AQA Paper 1 2022 June — Question 6 6 marks

Exam BoardAQA
ModulePaper 1 (Paper 1)
Year2022
SessionJune
Marks6
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicGeneralised Binomial Theorem
TypeApproximation for small x
DifficultyStandard +0.8 This question combines binomial expansion with small-angle approximations and requires algebraic manipulation of two different series. Part (a) is routine, but part (b) requires recognizing that √(cos x) ≈ √(1 - x²/2) and then substituting the binomial expansion, while also expanding sin 4x ≈ 4x. The multi-step nature and need to combine different approximation techniques elevates this above average difficulty.
Spec1.04c Extend binomial expansion: rational n, |x|<11.05e Small angle approximations: sin x ~ x, cos x ~ 1-x^2/2, tan x ~ x

6
  1. Find the first two terms, in ascending powers of \(x\), of the binomial expansion of $$\left( 1 - \frac { x } { 2 } \right) ^ { \frac { 1 } { 2 } }$$ 6
  2. Hence, for small values of \(x\), show that $$\sin 4 x + \sqrt { \cos x } \approx A + B x + C x ^ { 2 }$$ where \(A , B\) and \(C\) are constants to be found.

Question 6(a):
AnswerMarks Guidance
Answer/WorkingMarks Guidance
\(\left(1 - \frac{x}{2}\right)^{\frac{1}{2}} \approx 1 + \left(\frac{1}{2}\right)\left(-\frac{x}{2}\right)\)M1 Expands to obtain first two terms; can be unsimplified; condone sign error
\(\approx 1 - \frac{1}{4}x\)A1 Accept if listed as two separate terms; ignore any extra terms
Question 6(b):
AnswerMarks Guidance
Answer/WorkingMarks Guidance
\(\sin kx \approx kx\) or \(\sqrt{\cos x} \approx \sqrt{1 - \frac{x^2}{2}}\)M1 States or uses at least one small angle approximation correctly
\(\sin(4x) + \sqrt{\cos x} \approx 4x + \sqrt{1 - \frac{x^2}{2}} \approx 4x + \left(1 - \frac{x^2}{4}\right)\)A1 Uses both small angle approximations correctly; must eliminate all trig expressions; inconsistent variables for angles must eventually be consistent
Uses expansion from (a) with \(x\) replaced by \(x^2\); or applies binomial theorem correctly to \(\left(1 - \frac{x^2}{2}\right)^{\frac{1}{2}}\)M1 Must have replaced \(x\) with \(x^2\); ignore extra terms
\(\approx 1 + 4x - \frac{1}{4}x^2\)R1 Accept any order; ignore higher powers of \(x\); must be in terms of \(x\); do not ISW
## Question 6(a):

| Answer/Working | Marks | Guidance |
|---|---|---|
| $\left(1 - \frac{x}{2}\right)^{\frac{1}{2}} \approx 1 + \left(\frac{1}{2}\right)\left(-\frac{x}{2}\right)$ | M1 | Expands to obtain first two terms; can be unsimplified; condone sign error |
| $\approx 1 - \frac{1}{4}x$ | A1 | Accept if listed as two separate terms; ignore any extra terms |

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## Question 6(b):

| Answer/Working | Marks | Guidance |
|---|---|---|
| $\sin kx \approx kx$ or $\sqrt{\cos x} \approx \sqrt{1 - \frac{x^2}{2}}$ | M1 | States or uses at least one small angle approximation correctly |
| $\sin(4x) + \sqrt{\cos x} \approx 4x + \sqrt{1 - \frac{x^2}{2}} \approx 4x + \left(1 - \frac{x^2}{4}\right)$ | A1 | Uses both small angle approximations correctly; must eliminate all trig expressions; inconsistent variables for angles must eventually be consistent |
| Uses expansion from (a) with $x$ replaced by $x^2$; or applies binomial theorem correctly to $\left(1 - \frac{x^2}{2}\right)^{\frac{1}{2}}$ | M1 | Must have replaced $x$ with $x^2$; ignore extra terms |
| $\approx 1 + 4x - \frac{1}{4}x^2$ | R1 | Accept any order; ignore higher powers of $x$; must be in terms of $x$; do not ISW |

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6
\begin{enumerate}[label=(\alph*)]
\item Find the first two terms, in ascending powers of $x$, of the binomial expansion of

$$\left( 1 - \frac { x } { 2 } \right) ^ { \frac { 1 } { 2 } }$$

6
\item Hence, for small values of $x$, show that

$$\sin 4 x + \sqrt { \cos x } \approx A + B x + C x ^ { 2 }$$

where $A , B$ and $C$ are constants to be found.
\end{enumerate}

\hfill \mbox{\textit{AQA Paper 1 2022 Q6 [6]}}