OCR C2 2009 January — Question 7 9 marks

Exam BoardOCR
ModuleC2 (Core Mathematics 2)
Year2009
SessionJanuary
Marks9
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicBinomial Theorem (positive integer n)
TypeTwo equations from coefficients
DifficultyModerate -0.3 This is a straightforward application of the binomial theorem requiring students to set up equations from given coefficients and solve simultaneously. The algebra is routine (substitution and solving quadratic/linear systems), and the binomial expansion formula for n=4 is standard C2 content. Part (i) is guided, making the subsequent parts mechanical rather than requiring problem-solving insight.
Spec1.04a Binomial expansion: (a+b)^n for positive integer n

7 In the binomial expansion of \(( k + a x ) ^ { 4 }\) the coefficient of \(x ^ { 2 }\) is 24 .
  1. Given that \(a\) and \(k\) are both positive, show that \(a k = 2\).
  2. Given also that the coefficient of \(x\) in the expansion is 128 , find the values of \(a\) and \(k\).
  3. Hence find the coefficient of \(x ^ { 3 }\) in the expansion.

Question 7:
Part (i):
AnswerMarks Guidance
\(6k^2a^2 = 24\), \(k^2a^2 = 4\), \(ak = 2\)M1*, M1dep*, A1 3 Obtain at least two of \(6, k^2, a^2\); equate \(6k^ma^n\) to 24; show \(ak=2\) convincingly
Part (ii):
AnswerMarks Guidance
\(4k^3a = 128\); \(4k^3\left(\frac{2}{k}\right) = 128\)B1, M1 State or imply coeff of \(x\) is \(4k^3a\); equate to 128 and attempt to eliminate \(a\) or \(k\)
\(k^2 = 16\), \(k = 4\), \(a = \frac{1}{2}\)A1, A1 4 Obtain \(k=4\); obtain \(a = \frac{1}{2}\)
SR B1 for \(k = \pm 4\), \(a = \pm\frac{1}{2}\)
Part (iii):
AnswerMarks Guidance
\(4 \times 4 \times \left(\frac{1}{2}\right)^3 = 2\)M1, A1 2 Attempt \(4 \times k \times a^3\) following their \(a\) and \(k\); obtain 2 (allow \(2x^3\))
# Question 7:

## Part (i):
| $6k^2a^2 = 24$, $k^2a^2 = 4$, $ak = 2$ | M1*, M1dep*, A1 **3** | Obtain at least two of $6, k^2, a^2$; equate $6k^ma^n$ to 24; show $ak=2$ convincingly |
|---|---|---|

## Part (ii):
| $4k^3a = 128$; $4k^3\left(\frac{2}{k}\right) = 128$ | B1, M1 | State or imply coeff of $x$ is $4k^3a$; equate to 128 and attempt to eliminate $a$ or $k$ |
|---|---|---|
| $k^2 = 16$, $k = 4$, $a = \frac{1}{2}$ | A1, A1 **4** | Obtain $k=4$; obtain $a = \frac{1}{2}$ |
| | | **SR** B1 for $k = \pm 4$, $a = \pm\frac{1}{2}$ |

## Part (iii):
| $4 \times 4 \times \left(\frac{1}{2}\right)^3 = 2$ | M1, A1 **2** | Attempt $4 \times k \times a^3$ following their $a$ and $k$; obtain 2 (allow $2x^3$) |
|---|---|---|

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7 In the binomial expansion of $( k + a x ) ^ { 4 }$ the coefficient of $x ^ { 2 }$ is 24 .\\
(i) Given that $a$ and $k$ are both positive, show that $a k = 2$.\\
(ii) Given also that the coefficient of $x$ in the expansion is 128 , find the values of $a$ and $k$.\\
(iii) Hence find the coefficient of $x ^ { 3 }$ in the expansion.

\hfill \mbox{\textit{OCR C2 2009 Q7 [9]}}