| Exam Board | OCR MEI |
|---|---|
| Module | Paper 1 (Paper 1) |
| Year | 2018 |
| Session | June |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Generalised Binomial Theorem |
| Type | Factoring out constants first |
| Difficulty | Standard +0.8 This is a substantial multi-part question requiring differentiation of a composite function, binomial expansion with factoring (requiring recognition that 27-8x³ = 27(1-8x³/27)), validity of series approximation, trapezium rule application, and graphical interpretation. While individual techniques are A-level standard, the combination of skills, the need to factor out constants correctly for the binomial expansion, and the conceptual understanding required for parts (iii) and (v) make this moderately challenging but not exceptional. |
| Spec | 1.04c Extend binomial expansion: rational n, |x|<11.04d Binomial expansion validity: convergence conditions1.07l Derivative of ln(x): and related functions1.09f Trapezium rule: numerical integration |
| \(x\) | \(f ( x )\) | \(f ^ { \prime } ( x )\) |
| 0 | 3 | 0 |
| 0.25 | 2.9954 | - 0.056 |
| 0.5 | 2.9625 | - 0.228 |
| 0.75 | 2.8694 | - 0.547 |
| 1 | 2.6684 | - 1.124 |
| 1.25 | 2.2490 | - 1.977 |
| 1.5 | 0 | ERROR |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(f'(x) = \frac{1}{3}(27 - 8x^3)^{-2/3} \times (-24x^2)\) | M1 (AO 1.1a) | Using the chain rule |
| \(= \frac{-8x^2}{(27-8x^3)^{2/3}}\) | A1 (AO 1.1) | Allow unsimplified |
| \(f'(1.5) = -\frac{8 \times 1.5^2}{0}\) and dividing by zero gives the error | E1 [3] (AO 2.4) | Sufficient to say "can't divide by zero" oe |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \((27 - 8x^3)^{1/3} = 27^{1/3}\left(1 - \frac{8}{27}x^3\right)^{1/3}\) | B1 (AO 3.1a) | Dealing with the 27 correctly |
| \(= 3\left[1 + \left(\frac{1}{3}\right)\left(-\frac{8x^3}{27}\right) + \frac{\left(\frac{1}{3}\right)\left(-\frac{2}{3}\right)}{2!}\left(-\frac{8x^3}{27}\right)^2 + ...\right]\) | M1 (AO 1.1a) | Using the Binomial expansion substantially correctly |
| \(= 3 - \frac{8x^3}{27} - \frac{64x^6}{2187} + ...\) | A1 [3] (AO 1.1b) | Cao |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| The binomial expansion is valid for \(\left | -\frac{8x^3}{27}\right | < 1\) |
| \( | x | < 1.5\) and the limits of the integral are completely in this interval |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\frac{0.25}{2}(3 + 2.6684 + 2(2.9954 + 2.9625 + 2.8694))\) | B1 (AO 1.1a) | \(h = 0.25\) used |
| M1 (AO 1.1b) | For sum in the bracket — condone one slip. Allow for 2.92 or better | |
| \(= \frac{0.25}{2} \times 23.3224 = 2.9153\) | A1 [3] (AO 1.1b) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| There is area between the curve and the top of the trapezia, so some area is missing from the estimate | E1 [1] (AO 2.4) | Allow for any sensible explanation e.g. the trapezia are under the curve. "The curve is concave downwards" on its own is not quite enough |
# Question 13:
## Part (i):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $f'(x) = \frac{1}{3}(27 - 8x^3)^{-2/3} \times (-24x^2)$ | **M1** (AO 1.1a) | Using the chain rule |
| $= \frac{-8x^2}{(27-8x^3)^{2/3}}$ | **A1** (AO 1.1) | Allow unsimplified |
| $f'(1.5) = -\frac{8 \times 1.5^2}{0}$ and dividing by zero gives the error | **E1** [3] (AO 2.4) | Sufficient to say "can't divide by zero" oe |
