Pre-U Pre-U 9794/1 2010 June — Question 7 9 marks

Exam BoardPre-U
ModulePre-U 9794/1 (Pre-U Mathematics Paper 1)
Year2010
SessionJune
Marks9
TopicGeneralised Binomial Theorem
TypeProduct with quadratic or higher term
DifficultyStandard +0.3 This is a straightforward binomial expansion question with standard follow-through parts. Part (i) requires routine application of the binomial series to expand (4-3x)^(-1/2), part (ii) involves simple polynomial multiplication, and part (iii) is basic integration of a quadratic. All techniques are standard A-level fare with no novel insight required, making it slightly easier than average.
Spec1.04c Extend binomial expansion: rational n, |x|<11.04d Binomial expansion validity: convergence conditions1.08d Evaluate definite integrals: between limits

Let \(f(x) = \frac{1 + x^2}{\sqrt{4 - 3x}}\)
  1. Obtain in ascending powers of \(x\) the first three terms in the expansion of \(\frac{1}{\sqrt{4 - 3x}}\) and state the values of \(x\) for which this expansion is valid. [5]
  2. Hence obtain an approximation to \(f(x)\) in the form \(a + bx + cx^2\) where \(a\), \(b\) and \(c\) are constants. [2]
  3. Use your approximation to estimate \(\int_0^{0.1} f(x) dx\). [2]

AnswerMarks Guidance
(i) Attempt to find the third term of \(\left(1-\frac{3}{4}x\right)^{-\frac{1}{2}}\)M1
\(\frac{1}{2}\left(-\frac{1}{2}\right)\left(-\frac{3}{2}\right)\left(-\frac{3}{4}x\right)^2\) unsimplifiedA1
\(1+\frac{3}{8}x\)B1
Multiply their expansion by \(\frac{1}{2}\left(\frac{1}{2}+\frac{3x}{16}+\frac{27x^2}{256}\right)\)B1
\(-\frac{4}{3} < x < \frac{4}{3}\)B1 [5]
(ii) Attempt to multiply the expansion found in (i) by \(1 + x^2\) or multiply relevant terms.M1
\(\frac{1}{2} + \frac{3}{16}x + \frac{155}{256}x^2\) aefA1 [2]
(iii) Attempt to integrate their expansion and substitute at least 0.1M1
0.0511A1 [2]
(i) Attempt to find the third term of $\left(1-\frac{3}{4}x\right)^{-\frac{1}{2}}$ | M1 |
$\frac{1}{2}\left(-\frac{1}{2}\right)\left(-\frac{3}{2}\right)\left(-\frac{3}{4}x\right)^2$ unsimplified | A1 |
$1+\frac{3}{8}x$ | B1 |
Multiply their expansion by $\frac{1}{2}\left(\frac{1}{2}+\frac{3x}{16}+\frac{27x^2}{256}\right)$ | B1 |
$-\frac{4}{3} < x < \frac{4}{3}$ | B1 | [5]

(ii) Attempt to multiply the expansion found in (i) by $1 + x^2$ or multiply relevant terms. | M1 |
$\frac{1}{2} + \frac{3}{16}x + \frac{155}{256}x^2$ aef | A1 | [2]

(iii) Attempt to integrate their expansion and substitute at least 0.1 | M1 |
0.0511 | A1 | [2]
Let $f(x) = \frac{1 + x^2}{\sqrt{4 - 3x}}$

\begin{enumerate}[label=(\roman*)]
\item Obtain in ascending powers of $x$ the first three terms in the expansion of $\frac{1}{\sqrt{4 - 3x}}$ and state the values of $x$ for which this expansion is valid. [5]

\item Hence obtain an approximation to $f(x)$ in the form $a + bx + cx^2$ where $a$, $b$ and $c$ are constants. [2]

\item Use your approximation to estimate $\int_0^{0.1} f(x) dx$. [2]
\end{enumerate}

\hfill \mbox{\textit{Pre-U Pre-U 9794/1 2010 Q7 [9]}}