Edexcel P3 2022 June — Question 6 10 marks

Exam BoardEdexcel
ModuleP3 (Pure Mathematics 3)
Year2022
SessionJune
Marks10
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicProduct & Quotient Rules
TypeShow derivative equals given algebraic form
DifficultyStandard +0.3 This is a standard product rule differentiation question with algebraic manipulation. Part (a) requires applying the product rule to a product involving a square root (which students at this level should recognize as a power), then simplifying to match the given form. Parts (b)-(d) are routine follow-through questions involving solving a quadratic, substituting to find coordinates, and determining a range. The algebraic manipulation is slightly more involved than the most basic examples, but this is a typical P3/C3 exam question with clear structure and standard techniques throughout.
Spec1.07n Stationary points: find maxima, minima using derivatives1.07q Product and quotient rules: differentiation1.07r Chain rule: dy/dx = dy/du * du/dx and connected rates
6. \begin{figure}[h]
\includegraphics[alt={},max width=\textwidth]{44035bf8-f54c-472a-b969-b4fa4fa3d203-18_579_643_255_653} \captionsetup{labelformat=empty} \caption{Figure 3}
\end{figure} \section*{In this question you must show all stages of your working.} Solutions relying entirely on calculator technology are not acceptable. The function f is defined by $$f ( x ) = 5 \left( x ^ { 2 } - 2 \right) ( 4 x + 9 ) ^ { \frac { 1 } { 2 } } \quad x \geqslant - \frac { 9 } { 4 }$$
  1. Show that $$f ^ { \prime } ( x ) = \frac { k \left( 5 x ^ { 2 } + 9 x - 2 \right) } { ( 4 x + 9 ) ^ { \frac { 1 } { 2 } } }$$ where \(k\) is an integer to be found.
  2. Hence, find the values of \(x\) for which \(\mathrm { f } ^ { \prime } ( x ) = 0\) Figure 3 shows a sketch of the curve \(C\) with equation \(y = \mathrm { f } ( x )\). The curve has a local maximum at the point \(P\)
  3. Find the exact coordinates of \(P\) The function g is defined by $$g ( x ) = 2 f ( x ) + 4 \quad - \frac { 9 } { 4 } \leqslant x \leqslant 0$$
  4. Find the range of g
Question 6:
Part (a):
AnswerMarks Guidance
Answer/WorkingMark Guidance
\((f'(x) =) 5(x^2-2) \times \frac{1}{2} \times 4 \times (4x+9)^{-\frac{1}{2}} + 10x(4x+9)^{\frac{1}{2}}\)M1 Attempts product rule. Award for \(\pm A(5x^2-10)(4x+9)^{-\frac{1}{2}} \pm Bx(4x+9)^{\frac{1}{2}}\) where \(A, B \neq 0\). Indices do not need processing.
Correct unsimplified expressionA1 e.g. \(2(5x^2-10)(4x+9)^{-\frac{1}{2}} + 10x(4x+9)^{\frac{1}{2}}\) oe
\((f'(x) =) \frac{10(x^2-2)+10x(4x+9)}{(4x+9)^{\frac{1}{2}}} = \frac{50x^2+90x-20}{(4x+9)^{\frac{1}{2}}}\)dM1 Proceeds to single fraction, multiplies out brackets in numerator. Cannot proceed straight to final answer without working. Dependent on M1.
\((f'(x) =) \frac{10(5x^2+9x-2)}{(4x+9)^{\frac{1}{2}}}\)A1 No errors in main body of solution. Accept \(\frac{10(5x^2+9x-2)}{\sqrt{4x+9}}\)
(4 marks)
Part (b):
AnswerMarks Guidance
Answer/WorkingMark Guidance
Sets \(f'(x)=0\) to give \(x=-2, \frac{1}{5}\)B1 Withhold if one is rejected. Condone \((-2,0)\), \(\left(\frac{1}{5},0\right)\). May be seen in (c).
(1 mark)
Part (c):
AnswerMarks Guidance
Answer/WorkingMark Guidance
\(y = 5\left(("-2")^2-2\right)\left(4\times("-2")+9\right)^{\frac{1}{2}} = 10\)M1 Substitutes smaller root from solving \(5x^2+9x-2=0\) into \(y=5(x^2-2)(4x+9)^{\frac{1}{2}}\) and proceeds to find \(y\). Condone bracket omissions, miscopying or arithmetical slips.
\((-2, 10)\)A1cao \((-2,10)\) only. Allow \(x=-2, y=10\)
(2 marks)
Part (d):
AnswerMarks Guidance
Answer/WorkingMark Guidance
\(f(0) = -30\)B1 May be implied by sight of \(-56\)
Upper bound \(= 2\times"10"+4 = \ldots\) or Lower bound \(= 2\times"-30"+4 = \ldots\)M1 Correct attempt at lower or upper bound. Upper: attempts \(2\times"10"+4\), identifies as upper bound (\(<"24"\) or \(\leqslant"24"\) acceptable). Lower: attempts \(2\times"-30"+4\), identifies as lower bound (\(>"-56"\) or \(\geqslant"-56"\) acceptable). Cannot score for using \(2\times f\!\left(\frac{1}{5}\right)+4\) as lower bound.
\(-56 \leqslant g(x) \leqslant 24\)A1 oe e.g. \(\{g(x)\in\mathbb{R}: -56\leqslant g(x) \cap g(x)\leqslant 24\}\) or \(g\in[-56,24]\). Do NOT accept \(-56\leqslant g(x)\cup g(x)\leqslant 24\) or \(-56\leqslant g(x)\) OR \(g(x)\leqslant 24\)
(3 marks)
## Question 6:

