| Exam Board | OCR |
|---|---|
| Module | C2 (Core Mathematics 2) |
| Year | 2015 |
| Session | June |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Binomial Theorem (positive integer n) |
| Type | Product with unknown constant to determine |
| Difficulty | Moderate -0.3 This is a straightforward binomial expansion question requiring standard application of the formula for the first three terms, followed by a simple algebraic manipulation to find the unknown constant. The product in part (ii) only requires multiplying out two terms and solving a linear equation. While it has multiple steps, each is routine for C2 level with no conceptual challenges or novel problem-solving required. |
| Spec | 1.04a Binomial expansion: (a+b)^n for positive integer n |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \((2+ax)^6 = 64 + 192ax + 240a^2x^2\) | ||
| Obtain \(64\) | B1 | Allow \(2^6\) but not \(64x^0\) |
| Obtain \(192ax\) | B1 | Must be \(192ax\), not unsimplified equiv |
| Attempt \(3^{\text{rd}}\) term — product of \(15\), \(2^4\) and \((ax)^2\) | M1 | Allow M1 for \(ax^2\) rather than \((ax)^2\); binomial coeff must be \(15\) soi; \(240ax^2\) implies M1 even if no other method shown |
| Obtain \(240a^2x^2\) | A1 | Or \(240(ax)^2\) but A0 if this then becomes \(240ax^2\); A0 if expansion subsequently spoiled |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \((3 \times 192a) + (-5 \times 64)\) | M1 | Attempt both relevant terms; M0 if additional terms used; could be coefficients or terms still involving \(x\) but must be consistent for both terms |
| \(576a - 320 = 64\) | A1FT | Equate to \(64\), to obtain any correct equation, possibly still unsimplified |
| \(576a = 384\), \(a = \frac{2}{3}\) | A1 | Obtain \(a = \frac{2}{3}\) CWO; fraction must be simplified; allow exact decimal equiv only |
## Question 4:
### Part (i)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $(2+ax)^6 = 64 + 192ax + 240a^2x^2$ | | |
| Obtain $64$ | B1 | Allow $2^6$ but not $64x^0$ |
| Obtain $192ax$ | B1 | Must be $192ax$, not unsimplified equiv |
| Attempt $3^{\text{rd}}$ term — product of $15$, $2^4$ and $(ax)^2$ | M1 | Allow M1 for $ax^2$ rather than $(ax)^2$; binomial coeff must be $15$ soi; $240ax^2$ implies M1 even if no other method shown |
| Obtain $240a^2x^2$ | A1 | Or $240(ax)^2$ but A0 if this then becomes $240ax^2$; A0 if expansion subsequently spoiled |
### Part (ii)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $(3 \times 192a) + (-5 \times 64)$ | M1 | Attempt both relevant terms; M0 if additional terms used; could be coefficients or terms still involving $x$ but must be consistent for both terms |
| $576a - 320 = 64$ | A1FT | Equate to $64$, to obtain any correct equation, possibly still unsimplified |
| $576a = 384$, $a = \frac{2}{3}$ | A1 | Obtain $a = \frac{2}{3}$ CWO; fraction must be simplified; allow exact decimal equiv only |
---4 (i) Find and simplify the first three terms in the binomial expansion of $( 2 + a x ) ^ { 6 }$ in ascending powers of $x$.\\
(ii) In the expansion of $( 3 - 5 x ) ( 2 + a x ) ^ { 6 }$, the coefficient of $x$ is 64 . Find the value of $a$.
\hfill \mbox{\textit{OCR C2 2015 Q4 [7]}}