SPS SPS SM Pure 2025 February — Question 9 9 marks

Exam BoardSPS
ModuleSPS SM Pure (SPS SM Pure)
Year2025
SessionFebruary
Marks9
TopicProduct & Quotient Rules
TypeShow derivative equals given algebraic form
DifficultyStandard +0.3 This is a straightforward multi-part question testing product rule, chain rule, and basic transformations. Part (a) is routine differentiation with algebraic manipulation to match a given form. Parts (b)-(d) involve standard techniques (solving f'(x)=0, finding range, applying transformations) with no novel insights required. Slightly easier than average due to clear scaffolding and standard methods throughout.
Spec1.02w Graph transformations: simple transformations of f(x)1.07n Stationary points: find maxima, minima using derivatives1.07r Chain rule: dy/dx = dy/du * du/dx and connected rates

9. \begin{figure}[h]
\includegraphics[alt={},max width=\textwidth]{bc7fb499-9462-40ae-88f4-87fc60f6a005-18_542_551_212_790} \captionsetup{labelformat=empty} \caption{Figure 3}
\end{figure} The curve \(C\) has equation \(y = \mathrm { f } ( x )\), where $$f ( x ) = x ^ { 3 } \sqrt { 4 x + 7 } \quad x \geq - \frac { 7 } { 4 }$$
  1. Show that $$\mathrm { f } ^ { \prime } ( x ) = \frac { k x ^ { 2 } ( 2 x + 3 ) } { \sqrt { 4 x + 7 } }$$ where \(k\) is a constant to be found. The point \(P\), shown in Figure 3, is the minimum turning point on \(C\).
  2. Find the coordinates of \(P\).
  3. Hence find the range of the function g defined by $$\operatorname { g } ( x ) = - 4 \mathrm { f } ( x ) \quad x \geq - \frac { 7 } { 4 }$$ The point \(Q\) with coordinates \(\left( \frac { 1 } { 2 } , \frac { 3 } { 8 } \right)\) lies on \(C\).
  4. Find the coordinates of the point to which \(Q\) is mapped when \(C\) is transformed to the curve with equation $$y = 40 f \left( x - \frac { 3 } { 2 } \right) - 8$$

9.

\begin{figure}[h]
\begin{center}
  \includegraphics[alt={},max width=\textwidth]{bc7fb499-9462-40ae-88f4-87fc60f6a005-18_542_551_212_790}
\captionsetup{labelformat=empty}
\caption{Figure 3}
\end{center}
\end{figure}

The curve $C$ has equation $y = \mathrm { f } ( x )$, where

$$f ( x ) = x ^ { 3 } \sqrt { 4 x + 7 } \quad x \geq - \frac { 7 } { 4 }$$
\begin{enumerate}[label=(\alph*)]
\item Show that

$$\mathrm { f } ^ { \prime } ( x ) = \frac { k x ^ { 2 } ( 2 x + 3 ) } { \sqrt { 4 x + 7 } }$$

where $k$ is a constant to be found.

The point $P$, shown in Figure 3, is the minimum turning point on $C$.
\item Find the coordinates of $P$.
\item Hence find the range of the function g defined by

$$\operatorname { g } ( x ) = - 4 \mathrm { f } ( x ) \quad x \geq - \frac { 7 } { 4 }$$

The point $Q$ with coordinates $\left( \frac { 1 } { 2 } , \frac { 3 } { 8 } \right)$ lies on $C$.
\item Find the coordinates of the point to which $Q$ is mapped when $C$ is transformed to the curve with equation

$$y = 40 f \left( x - \frac { 3 } { 2 } \right) - 8$$
\end{enumerate}

\hfill \mbox{\textit{SPS SPS SM Pure 2025 Q9 [9]}}