SPS SPS SM Pure 2024 February — Question 6 11 marks

Exam BoardSPS
ModuleSPS SM Pure (SPS SM Pure)
Year2024
SessionFebruary
Marks11
TopicProduct & Quotient Rules
TypeShow derivative equals given algebraic form
DifficultyStandard +0.3 This is a multi-part question requiring product rule differentiation (straightforward algebra), sign change reasoning for root location, algebraic rearrangement of the derivative equation, and iterative calculation. Part (a) is routine differentiation with some algebraic simplification; parts (b)-(d) are standard A-level techniques (intermediate value theorem, rearranging equations, iteration). The product rule application and algebra are slightly more involved than minimal examples, but all techniques are standard curriculum material with no novel insight required.
Spec1.07q Product and quotient rules: differentiation1.09d Newton-Raphson method

6. Curve \(C\) has equation $$y = \left( x ^ { 2 } - 5 x + 8 \right) \mathrm { e } ^ { x ^ { 2 } } \quad x \in \mathbb { R }$$
  1. Show that $$\frac { \mathrm { d } y } { \mathrm {~d} x } = \left( 2 x ^ { 3 } - 10 x ^ { 2 } + 18 x - 5 \right) \mathrm { e } ^ { x ^ { 2 } }$$ Given that
    • \(C\) has only one stationary point
    • the stationary point has \(x\) coordinate \(\alpha\)
    • \(\frac { \mathrm { d } y } { \mathrm {~d} x } \approx - 0.5\) at \(x = 0.3\)
    • \(\frac { \mathrm { d } y } { \mathrm {~d} x } \approx 0.9\) at \(x = 0.4\)
    • explain why \(0.3 < \alpha < 0.4\)
    • Show that \(\alpha\) is a solution of the equation
    $$x = \frac { 5 \left( 2 x ^ { 2 } + 1 \right) } { 2 \left( x ^ { 2 } + 9 \right) }$$
  2. Using the iteration formula $$x _ { n + 1 } = \frac { 5 \left( 2 x _ { n } ^ { 2 } + 1 \right) } { 2 \left( x _ { n } ^ { 2 } + 9 \right) } \quad \text { with } x _ { 1 } = 0.3$$ find
    1. the value of \(x _ { 3 }\) to 4 decimal places,
    2. the value of \(\alpha\) to 4 decimal places.

6.

Curve $C$ has equation

$$y = \left( x ^ { 2 } - 5 x + 8 \right) \mathrm { e } ^ { x ^ { 2 } } \quad x \in \mathbb { R }$$
\begin{enumerate}[label=(\alph*)]
\item Show that

$$\frac { \mathrm { d } y } { \mathrm {~d} x } = \left( 2 x ^ { 3 } - 10 x ^ { 2 } + 18 x - 5 \right) \mathrm { e } ^ { x ^ { 2 } }$$

Given that

\begin{itemize}
  \item $C$ has only one stationary point
  \item the stationary point has $x$ coordinate $\alpha$
  \item $\frac { \mathrm { d } y } { \mathrm {~d} x } \approx - 0.5$ at $x = 0.3$
  \item $\frac { \mathrm { d } y } { \mathrm {~d} x } \approx 0.9$ at $x = 0.4$
\item explain why $0.3 < \alpha < 0.4$
\item Show that $\alpha$ is a solution of the equation
\end{itemize}

$$x = \frac { 5 \left( 2 x ^ { 2 } + 1 \right) } { 2 \left( x ^ { 2 } + 9 \right) }$$
\item Using the iteration formula

$$x _ { n + 1 } = \frac { 5 \left( 2 x _ { n } ^ { 2 } + 1 \right) } { 2 \left( x _ { n } ^ { 2 } + 9 \right) } \quad \text { with } x _ { 1 } = 0.3$$

find
\begin{enumerate}[label=(\roman*)]
\item the value of $x _ { 3 }$ to 4 decimal places,
\item the value of $\alpha$ to 4 decimal places.
\end{enumerate}\end{enumerate}

\hfill \mbox{\textit{SPS SPS SM Pure 2024 Q6 [11]}}