Moderate -0.3 This is a straightforward application of a given iterative formula requiring only calculator substitution for two iterations. The rearrangement is already provided, and students simply substitute x₁=1.2 into the formula twice. This is slightly easier than average as it's purely mechanical with no problem-solving, though the logarithmic expression requires careful calculator work.
Use Simpson's rule with 7 ordinates (6 strips) to find an estimate for \(\int _ { 0 } ^ { 3 } 4 ^ { x } \mathrm {~d} x\).
A curve is defined by the equation \(y = 4 ^ { x }\). The curve intersects the line \(y = 8 - 2 x\) at a single point where \(x = \alpha\).
Show that \(\alpha\) lies between 1.2 and 1.3.
The equation \(4 ^ { x } = 8 - 2 x\) can be rearranged into the form \(x = \frac { \ln ( 8 - 2 x ) } { \ln 4 }\).
Use the iterative formula \(x _ { n + 1 } = \frac { \ln \left( 8 - 2 x _ { n } \right) } { \ln 4 }\) with \(x _ { 1 } = 1.2\) to find the values of \(x _ { 2 }\) and \(x _ { 3 }\), giving your answers to three decimal places.
(2 marks)
1
\begin{enumerate}[label=(\alph*)]
\item Use Simpson's rule with 7 ordinates (6 strips) to find an estimate for $\int _ { 0 } ^ { 3 } 4 ^ { x } \mathrm {~d} x$.
\item A curve is defined by the equation $y = 4 ^ { x }$. The curve intersects the line $y = 8 - 2 x$ at a single point where $x = \alpha$.
\begin{enumerate}[label=(\roman*)]
\item Show that $\alpha$ lies between 1.2 and 1.3.
\item The equation $4 ^ { x } = 8 - 2 x$ can be rearranged into the form $x = \frac { \ln ( 8 - 2 x ) } { \ln 4 }$.
Use the iterative formula $x _ { n + 1 } = \frac { \ln \left( 8 - 2 x _ { n } \right) } { \ln 4 }$ with $x _ { 1 } = 1.2$ to find the values of $x _ { 2 }$ and $x _ { 3 }$, giving your answers to three decimal places.\\
(2 marks)
\end{enumerate}\end{enumerate}
\hfill \mbox{\textit{AQA C3 2012 Q1 [8]}}