Standard +0.3 This is a straightforward multi-part question covering standard C3 techniques: showing a root exists in an interval (simple substitution), algebraic rearrangement, performing a few iterations with a calculator, quotient rule differentiation, and finding a normal gradient. All parts are routine applications of well-practiced methods with no novel problem-solving required, making it slightly easier than average.
The curve with equation
$$y = \frac { \cos x } { 2 x + 1 } , \quad x > - \frac { 1 } { 2 }$$
intersects the line \(y = \frac { 1 } { 2 }\) at the point where \(x = \alpha\).
Show that \(\alpha\) lies between 0 and \(\frac { \pi } { 2 }\).
Show that the equation \(\frac { \cos x } { 2 x + 1 } = \frac { 1 } { 2 }\) can be rearranged into the form
$$x = \cos x - \frac { 1 } { 2 }$$
Use the iteration \(x _ { n + 1 } = \cos x _ { n } - \frac { 1 } { 2 }\) with \(x _ { 1 } = 0\) to find \(x _ { 3 }\), giving your answer to three decimal places.
Given that \(y = \frac { \cos x } { 2 x + 1 }\), use the quotient rule to find an expression for \(\frac { \mathrm { d } y } { \mathrm {~d} x }\).
Hence find the gradient of the normal to the curve \(y = \frac { \cos x } { 2 x + 1 }\) at the point on the curve where \(x = 0\).
1
\begin{enumerate}[label=(\alph*)]
\item The curve with equation
$$y = \frac { \cos x } { 2 x + 1 } , \quad x > - \frac { 1 } { 2 }$$
intersects the line $y = \frac { 1 } { 2 }$ at the point where $x = \alpha$.
\begin{enumerate}[label=(\roman*)]
\item Show that $\alpha$ lies between 0 and $\frac { \pi } { 2 }$.
\item Show that the equation $\frac { \cos x } { 2 x + 1 } = \frac { 1 } { 2 }$ can be rearranged into the form
$$x = \cos x - \frac { 1 } { 2 }$$
\item Use the iteration $x _ { n + 1 } = \cos x _ { n } - \frac { 1 } { 2 }$ with $x _ { 1 } = 0$ to find $x _ { 3 }$, giving your answer to three decimal places.
\end{enumerate}\item \begin{enumerate}[label=(\roman*)]
\item Given that $y = \frac { \cos x } { 2 x + 1 }$, use the quotient rule to find an expression for $\frac { \mathrm { d } y } { \mathrm {~d} x }$.
\item Hence find the gradient of the normal to the curve $y = \frac { \cos x } { 2 x + 1 }$ at the point on the curve where $x = 0$.
\end{enumerate}\end{enumerate}
\hfill \mbox{\textit{AQA C3 2009 Q1 [10]}}