### Part (a):

| Answer/Working | Mark | Guidance |
|---|---|---|
| $(f'(x) =) 5(x^2-2) \times \frac{1}{2} \times 4 \times (4x+9)^{-\frac{1}{2}} + 10x(4x+9)^{\frac{1}{2}}$ | M1 | Attempts product rule. Award for $\pm A(5x^2-10)(4x+9)^{-\frac{1}{2}} \pm Bx(4x+9)^{\frac{1}{2}}$ where $A, B \neq 0$. Indices do not need processing. |
| Correct unsimplified expression | A1 | e.g. $2(5x^2-10)(4x+9)^{-\frac{1}{2}} + 10x(4x+9)^{\frac{1}{2}}$ oe |
| $(f'(x) =) \frac{10(x^2-2)+10x(4x+9)}{(4x+9)^{\frac{1}{2}}} = \frac{50x^2+90x-20}{(4x+9)^{\frac{1}{2}}}$ | dM1 | Proceeds to single fraction, multiplies out brackets in numerator. Cannot proceed straight to final answer without working. Dependent on M1. |
| $(f'(x) =) \frac{10(5x^2+9x-2)}{(4x+9)^{\frac{1}{2}}}$ | A1 | No errors in main body of solution. Accept $\frac{10(5x^2+9x-2)}{\sqrt{4x+9}}$ |

**(4 marks)**

---

### Part (b):

| Answer/Working | Mark | Guidance |
|---|---|---|
| Sets $f'(x)=0$ to give $x=-2, \frac{1}{5}$ | B1 | Withhold if one is rejected. Condone $(-2,0)$, $\left(\frac{1}{5},0\right)$. May be seen in (c). |

**(1 mark)**

---

### Part (c):

| Answer/Working | Mark | Guidance |
|---|---|---|
| $y = 5\left(("-2")^2-2\right)\left(4\times("-2")+9\right)^{\frac{1}{2}} = 10$ | M1 | Substitutes smaller root from solving $5x^2+9x-2=0$ into $y=5(x^2-2)(4x+9)^{\frac{1}{2}}$ and proceeds to find $y$. Condone bracket omissions, miscopying or arithmetical slips. |
| $(-2, 10)$ | A1cao | $(-2,10)$ only. Allow $x=-2, y=10$ |

**(2 marks)**

---

### Part (d):

| Answer/Working | Mark | Guidance |
|---|---|---|
| $f(0) = -30$ | B1 | May be implied by sight of $-56$ |
| Upper bound $= 2\times"10"+4 = \ldots$ or Lower bound $= 2\times"-30"+4 = \ldots$ | M1 | Correct attempt at lower or upper bound. Upper: attempts $2\times"10"+4$, identifies as upper bound ($<"24"$ or $\leqslant"24"$ acceptable). Lower: attempts $2\times"-30"+4$, identifies as lower bound ($>"-56"$ or $\geqslant"-56"$ acceptable). Cannot score for using $2\times f\!\left(\frac{1}{5}\right)+4$ as lower bound. |
| $-56 \leqslant g(x) \leqslant 24$ | A1 | oe e.g. $\{g(x)\in\mathbb{R}: -56\leqslant g(x) \cap g(x)\leqslant 24\}$ or $g\in[-56,24]$. Do NOT accept $-56\leqslant g(x)\cup g(x)\leqslant 24$ or $-56\leqslant g(x)$ OR $g(x)\leqslant 24$ |

**(3 marks)**

---
6.

\begin{figure}[h]
\begin{center}
  \includegraphics[alt={},max width=\textwidth]{44035bf8-f54c-472a-b969-b4fa4fa3d203-18_579_643_255_653}
\captionsetup{labelformat=empty}
\caption{Figure 3}
\end{center}
\end{figure}

\section*{In this question you must show all stages of your working.}
Solutions relying entirely on calculator technology are not acceptable.

The function f is defined by

$$f ( x ) = 5 \left( x ^ { 2 } - 2 \right) ( 4 x + 9 ) ^ { \frac { 1 } { 2 } } \quad x \geqslant - \frac { 9 } { 4 }$$
\begin{enumerate}[label=(\alph*)]
\item Show that

$$f ^ { \prime } ( x ) = \frac { k \left( 5 x ^ { 2 } + 9 x - 2 \right) } { ( 4 x + 9 ) ^ { \frac { 1 } { 2 } } }$$

where $k$ is an integer to be found.
\item Hence, find the values of $x$ for which $\mathrm { f } ^ { \prime } ( x ) = 0$

Figure 3 shows a sketch of the curve $C$ with equation $y = \mathrm { f } ( x )$.

The curve has a local maximum at the point $P$
\item Find the exact coordinates of $P$

The function g is defined by

$$g ( x ) = 2 f ( x ) + 4 \quad - \frac { 9 } { 4 } \leqslant x \leqslant 0$$
\item Find the range of g
\end{enumerate}

\hfill \mbox{\textit{Edexcel P3 2022 Q6 [10]}}
This paper (9 questions)
View full paper
Q1 6 Q2 9 Q3 6 Q4 7 Q5 8 Q6 10 Q7 8 Q8 12 Q9 9