| Exam Board | Edexcel |
|---|---|
| Module | P1 (Pure Mathematics 1) |
| Year | 2021 |
| Session | January |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Chain Rule |
| Type | Find stationary points and nature |
| Difficulty | Moderate -0.3 This is a straightforward P1 question requiring basic differentiation of powers (rewriting √x as x^(1/2)), solving a quadratic equation, and finding the second derivative. While it involves multiple steps and requires showing working, all techniques are standard and routine for this level with no conceptual challenges or novel problem-solving required. |
| Spec | 1.02b Surds: manipulation and rationalising denominators1.02f Solve quadratic equations: including in a function of unknown1.07i Differentiate x^n: for rational n and sums |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(2x-3\sqrt{x}-5=9 \Rightarrow 2x-3\sqrt{x}-14=0\) and treats as quadratic equation | M1 | Sets \(u=\sqrt{x}\) giving \(2u^2-3u-14=0\); condone use of other variables |
| \((2\sqrt{x}-7)(\sqrt{x}+2)=0 \Rightarrow \sqrt{x}=\frac{7}{2}, (-2)\) | A1 | Ignore reference to \(-2\); condone \(x=\frac{7}{2}\) |
| \(x=\left(\frac{7}{2}\right)^2=\frac{49}{4}\) | dM1 A1 (4) | dM1: attempts one value for \(x\); A1: \(x=\frac{49}{4}\) only; if 4 found it must be rejected |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(f'(x)= 2-\frac{3}{2}x^{-\frac{1}{2}}\) | B1 | Two correct terms, may be unsimplified; index must be processed |
| \(f''(x)=\frac{3}{4}x^{-\frac{3}{2}}\) | M1 A1 | M1: any index of \(f'(x)\) reduced by one; A1: exact equivalent |
| \(\frac{3}{4}x^{-\frac{3}{2}}=6 \Rightarrow x^{-\frac{3}{2}}=8 \Rightarrow x=\frac{1}{4}\) | dM1 A1 (5) | dM1: sets \(f''(x)=6\) of form \(Bx^k=6\), achieves \(x^m=A\) (\(m\neq1\)); A1: \(\frac{1}{4}\) cso with intermediate working shown |
## Question 7(a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $2x-3\sqrt{x}-5=9 \Rightarrow 2x-3\sqrt{x}-14=0$ and treats as quadratic equation | M1 | Sets $u=\sqrt{x}$ giving $2u^2-3u-14=0$; condone use of other variables |
| $(2\sqrt{x}-7)(\sqrt{x}+2)=0 \Rightarrow \sqrt{x}=\frac{7}{2}, (-2)$ | A1 | Ignore reference to $-2$; condone $x=\frac{7}{2}$ |
| $x=\left(\frac{7}{2}\right)^2=\frac{49}{4}$ | dM1 A1 **(4)** | dM1: attempts one value for $x$; A1: $x=\frac{49}{4}$ only; if 4 found it must be rejected |
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## Question 7(b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $f'(x)= 2-\frac{3}{2}x^{-\frac{1}{2}}$ | B1 | Two correct terms, may be unsimplified; index must be processed |
| $f''(x)=\frac{3}{4}x^{-\frac{3}{2}}$ | M1 A1 | M1: any index of $f'(x)$ reduced by one; A1: exact equivalent |
| $\frac{3}{4}x^{-\frac{3}{2}}=6 \Rightarrow x^{-\frac{3}{2}}=8 \Rightarrow x=\frac{1}{4}$ | dM1 A1 **(5)** | dM1: sets $f''(x)=6$ of form $Bx^k=6$, achieves $x^m=A$ ($m\neq1$); A1: $\frac{1}{4}$ cso with intermediate working shown |
---7. In this question you must show all stages of your working.
Solutions relying on calculator technology are not acceptable.
$$f ( x ) = 2 x - 3 \sqrt { x } - 5 \quad x > 0$$
\begin{enumerate}[label=(\alph*)]
\item Solve the equation
$$f ( x ) = 9$$
\item Solve the equation
$$\mathrm { f } ^ { \prime \prime } ( x ) = 6$$
\begin{center}
\end{center}
\end{enumerate}
\hfill \mbox{\textit{Edexcel P1 2021 Q7 [9]}}