| Exam Board | OCR |
|---|---|
| Module | C2 (Core Mathematics 2) |
| Year | 2008 |
| Session | January |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Binomial Theorem (positive integer n) |
| Type | Sum/difference of two binomials simplification |
| Difficulty | Moderate -0.3 This is a straightforward binomial expansion question requiring routine application of the binomial theorem for n=4, followed by algebraic manipulation. Part (i) is standard recall, part (ii) involves simple subtraction showing odd terms cancel, and part (iii) is basic equation solving. While multi-part, each step follows directly from the previous with no novel insight required, making it slightly easier than average. |
| Spec | 1.02f Solve quadratic equations: including in a function of unknown1.04a Binomial expansion: (a+b)^n for positive integer n |
| Answer | Marks | Guidance |
|---|---|---|
| \((2x+5)^4 = (2x)^4 + 4(2x)^3 5 + 6(2x)^2 5^2 + 4(2x)5^3 + 5^4\) | M1* | Attempt expansion involving powers of \(2x\) and 5 (at least 4 terms) |
| \(= 16x^4 + 160x^3 + 600x^2 + 1000x + 625\) | M1*, A1dep*, A1 [4] | Attempt coefficients of 1, 4, 6, 4, 1; Obtain two correct terms; Obtain a fully correct expansion |
| Answer | Marks | Guidance |
|---|---|---|
| \((2x+5)^4 - (2x-5)^4 = 320x^3 + 2000x\) | M1, A1 [2] | Identify relevant terms (and no others) by sign change; Obtain \(320x^3 + 2000x\) cwo |
| Answer | Marks | Guidance |
|---|---|---|
| \(9^4 - (-1)^4 = 6560\) and \(7360 - 800 = 6560\) A.G. | B1 | Confirm root, at any point |
| \(320x^3 - 1680x + 800 = 0\) | M1 | Attempt complete division by \((x-2)\) or equiv |
| \(4x^3 - 21x + 10 = 0\) | A1\(\sqrt{}\) | Obtain quotient of \(ax^2 + 2ax + k\), where \(a\) is their coeff of \(x^3\) |
| \((x-2)(4x^2 + 8x - 5) = 0\) | A1 | Obtain \((4x^2 + 8x - 5)\) (or multiple thereof) |
| \((x-2)(2x-1)(2x+5) = 0\) | M1 | Attempt to solve quadratic |
| Hence \(x = \frac{1}{2}\), \(x = -2\frac{1}{2}\) | A1 [6] | Obtain \(x = \frac{1}{2}\), \(x = -2\frac{1}{2}\). SR: answer only is B1 B1 |
# Question 10:
## Part (i)
| $(2x+5)^4 = (2x)^4 + 4(2x)^3 5 + 6(2x)^2 5^2 + 4(2x)5^3 + 5^4$ | M1* | Attempt expansion involving powers of $2x$ and 5 (at least 4 terms) |
| $= 16x^4 + 160x^3 + 600x^2 + 1000x + 625$ | M1*, A1dep*, A1 **[4]** | Attempt coefficients of 1, 4, 6, 4, 1; Obtain two correct terms; Obtain a fully correct expansion |
## Part (ii)
| $(2x+5)^4 - (2x-5)^4 = 320x^3 + 2000x$ | M1, A1 **[2]** | Identify relevant terms (and no others) by sign change; Obtain $320x^3 + 2000x$ cwo |
## Part (iii)
| $9^4 - (-1)^4 = 6560$ and $7360 - 800 = 6560$ **A.G.** | B1 | Confirm root, at any point |
| $320x^3 - 1680x + 800 = 0$ | M1 | Attempt complete division by $(x-2)$ or equiv |
| $4x^3 - 21x + 10 = 0$ | A1$\sqrt{}$ | Obtain quotient of $ax^2 + 2ax + k$, where $a$ is their coeff of $x^3$ |
| $(x-2)(4x^2 + 8x - 5) = 0$ | A1 | Obtain $(4x^2 + 8x - 5)$ (or multiple thereof) |
| $(x-2)(2x-1)(2x+5) = 0$ | M1 | Attempt to solve quadratic |
| Hence $x = \frac{1}{2}$, $x = -2\frac{1}{2}$ | A1 **[6]** | Obtain $x = \frac{1}{2}$, $x = -2\frac{1}{2}$. SR: answer only is B1 B1 |10 (i) Find the binomial expansion of $( 2 x + 5 ) ^ { 4 }$, simplifying the terms.\\
(ii) Hence show that $( 2 x + 5 ) ^ { 4 } - ( 2 x - 5 ) ^ { 4 }$ can be written as
$$320 x ^ { 3 } + k x$$
where the value of the constant $k$ is to be stated.\\
(iii) Verify that $x = 2$ is a root of the equation
$$( 2 x + 5 ) ^ { 4 } - ( 2 x - 5 ) ^ { 4 } = 3680 x - 800$$
and find the other possible values of $x$.
\hfill \mbox{\textit{OCR C2 2008 Q10 [12]}}