| Exam Board | OCR |
|---|---|
| Module | C4 (Core Mathematics 4) |
| Year | 2012 |
| Session | June |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Generalised Binomial Theorem |
| Type | Expand and state validity |
| Difficulty | Standard +0.3 This is a standard C4 binomial expansion question requiring the product of two expansions: (1+x²) and (1+4x)^(-1/2). It involves routine application of the generalised binomial theorem with a fractional/negative index, multiplication of series, and stating the validity condition |4x|<1. While it requires careful algebraic manipulation across multiple steps, it follows a well-practiced template with no novel problem-solving required. |
| Spec | 1.04c Extend binomial expansion: rational n, |x|<11.04d Binomial expansion validity: convergence conditions |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Expansion of \((1+4x)^{-\frac{1}{2}}\); first 2 terms \(= 1-2x\) | B1 | Or \((1+4x)^{\frac{1}{2}} = 1+2x\ldots\) |
| 3rd term \(= \frac{-\frac{1}{2}\cdot(-\frac{1}{2}-1)}{2}\cdot 16x^2\) [Accept \(4x^2\) for \(16x^2\)] | M1 | 3rd term \(= \frac{\frac{1}{2}\cdot(-\frac{1}{2})}{2}\cdot 16x^2\) |
| \(= +6x^2\) | A1 | \(= -2x^2\) |
| 4th term \(= \frac{-\frac{1}{2}\cdot(-\frac{1}{2}-1)\cdot(-\frac{1}{2}-2)}{2\cdot3}\cdot 64x^3\) | M1 | 4th term \(= \frac{\frac{1}{2}\cdot(-\frac{1}{2})\cdot(-\frac{3}{2})}{2\cdot3}\cdot 64x^3\) |
| \(= -20x^3\) | A1 | \(= +4x^3\) |
| \(1-2x+7x^2-22x^3\); \(1+ax+(b+1)x^2+(a+c)x^3\) | A1 ft | ft only \((1+4x)^{-\frac{1}{2}}=1+ax+bx^2+cx^3\) provided \(a,b,c\) attempted and at least one M1 obtained |
| Total: [6] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \( | x | <\frac{1}{4}\); \(-\frac{1}{4} |
| Total: [1] |
# Question 3:
## Part (i)
| Answer | Marks | Guidance |
|--------|-------|----------|
| Expansion of $(1+4x)^{-\frac{1}{2}}$; first 2 terms $= 1-2x$ | B1 | Or $(1+4x)^{\frac{1}{2}} = 1+2x\ldots$ |
| 3rd term $= \frac{-\frac{1}{2}\cdot(-\frac{1}{2}-1)}{2}\cdot 16x^2$ [Accept $4x^2$ for $16x^2$] | M1 | 3rd term $= \frac{\frac{1}{2}\cdot(-\frac{1}{2})}{2}\cdot 16x^2$ |
| $= +6x^2$ | A1 | $= -2x^2$ |
| 4th term $= \frac{-\frac{1}{2}\cdot(-\frac{1}{2}-1)\cdot(-\frac{1}{2}-2)}{2\cdot3}\cdot 64x^3$ | M1 | 4th term $= \frac{\frac{1}{2}\cdot(-\frac{1}{2})\cdot(-\frac{3}{2})}{2\cdot3}\cdot 64x^3$ |
| $= -20x^3$ | A1 | $= +4x^3$ |
| $1-2x+7x^2-22x^3$; $1+ax+(b+1)x^2+(a+c)x^3$ | A1 ft | ft only $(1+4x)^{-\frac{1}{2}}=1+ax+bx^2+cx^3$ provided $a,b,c$ attempted and at least one M1 obtained |
| **Total: [6]** | | |
## Part (ii)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $|x|<\frac{1}{4}$; $-\frac{1}{4}<x<\frac{1}{4}$; $\{-\frac{1}{4}<x, x<\frac{1}{4}\}$ no equality | B1 | But not $\{-\frac{1}{4}<x$ **OR** $x<\frac{1}{4}\}$. If choice mark what appears to be the final answer. |
| **Total: [1]** | | |
---3 (i) Expand $\frac { 1 + x ^ { 2 } } { \sqrt { 1 + 4 x } }$ in ascending powers of $x$, up to and including the term in $x ^ { 3 }$.\\
(ii) State the set of values of $x$ for which this expansion is valid.
\hfill \mbox{\textit{OCR C4 2012 Q3 [7]}}