## Part (ii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $(27 - 8x^3)^{1/3} = 27^{1/3}\left(1 - \frac{8}{27}x^3\right)^{1/3}$ | **B1** (AO 3.1a) | Dealing with the 27 correctly |
| $= 3\left[1 + \left(\frac{1}{3}\right)\left(-\frac{8x^3}{27}\right) + \frac{\left(\frac{1}{3}\right)\left(-\frac{2}{3}\right)}{2!}\left(-\frac{8x^3}{27}\right)^2 + ...\right]$ | **M1** (AO 1.1a) | Using the Binomial expansion substantially correctly |
| $= 3 - \frac{8x^3}{27} - \frac{64x^6}{2187} + ...$ | **A1** [3] (AO 1.1b) | Cao |
## Part (iii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| The binomial expansion is valid for $\left|-\frac{8x^3}{27}\right| < 1$ | **B1** (AO 2.4) | Allow unsimplified but must use correct modulus notation or equivalent |
| $|x| < 1.5$ and the limits of the integral are completely in this interval | **E1** [2] (AO 2.3) | Must indicate that the limits of the integral lie in their interval for which the expansion is valid |
## Part (iv):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\frac{0.25}{2}(3 + 2.6684 + 2(2.9954 + 2.9625 + 2.8694))$ | **B1** (AO 1.1a) | $h = 0.25$ used |
| | **M1** (AO 1.1b) | For sum in the bracket — condone one slip. Allow for 2.92 or better |
| $= \frac{0.25}{2} \times 23.3224 = 2.9153$ | **A1** [3] (AO 1.1b) | |
## Part (v):
| Answer | Marks | Guidance |
|--------|-------|----------|
| There is area between the curve and the top of the trapezia, so some area is missing from the estimate | **E1** [1] (AO 2.4) | Allow for any sensible explanation e.g. the trapezia are under the curve. "The curve is concave downwards" on its own is not quite enough |
---13 The function $\mathrm { f } ( x )$ is defined by $\mathrm { f } ( x ) = \sqrt [ 3 ] { 27 - 8 x ^ { 3 } }$. Jenny uses her scientific calculator to create a table of values for $\mathrm { f } ( x )$ and $\mathrm { f } ^ { \prime } ( x )$.
\begin{center}
\begin{tabular}{ | r | r | r | }
\hline
\multicolumn{1}{|c|}{$x$} & \multicolumn{1}{c|}{$f ( x )$} & \multicolumn{1}{c|}{$f ^ { \prime } ( x )$} \\
\hline
0 & 3 & 0 \\
0.25 & 2.9954 & - 0.056 \\
0.5 & 2.9625 & - 0.228 \\
0.75 & 2.8694 & - 0.547 \\
1 & 2.6684 & - 1.124 \\
1.25 & 2.2490 & - 1.977 \\
1.5 & 0 & ERROR \\
\hline
\end{tabular}
\end{center}
(i) Use calculus to find an expression for $\mathrm { f } ^ { \prime } ( x )$ and hence explain why the calculator gives an error for $\mathrm { f } ^ { \prime } ( 1.5 )$.\\
(ii) Find the first three terms of the binomial expansion of $\mathrm { f } ( x )$.\\
(iii) Jenny integrates the first three terms of the binomial expansion of $\mathrm { f } ( x )$ to estimate the value of $\int _ { 0 } ^ { 1 } \sqrt [ 3 ] { 27 - 8 x ^ { 3 } } \mathrm {~d} x$. Explain why Jenny's method is valid in this case. (You do not need to evaluate Jenny's approximation.)\\
(iv) Use the trapezium rule with 4 strips to obtain an estimate for $\int _ { 0 } ^ { 1 } \sqrt [ 3 ] { 27 - 8 x ^ { 3 } } \mathrm {~d} x$.
The calculator gives 2.92117438 for $\int _ { 0 } ^ { 1 } \sqrt [ 3 ] { 27 - 8 x ^ { 3 } } \mathrm {~d} x$. The graph of $y = \mathrm { f } ( x )$ is shown in Fig. 13.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{904025c9-6d68-4344-bd41-8c0fccfcf92f-08_490_906_1505_568}
\captionsetup{labelformat=empty}
\caption{Fig. 13}
\end{center}
\end{figure}
(v) Explain why the trapezium rule gives an underestimate.
\hfill \mbox{\textit{OCR MEI Paper 1 2018 Q13 [12]}}