| Exam Board | Edexcel |
|---|---|
| Module | P3 (Pure Mathematics 3) |
| Year | 2024 |
| Session | January |
| Marks | 13 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Partial Fractions |
| Type | Simplify then show identity |
| Difficulty | Standard +0.3 Part (a) requires algebraic manipulation to combine fractions over a common denominator and simplify—straightforward but multi-step. Part (b) is routine differentiation using quotient rule. Parts (c) and (d) involve standard inverse function work and composition. All techniques are standard P3/C3 material with no novel insight required, making this slightly easier than average. |
| Spec | 1.02k Simplify rational expressions: factorising, cancelling, algebraic division1.02v Inverse and composite functions: graphs and conditions for existence1.06d Natural logarithm: ln(x) function and properties1.07h Differentiation from first principles: for sin(x) and cos(x)1.07n Stationary points: find maxima, minima using derivatives |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(3x^2 + 7x - 20 = (3x-5)(x+4)\) seen or used | B1 | Must be seen or used |
| Combines fractions with correct common denominator, e.g. \(\frac{2x^2-32}{(3x-5)(x+4)} + \frac{8}{3x-5} = \frac{2x^2-32+8(x+4)}{(3x-5)(x+4)}\) | M1 | Order of terms in numerator consistent with common denominator |
| \(= \frac{2x(x+4)}{(3x-5)(x+4)} = \frac{2x}{3x-5}\) * | A1* | No errors seen; condone missing trailing bracket if "recovered" |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(f'(x) = \frac{2(3x-5) - 3\times2x}{(3x-5)^2}\) | M1A1 | M1 for attempt using quotient rule giving form \(\frac{\alpha(3x-5)-\beta x}{(3x-5)^2}\); A1 correct derivative |
| \(f'(x) = \frac{-10}{(3x-5)^2}\) | A1cso* | Requires correct simplified derivative \(\frac{-10}{(3x-5)^2}\) |
| As \((3x-5)^2 > 0\) (for \(x>2\)) then \(f'(x) < 0\), hence f is decreasing | A1cso* | Must state \((3x-5)^2 > 0\), so \(f'(x)<0\), and conclude f is decreasing |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(g^{-1}(x) = e^{\frac{x-3}{2}}\) | M1A1 | M1 for rearranging \(y=3+2\ln x\) to \(x=e^{f(y)}\) or \(x=3+2\ln y\) to \(y=e^{f(x)}\); A1 correct answer |
| Domain: \(x \ldots 3\) | B1 | Correct domain \(x \geqslant 3\), accept \([3,\infty)\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(g\!\left(\frac{2a}{3a-5}\right)=5 \Rightarrow 3+2\ln\!\left(\frac{2a}{3a-5}\right)=5\) | B1 | Sets up valid equation in \(a\) |
| \(\ln\!\left(\frac{2a}{3a-5}\right)=1 \Rightarrow \frac{2a}{3a-5}=e\) | M1 | Rearranges to \(\ln(\cdots)=k\) and removes ln correctly |
| \(\frac{2a}{3a-5}=e \Rightarrow a=\ldots\) | dM1 | Rearranges; depends on previous M mark |
| \(a = \frac{5e}{3e-2}\) | A1 | Accept \(\frac{-5e}{2-3e}\); accept \(e^1\) for \(e\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(gf(a)=5 \Rightarrow \frac{2a}{3a-5}=g^{-1}(5)\) | B1 | Sets up valid equation using \(g^{-1}(5)\) |
| \(\frac{2a}{3a-5}=g^{-1}(5) \Rightarrow \frac{2a}{3a-5}=e\) | M1 | Attempts \(g^{-1}(5)\) and reaches \(\frac{2a}{3a-5}=e^k\) |
