Question 1 - A-Level Maths

Edexcel C4 — Question 1 6 marks

Exam Board	Edexcel
Module	C4 (Core Mathematics 4)
Marks	6
Paper	Download PDF ↗
Mark scheme	Download PDF ↗
Topic	Partial Fractions
Type	Partial fractions then differentiate
Difficulty	Standard +0.3 This is a straightforward C4 partial fractions question with standard decomposition followed by routine differentiation. Part (a) is mechanical algebra, and part (b) requires differentiating two simple terms then showing negativity—slightly above average only because it combines two techniques and requires proving an inequality rather than just computing a derivative.
Spec	1.02y Partial fractions: decompose rational functions 1.07i Differentiate x^n: for rational n and sums

The function $f$ is given by

$$f ( x ) = \frac { 3 ( x + 1 ) } { ( x + 2 ) ( x - 1 ) } , x \in \mathbb { R } , x \neq - 2 , x \neq 1$$

Express $\mathrm { f } ( x )$ in partial fractions.
Hence, or otherwise, prove that $\mathrm { f } ^ { \prime } ( x ) < 0$ for all values of $x$ in the domain.

Show mark scheme Show mark scheme source

Question 1:

Part (a)

Answer	Marks	Guidance
$\frac{3(x+1)}{(x+2)(x-1)} \equiv \frac{A}{x+2} + \frac{B}{x-1}$	M1	Setting up partial fractions
$3(x+1) = A(x-1) + B(x+2)$		Multiplying through
$x=1$: $6 = 3B \Rightarrow B = 2$	A1	Correct value of B
$x=-2$: $-3 = -3A \Rightarrow A = 1$	A1	Correct value of A
$f(x) = \frac{1}{x+2} + \frac{2}{x-1}$		Final answer

Part (b)

Answer	Marks	Guidance
$f'(x) = -\frac{1}{(x+2)^2} - \frac{2}{(x-1)^2}$	M1	Differentiating their partial fractions
Both terms $-\frac{1}{(x+2)^2}$ and $-\frac{2}{(x-1)^2}$ are always $\leq 0$	M1	Considering signs of both terms
Each fraction is strictly negative for all $x$ in domain, so $f'(x) < 0$ for all $x$ in domain	A1	Complete argument, conclusion stated

# Question 1:

## Part (a)
| $\frac{3(x+1)}{(x+2)(x-1)} \equiv \frac{A}{x+2} + \frac{B}{x-1}$ | M1 | Setting up partial fractions |
|---|---|---|
| $3(x+1) = A(x-1) + B(x+2)$ | | Multiplying through |
| $x=1$: $6 = 3B \Rightarrow B = 2$ | A1 | Correct value of B |
| $x=-2$: $-3 = -3A \Rightarrow A = 1$ | A1 | Correct value of A |
| $f(x) = \frac{1}{x+2} + \frac{2}{x-1}$ | | Final answer |

## Part (b)
| $f'(x) = -\frac{1}{(x+2)^2} - \frac{2}{(x-1)^2}$ | M1 | Differentiating their partial fractions |
| Both terms $-\frac{1}{(x+2)^2}$ and $-\frac{2}{(x-1)^2}$ are always $\leq 0$ | M1 | Considering signs of both terms |
| Each fraction is strictly negative for all $x$ in domain, so $f'(x) < 0$ for all $x$ in domain | A1 | Complete argument, conclusion stated |

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Show LaTeX source

\begin{enumerate}
  \item The function $f$ is given by
\end{enumerate}

$$f ( x ) = \frac { 3 ( x + 1 ) } { ( x + 2 ) ( x - 1 ) } , x \in \mathbb { R } , x \neq - 2 , x \neq 1$$

(a) Express $\mathrm { f } ( x )$ in partial fractions.\\
(b) Hence, or otherwise, prove that $\mathrm { f } ^ { \prime } ( x ) < 0$ for all values of $x$ in the domain.\\

\hfill \mbox{\textit{Edexcel C4  Q1 [6]}}

This paper (9 questions)

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Q1 6 Q2 4 Q3 6 Q4 6 Q5 8 Q6 11 Q7 11 Q8 13 Q9 14

Answer	Marks	Guidance
\(\frac{3(x+1)}{(x+2)(x-1)} \equiv \frac{A}{x+2} + \frac{B}{x-1}\)	M1	Setting up partial fractions
\(3(x+1) = A(x-1) + B(x+2)\)		Multiplying through
\(x=1\): \(6 = 3B \Rightarrow B = 2\)	A1	Correct value of B
\(x=-2\): \(-3 = -3A \Rightarrow A = 1\)	A1	Correct value of A
\(f(x) = \frac{1}{x+2} + \frac{2}{x-1}\)		Final answer

Answer	Marks	Guidance
\(f'(x) = -\frac{1}{(x+2)^2} - \frac{2}{(x-1)^2}\)	M1	Differentiating their partial fractions
Both terms \(-\frac{1}{(x+2)^2}\) and \(-\frac{2}{(x-1)^2}\) are always \(\leq 0\)	M1	Considering signs of both terms
Each fraction is strictly negative for all \(x\) in domain, so \(f'(x) < 0\) for all \(x\) in domain	A1	Complete argument, conclusion stated