ and $\frac{dy}{dt}$ correct |
| $\frac{dy}{dx} = \frac{2^t\ln 2}{-\frac{1}{2}}$ | M1 | Attempts $\frac{dy}{dt}$ divided by $\frac{dx}{dt}$ |
| At $A$, $t=2$, so $m(\text{T})=-8\ln 2 \Rightarrow m(\text{N})=\frac{1}{8\ln 2}$ | M1 | Applies $t=2$ and $m(\text{N})=\frac{-1}{m(\text{T})}$ |
| $y-3=\frac{1}{8\ln 2}(x-0)$ or $y=3+\frac{1}{8\ln 2}x$ | M1 A1 cso | Correct normal equation |
## Part (d):
| $\text{Area}(R)=\int(2^t-1)\cdot\left(-\frac{1}{2}\right)dt$ | M1 | Complete substitution for both $y$ and $dx$ |
| $x=-1\to t=4$ and $x=1\to t=0$ | B1 | Correct changed limits |
| $=\left\{-\frac{1}{2}\right\}\left(\frac{2^t}{\ln 2}-t\right)$ | M1* A1 | Integrates $2^t$ correctly to give $\frac{2^t}{\ln 2}$; correct integration of $(2^t-1)$ |
| $\left\{-\frac{1}{2}\left[\frac{2^t}{\ln 2}-t\right]_4^0\right\} = -\frac{1}{2}\left(\left(\frac{1}{\ln 2}\right)-\left(\frac{16}{\ln 2}-4\right)\right)$ | dM1* | Substitutes changed limits and subtracts either way round |
| $=\frac{15}{2\ln 2}-2$ | A1 | $\frac{15}{2\ln 2}-2$ or equivalent |
# Question 5:
## Part (a):
| $\{x=0 \Rightarrow\}\ y = 2^2 - 1 = 3$ | M1, A1 | M1: Applies $x=0$ in their Cartesian equation. A1: Correct answer of 3. **[2]** |
## Part (b):
| $\{y=0 \Rightarrow\}\ 0 = 2^{2-2x} - 1 \Rightarrow 0 = 2-2x \Rightarrow x=1$ | M1, A1 | M1: Applies $y=0$ to obtain a value for $x$. A1: $x=1$. Must be seen in part (b). **[2]** |
## Part (c):
| $\frac{dy}{dx} = -2(2^{2-2x})\ln 2$ | M1, A1 | M1: $\pm\lambda 2^{2-2x},\ \lambda \neq 1$. A1: $-2(2^{2-2x})\ln 2$ or equivalent. |
| At $A$, $x=0$, so $m(\mathbf{T}) = -8\ln 2 \Rightarrow m(\mathbf{N}) = \frac{1}{8\ln 2}$ | M1 | Applies $x=0$ and $m(\mathbf{N}) = \frac{-1}{m(\mathbf{T})}$ |
| $y - 3 = \frac{1}{8\ln 2}(x-0)$ or $y = 3 + \frac{1}{8\ln 2}x$ or equivalent | M1 A1 oe | As in the original scheme. **[5]** |
## Part (d):
| $\text{Area}(R) = \int(2^{2-2x}-1)\,dx$ | M1 | Form the integral of their Cartesian equation of $C$. |
| $= \int_{-1}^{1}(2^{2-2x}-1)\,dx$ | B1 | For $2^{2-2x}-1$ with limits $x=-1$ and $x=1$. |
| $= \left(\frac{2^{2-2x}}{-2\ln 2} - x\right)$ | M1* | Either $2^{2-2x} \to \frac{2^{2-2x}}{-2\ln 2}$, or $(2^{2-2x}-1) \to \frac{2^{2-2x}}{\pm\alpha(\ln 2)} - x$, or $(2^{2-2x}-1) \to \pm\alpha(\ln 2)(2^{2-2x}) - x$ |
| $\left(\frac{2^{2-2x}}{-2\ln 2} - x\right)$ | A1 | $(2^{2-2x}-1) \to \frac{2^{2-2x}}{-2\ln 2} - x$ |
| $= \left(\frac{1}{-2\ln 2}-1\right) - \left(\frac{16}{-2\ln 2}+1\right) = \frac{15}{2\ln 2} - 2$ | dM1*, A1 | dM1*: Depends on previous method mark. Substitutes limits of $-1$ and their $x_B$ and subtracts either way round. A1: $\frac{15}{2\ln 2} - 2$ or equivalent. **[6]** |
---\begin{enumerate}[label=(\alph*)]
\item Show that $A$ has coordinates $( 0,3 )$.
\item Find the $x$ coordinate of the point $B$.
\item Find an equation of the normal to $C$ at the point $A$.
The region $R$, as shown shaded in Figure 2, is bounded by the curve $C$, the line $x = - 1$ and the $x$-axis.
\item Use integration to find the exact area of $R$.
\end{enumerate}
\hfill \mbox{\textit{Edexcel C4 2013 Q5 [15]}}