| Exam Board | OCR |
|---|---|
| Module | H240/01 (Pure Mathematics) |
| Year | 2023 |
| Session | June |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Topic | Generalised Binomial Theorem |
| Type | Finding unknown constant from coefficient |
| Difficulty | Standard +0.3 This is a straightforward application of the binomial theorem with a fractional exponent followed by algebraic manipulation. Part (a) requires routine use of the formula, part (b) is standard recall of validity conditions, and part (c) involves expanding two binomials, collecting x² terms, and solving a quadratic equation. While it requires careful algebra across multiple steps, it follows a predictable pattern with no novel insight needed—slightly easier than average. |
| Spec | 1.04c Extend binomial expansion: rational n, |x|<11.04d Binomial expansion validity: convergence conditions |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\left(1+\frac{3}{4}x\right)^{\frac{3}{2}} = 1 + \left(\frac{3}{2}\right)\left(\frac{3}{4}x\right)\) | B1 | Correct first two terms. Allow unsimplified. Expect \(1 + \frac{9}{8}x\) |
| \(+\frac{\left(\frac{3}{2}\right)\left(\frac{1}{2}\right)}{2}\left(\frac{3}{4}x\right)^2\) | M1 | Attempt third term. Condone lack of brackets when attempting to square \(\frac{3}{4}x\). Coefficient must be \(\frac{(\frac{3}{2})(\frac{1}{2})}{2}\) or equiv |
| Correct third term | A1 | Allow unsimplified. \(\frac{3}{4}x^2\) is A0 unless recovered by later work. Expect \(\frac{27}{128}x^2\) |
| \((4+3x)^{\frac{3}{2}} = 8\left(1+\frac{3}{4}x\right)^{\frac{3}{2}} = 8 + 9x + \frac{27}{16}x^2\) | B1FT | Multiply their 3 term expansion by 8. Bracket expanded and coefficients simplified. If B1M1A1 awarded but attempt to simplify goes wrong, B1FT is not also awarded |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\ | x\ | < \frac{4}{3}\) or \(-\frac{4}{3} < x < \frac{4}{3}\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\left(8+9x+\frac{27}{16}x^2\right)\left(1+2ax+a^2x^2\right)\); coeff of \(x^2\) is \(8a^2 + 18a + \frac{27}{16}\) | M1 | Expand \((1+ax)^2\) and attempt at least one coeff of \(x^2\). Allow \(ax\) as middle term and/or \(a^2x^2\) as third term |
| \(8a^2 + 18a + \frac{27}{16} = \frac{107}{16}\), so \(8a^2 + 18a - 5 = 0\) | A1 | Equate to \(\frac{107}{16}\) to obtain correct quadratic. aef, including unsimplified. A0 if mix of terms and coefficients, but can be recovered |
| \((2a+5)(4a-1) = 0\); \(a = -\frac{5}{2}\) and \(a = \frac{1}{4}\) | A1 | Solve quadratic, possibly BC, to obtain \(a = -\frac{5}{2}\) and \(a = \frac{1}{4}\) |
## Question 8:
### Part (a):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\left(1+\frac{3}{4}x\right)^{\frac{3}{2}} = 1 + \left(\frac{3}{2}\right)\left(\frac{3}{4}x\right)$ | B1 | Correct first two terms. Allow unsimplified. Expect $1 + \frac{9}{8}x$ |
| $+\frac{\left(\frac{3}{2}\right)\left(\frac{1}{2}\right)}{2}\left(\frac{3}{4}x\right)^2$ | M1 | Attempt third term. Condone lack of brackets when attempting to square $\frac{3}{4}x$. Coefficient must be $\frac{(\frac{3}{2})(\frac{1}{2})}{2}$ or equiv |
| Correct third term | A1 | Allow unsimplified. $\frac{3}{4}x^2$ is **A0** unless recovered by later work. Expect $\frac{27}{128}x^2$ |
| $(4+3x)^{\frac{3}{2}} = 8\left(1+\frac{3}{4}x\right)^{\frac{3}{2}} = 8 + 9x + \frac{27}{16}x^2$ | B1FT | Multiply their 3 term expansion by 8. Bracket expanded and coefficients simplified. If **B1M1A1** awarded but attempt to simplify goes wrong, **B1FT** is **not** also awarded |
### Part (b):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\|x\| < \frac{4}{3}$ or $-\frac{4}{3} < x < \frac{4}{3}$ | B1 | Could also be $\|x\| \leq \frac{4}{3}$ or $-\frac{4}{3} \leq x \leq \frac{4}{3}$, as $n > 0$. Must be condition for $x$, not $kx$ |
### Part (c):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\left(8+9x+\frac{27}{16}x^2\right)\left(1+2ax+a^2x^2\right)$; coeff of $x^2$ is $8a^2 + 18a + \frac{27}{16}$ | M1 | Expand $(1+ax)^2$ and attempt at least one coeff of $x^2$. Allow $ax$ as middle term and/or $a^2x^2$ as third term |
| $8a^2 + 18a + \frac{27}{16} = \frac{107}{16}$, so $8a^2 + 18a - 5 = 0$ | A1 | Equate to $\frac{107}{16}$ to obtain correct quadratic. aef, including unsimplified. **A0** if mix of terms and coefficients, but can be recovered |
| $(2a+5)(4a-1) = 0$; $a = -\frac{5}{2}$ and $a = \frac{1}{4}$ | A1 | Solve quadratic, possibly **BC**, to obtain $a = -\frac{5}{2}$ and $a = \frac{1}{4}$ |8
\begin{enumerate}[label=(\alph*)]
\item Find the first three terms in the expansion of $( 4 + 3 x ) ^ { \frac { 3 } { 2 } }$ in ascending powers of $x$.
\item State the range of values of $x$ for which the expansion in part (a) is valid.
\item In the expansion of $( 4 + 3 x ) ^ { \frac { 3 } { 2 } } ( 1 + a x ) ^ { 2 }$ the coefficient of $x ^ { 2 }$ is $\frac { 107 } { 16 }$.
Determine the possible values of the constant $a$.
\end{enumerate}
\hfill \mbox{\textit{OCR H240/01 2023 Q8 [9]}}