| \(\frac{2a}{3a-5}=e \Rightarrow a=\ldots\) | dM1 | Rearranges; depends on previous M mark |
| \(a = \frac{5e}{3e-2}\) | A1 | Accept \(\frac{-5e}{2-3e}\) |
## Question 4:
### Part (a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $3x^2 + 7x - 20 = (3x-5)(x+4)$ seen or used | B1 | Must be seen or used |
| Combines fractions with correct common denominator, e.g. $\frac{2x^2-32}{(3x-5)(x+4)} + \frac{8}{3x-5} = \frac{2x^2-32+8(x+4)}{(3x-5)(x+4)}$ | M1 | Order of terms in numerator consistent with common denominator |
| $= \frac{2x(x+4)}{(3x-5)(x+4)} = \frac{2x}{3x-5}$ * | A1* | No errors seen; condone missing trailing bracket if "recovered" |
### Part (b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $f'(x) = \frac{2(3x-5) - 3\times2x}{(3x-5)^2}$ | M1A1 | M1 for attempt using quotient rule giving form $\frac{\alpha(3x-5)-\beta x}{(3x-5)^2}$; A1 correct derivative |
| $f'(x) = \frac{-10}{(3x-5)^2}$ | A1cso* | Requires correct simplified derivative $\frac{-10}{(3x-5)^2}$ |
| As $(3x-5)^2 > 0$ (for $x>2$) then $f'(x) < 0$, hence f is decreasing | A1cso* | Must state $(3x-5)^2 > 0$, so $f'(x)<0$, and conclude f is decreasing |
### Part (c):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $g^{-1}(x) = e^{\frac{x-3}{2}}$ | M1A1 | M1 for rearranging $y=3+2\ln x$ to $x=e^{f(y)}$ or $x=3+2\ln y$ to $y=e^{f(x)}$; A1 correct answer |
| Domain: $x \ldots 3$ | B1 | Correct domain $x \geqslant 3$, accept $[3,\infty)$ |
### Part (d) — Way 1:
| Answer/Working | Mark | Guidance |
|---|---|---|
| $g\!\left(\frac{2a}{3a-5}\right)=5 \Rightarrow 3+2\ln\!\left(\frac{2a}{3a-5}\right)=5$ | B1 | Sets up valid equation in $a$ |
| $\ln\!\left(\frac{2a}{3a-5}\right)=1 \Rightarrow \frac{2a}{3a-5}=e$ | M1 | Rearranges to $\ln(\cdots)=k$ and removes ln correctly |
| $\frac{2a}{3a-5}=e \Rightarrow a=\ldots$ | dM1 | Rearranges; depends on previous M mark |
| $a = \frac{5e}{3e-2}$ | A1 | Accept $\frac{-5e}{2-3e}$; accept $e^1$ for $e$ |
### Part (d) — Way 2:
| Answer/Working | Mark | Guidance |
|---|---|---|
| $gf(a)=5 \Rightarrow \frac{2a}{3a-5}=g^{-1}(5)$ | B1 | Sets up valid equation using $g^{-1}(5)$ |
| $\frac{2a}{3a-5}=g^{-1}(5) \Rightarrow \frac{2a}{3a-5}=e$ | M1 | Attempts $g^{-1}(5)$ and reaches $\frac{2a}{3a-5}=e^k$ |
| $\frac{2a}{3a-5}=e \Rightarrow a=\ldots$ | dM1 | Rearranges; depends on previous M mark |
| $a = \frac{5e}{3e-2}$ | A1 | Accept $\frac{-5e}{2-3e}$ |
---\begin{enumerate}
\item The function f is defined by
\end{enumerate}
$$f ( x ) = \frac { 2 x ^ { 2 } - 32 } { 3 x ^ { 2 } + 7 x - 20 } + \frac { 8 } { 3 x - 5 } \quad x \in \mathbb { R } \quad x > 2$$
(a) Show that $\mathrm { f } ( x ) = \frac { 2 x } { 3 x - 5 }$\\
(b) Show, using calculus, that f is a decreasing function.
You must make your reasoning clear.
The function g is defined by
$$g ( x ) = 3 + 2 \ln x \quad x \geqslant 1$$
(c) Find $\mathrm { g } ^ { - 1 }$\\
(d) Find the exact value of $a$ for which
$$\operatorname { gf } ( a ) = 5$$
\hfill \mbox{\textit{Edexcel P3 2024 Q4 [13]}}