| Exam Board | Edexcel |
|---|---|
| Module | Paper 1 (Paper 1) |
| Year | 2024 |
| Session | June |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Product & Quotient Rules |
| Type | Show derivative equals given algebraic form |
| Difficulty | Moderate -0.3 Part (a) is a straightforward quotient rule application with algebraic simplification to match a given form—standard A-level technique. Part (b) requires solving a quadratic inequality, which adds minimal complexity. This is a typical textbook-style question testing routine differentiation and inequality solving, slightly easier than average due to its predictable structure. |
| Spec | 1.07o Increasing/decreasing: functions using sign of dy/dx1.07q Product and quotient rules: differentiation |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(f(x) = \frac{2x-3}{x^2+4} \Rightarrow f'(x) = \frac{2(x^2+4)-2x(2x-3)}{(x^2+4)^2}\) | M1 | Attempts quotient rule of form \(\frac{P(x^2+4)-Qx(2x-3)}{(x^2+4)^2}\), \(P,Q>0\); condone bracketing errors |
| Fully correct unsimplified derivative | A1 | Fully correct differentiation in any form with correct bracketing |
| \(f'(x) = \frac{-2x^2+6x+8}{(x^2+4)^2}\) | A1 | Simplified form; allow numerator terms in different order |
| Total | (3) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(-2x^2+6x+8=0 \Rightarrow -2(x+1)(x-4)=0 \Rightarrow x=-1, 4\) | B1ft | Uses algebra to solve their \(ax^2+bx+c = 0\); must show working (factorisation, formula, or completing the square); (M1 on EPEN) |
| Chooses correct region for their numerator and critical values | M1 | Selects correct region; must be \(x\) not \(f(x)\); for \(a<0\) with roots \(\alpha < \beta\), need \(x<\alpha\) or \(x>\beta\) |
| \(x < -1\) or \(x > 4\) | A1 | Correct solution from correct numerator; do not accept \(4 < x < -1\) |
| Total | (3) |
# Question 5(a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $f(x) = \frac{2x-3}{x^2+4} \Rightarrow f'(x) = \frac{2(x^2+4)-2x(2x-3)}{(x^2+4)^2}$ | M1 | Attempts quotient rule of form $\frac{P(x^2+4)-Qx(2x-3)}{(x^2+4)^2}$, $P,Q>0$; condone bracketing errors |
| Fully correct unsimplified derivative | A1 | Fully correct differentiation in any form with correct bracketing |
| $f'(x) = \frac{-2x^2+6x+8}{(x^2+4)^2}$ | A1 | Simplified form; allow numerator terms in different order |
| **Total** | **(3)** | |
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# Question 5(b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $-2x^2+6x+8=0 \Rightarrow -2(x+1)(x-4)=0 \Rightarrow x=-1, 4$ | B1ft | Uses algebra to solve their $ax^2+bx+c = 0$; must show working (factorisation, formula, or completing the square); (M1 on EPEN) |
| Chooses correct region for their numerator and critical values | M1 | Selects correct region; must be $x$ not $f(x)$; for $a<0$ with roots $\alpha < \beta$, need $x<\alpha$ or $x>\beta$ |
| $x < -1$ or $x > 4$ | A1 | Correct solution from correct numerator; do not accept $4 < x < -1$ |
| **Total** | **(3)** | |
---\begin{enumerate}
\item The function f is defined by
\end{enumerate}
$$f ( x ) = \frac { 2 x - 3 } { x ^ { 2 } + 4 } \quad x \in \mathbb { R }$$
(a) Show that
$$\mathrm { f } ^ { \prime } ( x ) = \frac { a x ^ { 2 } + b x + c } { \left( x ^ { 2 } + 4 \right) ^ { 2 } }$$
where $a$, $b$ and $c$ are constants to be found.\\
(b) Hence, using algebra, find the values of $x$ for which f is decreasing. You must show each step in your working.
\hfill \mbox{\textit{Edexcel Paper 1 2024 Q5 [6]}}