| Exam Board | OCR |
|---|---|
| Module | H240/02 (Pure Mathematics and Statistics) |
| Year | 2019 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Topic | Generalised Binomial Theorem |
| Type | Expand and state validity |
| Difficulty | Moderate -0.3 This is a straightforward application of the binomial theorem with both positive integer and fractional powers. Part (a) is routine calculation, part (b)(i) requires standard binomial expansion for fractional powers, part (b)(ii) tests knowledge of validity conditions (|3x| < 1), and part (b)(iii) involves substitution and arithmetic. All parts are textbook-standard with no novel problem-solving required, making it slightly easier than average. |
| Spec | 1.04a Binomial expansion: (a+b)^n for positive integer n1.04c Extend binomial expansion: rational n, |x|<11.04d Binomial expansion validity: convergence conditions |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(= -48384\) or \(-48400\) | B1 | Allow \(-48384x^5\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(1 + 0.5 \times 3x + \frac{0.5\times(-0.5)}{2}\times(3x)^2 + \frac{0.5\times(-0.5)(-1.5)}{3!}\times(3x)^3\) | M1 | M1 for at least 3 terms correct; Condone any missing brackets; SC \(1+\frac{3}{2}x - \frac{3}{8}x^2 + \frac{3}{16}x^3\): M1 |
| \(= 1 + \frac{3}{2}x - \frac{9}{8}x^2 + \frac{27}{16}x^3\) or \(1 + 1.5x - 1.125x^2 + 1.6875x^3\) | A1, A1 | A1 for 3 terms correct; A1 for all correct |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(-\frac{1}{3} < x < \frac{1}{3}\) | B1 | Allow \( |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Sub \(x = 0.01\) in their expansion | M1 | Other correct methods may be seen, eg subst \(x = 0.2\) & \(\sqrt{1.6}\) |
| gives \(\sqrt{1.03} = 1.014889\ldots\) | A1 | Allow 1.01489 here (5 dps for series) |
| From series: \(\sqrt{103} = 10.14889(188\ldots)\) | If no working seen, 10.1488919 or better must be seen as evidence that series has been used | |
| From calculator \(\sqrt{103} = 10.14889(157\ldots)\) | A1 | Both must be seen for A1; Allow without statement; 5 dps for \(\sqrt{103}\) in both |
# Question 2:
## Part (a)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $= -48384$ or $-48400$ | B1 | Allow $-48384x^5$ |
## Part (b)(i)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $1 + 0.5 \times 3x + \frac{0.5\times(-0.5)}{2}\times(3x)^2 + \frac{0.5\times(-0.5)(-1.5)}{3!}\times(3x)^3$ | M1 | M1 for at least 3 terms correct; Condone any missing brackets; SC $1+\frac{3}{2}x - \frac{3}{8}x^2 + \frac{3}{16}x^3$: M1 |
| $= 1 + \frac{3}{2}x - \frac{9}{8}x^2 + \frac{27}{16}x^3$ or $1 + 1.5x - 1.125x^2 + 1.6875x^3$ | A1, A1 | A1 for 3 terms correct; A1 for all correct |
## Part (b)(ii)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $-\frac{1}{3} < x < \frac{1}{3}$ | B1 | Allow $|x| < \frac{1}{3}$ |
## Part (b)(iii)
| Answer/Working | Mark | Guidance |
|---|---|---|
| Sub $x = 0.01$ in their expansion | M1 | Other correct methods may be seen, eg subst $x = 0.2$ & $\sqrt{1.6}$ |
| gives $\sqrt{1.03} = 1.014889\ldots$ | A1 | Allow 1.01489 here (5 dps for series) |
| From series: $\sqrt{103} = 10.14889(188\ldots)$ | | If no working seen, 10.1488919 or better must be seen as evidence that series has been used |
| From calculator $\sqrt{103} = 10.14889(157\ldots)$ | A1 | Both must be seen for A1; Allow without statement; 5 dps for $\sqrt{103}$ in both |
---2
\begin{enumerate}[label=(\alph*)]
\item Find the coefficient of $x ^ { 5 }$ in the expansion of $( 3 - 2 x ) ^ { 8 }$.
\item \begin{enumerate}[label=(\roman*)]
\item Expand $( 1 + 3 x ) ^ { 0.5 }$ as far as the term in $x ^ { 3 }$.
\item State the range of values of $x$ for which your expansion is valid.
A student suggests the following check to determine whether the expansion obtained in part (b)(i) may be correct.\\
"Use the expansion to find an estimate for $\sqrt { 103 }$, correct to five decimal places, and compare this with the value of $\sqrt { 103 }$ given by your calculator."
\item Showing your working, carry out this check on your expansion from part (b)(i).
\end{enumerate}\end{enumerate}
\hfill \mbox{\textit{OCR H240/02 2019 Q2 [8]}}