| Exam Board | Edexcel |
|---|---|
| Module | P3 (Pure Mathematics 3) |
| Year | 2020 |
| Session | October |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Product & Quotient Rules |
| Type | Find range using calculus |
| Difficulty | Standard +0.3 This is a straightforward application of the quotient rule to find a derivative, followed by standard optimization to find the range. The quotient rule application is routine, and finding the minimum by setting f'(x)=0 is a standard technique. Slightly easier than average due to the mechanical nature of the steps. |
| Spec | 1.07o Increasing/decreasing: functions using sign of dy/dx1.07r Chain rule: dy/dx = dy/du * du/dx and connected rates |
| VIXV SIHIANI III IM IONOO | VIAV SIHI NI JYHAM ION OO | VI4V SIHI NI JLIYM ION OO |
(a) $\frac{dy}{dx} = \frac{(4x-1)^{1/2} \times 2 - (2x+3) \times 2(4x-1)^{-1/2}}{(4x-1)}$ M1 A1
$\frac{(4x-1)^{1/2} \times 2 - (2x+3) \times 2(4x-1)^{-1/2}}{(4x-1)} \times \frac{(4x-1)^{1/2}}{(4x-1)^{1/2}} = \frac{4x-8}{(4x-1)^{3/2}}$ dM1 A1
(4 marks)
(b) Turning point where $\frac{dy}{dx} = 0 \Rightarrow x = 2$ M1
Find value of $f$ at $x = 2 \Rightarrow f(x) = 7$ dM1 A1
Hence range is $f \geq 7$
(3 marks)
(7 marks)3.
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\caption{Figure 1}
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Figure 1 shows a sketch of a curve with equation $y = \mathrm { f } ( x )$ where
$$\mathrm { f } ( x ) = \frac { 2 x + 3 } { \sqrt { 4 x - 1 } } \quad x > \frac { 1 } { 4 }$$
\begin{enumerate}[label=(\alph*)]
\item Find, in simplest form, $\mathrm { f } ^ { \prime } ( x )$.
\item Hence find the range of f.
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VIXV SIHIANI III IM IONOO & VIAV SIHI NI JYHAM ION OO & VI4V SIHI NI JLIYM ION OO \\
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\hfill \mbox{\textit{Edexcel P3 2020 Q3 [7]}}