Question 5 - A-Level Maths

Edexcel C2 — Question 5 7 marks

Exam Board	Edexcel
Module	C2 (Core Mathematics 2)
Marks	7
Paper	Download PDF ↗
Mark scheme	Download PDF ↗
Topic	Binomial Theorem (positive integer n)
Type	Find n and constants from given terms
Difficulty	Moderate -0.3 This is a straightforward binomial expansion question requiring students to apply the binomial theorem formula to match coefficients. While it involves two unknowns and requires careful algebraic manipulation, the method is standard and well-practiced in C2. The question is slightly easier than average because it's a direct application of a single technique with no conceptual surprises, though the algebra requires some care.
Spec	1.04a Binomial expansion: (a+b)^n for positive integer n

The expansion of $(2 - px)^6$ in ascending powers of $x$, as far as the term in $x^2$, is $$64 + Ax + 135x^2.$$ Given that $p > 0$, find the value of $p$ and the value of $A$. [7]

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Answer	Marks	Guidance
$(2-px)^6 = 2^6 + \binom{6}{1}2^5(-px) + \binom{6}{2}2^4(-px)^2$	M1	Coeff. of $x$ or $\binom{n}{r}$ okay
$x^2$
$= 64 + 6 \times 2^5(-px) + 15 \times 2^4(-px)^2$	A1; A1	No
$\binom{n}{r}$
$15 \times 16p^2 = 135 \Rightarrow p^2 = \frac{9}{16}$ or $p = \frac{3}{4}$ (only)	M1, A1
$-6.32p = A$	M1
$\Rightarrow A = -144$	A1 ft	(their $p$ ($> 0$))

$(2-px)^6 = 2^6 + \binom{6}{1}2^5(-px) + \binom{6}{2}2^4(-px)^2$ | M1 | Coeff. of $x$ or $\binom{n}{r}$ okay
$x^2$ | |
$= 64 + 6 \times 2^5(-px) + 15 \times 2^4(-px)^2$ | A1; A1 | No

$\binom{n}{r}$ | |

$15 \times 16p^2 = 135 \Rightarrow p^2 = \frac{9}{16}$ or $p = \frac{3}{4}$ (only) | M1, A1 | 

$-6.32p = A$ | M1 | 

$\Rightarrow A = -144$ | A1 ft | (their $p$ ($> 0$))

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The expansion of $(2 - px)^6$ in ascending powers of $x$, as far as the term in $x^2$, is
$$64 + Ax + 135x^2.$$

Given that $p > 0$, find the value of $p$ and the value of $A$. [7]

\hfill \mbox{\textit{Edexcel C2  Q5 [7]}}

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Q1 4 Q2 6 Q3 7 Q4 8 Q5 7 Q6 8 Q7 8 Q8 12 Q9 15

Answer	Marks	Guidance
\((2-px)^6 = 2^6 + \binom{6}{1}2^5(-px) + \binom{6}{2}2^4(-px)^2\)	M1	Coeff. of \(x\) or \(\binom{n}{r}\) okay
\(x^2\)
\(= 64 + 6 \times 2^5(-px) + 15 \times 2^4(-px)^2\)	A1; A1	No
\(\binom{n}{r}\)
\(15 \times 16p^2 = 135 \Rightarrow p^2 = \frac{9}{16}\) or \(p = \frac{3}{4}\) (only)	M1, A1
\(-6.32p = A\)	M1
\(\Rightarrow A = -144\)	A1 ft	(their \(p\) (\(> 0\)))