| Exam Board | OCR |
|---|---|
| Module | C4 (Core Mathematics 4) |
| Year | 2012 |
| Session | January |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Generalised Binomial Theorem |
| Type | Two unknowns from two coefficient conditions |
| Difficulty | Standard +0.8 This question combines binomial expansion with fractional powers (part i is routine) with a more challenging algebraic problem requiring students to expand two expressions, equate coefficients systematically for x, x², and x³ terms to zero, then solve simultaneous equations for two unknowns. The conceptual leap of understanding what 'lowest degree term is x³' means and the algebraic manipulation required elevates this above standard C4 fare. |
| Spec | 1.04c Extend binomial expansion: rational n, |x|<11.04d Binomial expansion validity: convergence conditions |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| First two terms in expansion \(= 1 - x\) | B1 | Simplify to this, now or later |
| Third term shown as \(\frac{\frac{1}{4} \cdot -\frac{3}{4}}{2}(-4x)^2\) | M1 | \(-\frac{3}{4}\) can be \(\frac{1}{4}-1\); \((-4x)^2\) can be \(-4x^2\) or \(-16x^2\) |
| \(= -\frac{3}{2}x^2\) | A1 | |
| Fourth term shown as \(\frac{\frac{1}{4} \cdot -\frac{3}{4} \cdot -\frac{7}{4}}{2 \cdot 3}(-4x)^3\) | M1 | Similar allowances as for first M1 |
| \(= -\frac{7}{2}x^3\) | A1 | Complete expansion is \(1 - x - \frac{3}{2}x^2 - \frac{7}{2}x^3\ldots\) |
| [5] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \((1+bx^2)^7\) shown (implied) as \(1 + 7bx^2 + \ldots\) | B1 | |
| Clear indication that terms involving \(x\) and \(x^2\) must cancel | M1 | |
| \(a = -1\) | A1 FT | If (i) \(= 1 + \lambda x + \mu x^2\), \(a = \lambda\) |
| \(b = -\frac{3}{14}\) | A1 FT | If (i) \(= 1 + \lambda x + \mu x^2\), \(b = \frac{1}{7}\mu\). FT from wrong (i) only, not wrong \((1+bx^2)^7\) |
| [4] |
## Question 4(i):
| Answer | Marks | Guidance |
|--------|-------|----------|
| First two terms in expansion $= 1 - x$ | B1 | Simplify to this, now or later |
| Third term shown as $\frac{\frac{1}{4} \cdot -\frac{3}{4}}{2}(-4x)^2$ | M1 | $-\frac{3}{4}$ can be $\frac{1}{4}-1$; $(-4x)^2$ can be $-4x^2$ or $-16x^2$ |
| $= -\frac{3}{2}x^2$ | A1 | |
| Fourth term shown as $\frac{\frac{1}{4} \cdot -\frac{3}{4} \cdot -\frac{7}{4}}{2 \cdot 3}(-4x)^3$ | M1 | Similar allowances as for first M1 |
| $= -\frac{7}{2}x^3$ | A1 | Complete expansion is $1 - x - \frac{3}{2}x^2 - \frac{7}{2}x^3\ldots$ |
| **[5]** | | |
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## Question 4(ii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $(1+bx^2)^7$ shown (implied) as $1 + 7bx^2 + \ldots$ | B1 | |
| Clear indication that terms involving $x$ and $x^2$ must cancel | M1 | |
| $a = -1$ | A1 FT | If (i) $= 1 + \lambda x + \mu x^2$, $a = \lambda$ |
| $b = -\frac{3}{14}$ | A1 FT | If (i) $= 1 + \lambda x + \mu x^2$, $b = \frac{1}{7}\mu$. FT from wrong (i) only, not wrong $(1+bx^2)^7$ |
| **[4]** | | |
---4 (i) Expand $( 1 - 4 x ) ^ { \frac { 1 } { 4 } }$ in ascending powers of $x$, up to and including the term in $x ^ { 3 }$.\\
(ii) The term of lowest degree in the expansion of
$$( 1 + a x ) \left( 1 + b x ^ { 2 } \right) ^ { 7 } - ( 1 - 4 x ) ^ { \frac { 1 } { 4 } }$$
in ascending powers of $x$ is the term in $x ^ { 3 }$. Find the values of the constants $a$ and $b$.
\hfill \mbox{\textit{OCR C4 2012 Q4 [9]}}