Standard +0.3 This is a straightforward partial fractions question with an improper fraction (requiring long division to find constant A), followed by routine differentiation and a simple proof using the derivative. All steps are standard A-level techniques with no novel insight required, making it slightly easier than average.
11. \(\frac { - 6 x ^ { 2 } + 24 x - 9 } { ( x - 2 ) ( 1 - 3 x ) } \equiv A + \frac { B } { x - 2 } + \frac { C } { 1 - 3 x }\)
a. Find the values of the constants \(A , B\) and \(C\).
b. Using part (a), find \(\mathrm { f } ^ { \prime } ( x )\).
c. Prove that \(\mathrm { f } ( x )\) is an increasing function.
Guidance: M1 for uses a correct identity; Either \(-6x^2 + 24x - 9 \equiv A(x-2)(1-3x) + B(1-3x) + C(x-2)\) OR \(10x - 5 \equiv B(1-3x) + C(x-2)\) in a complete method to find values for \(B\) and \(C\); B1 for \(A = 2\); M1 for attempts to find the value of either \(B\) or \(C\) from their identity; A1 for \(B = -3\) and \(C = 1\)
Part b
Answer
Marks
Guidance
Answer: \(f'(x) = 3(x-2)^{-2} + 3(1-3x)^{-2}\) or \(3(x-2)^{-2} + 3(1-3x)^{-2}\)
Marks: 2
Guidance: M1 for differentiates \(2 - 3(x-2)^{-1} + (1-3x)^{-1}\) to give \(A(x-2)^{-1} + B(1-3x)^{-1}\); A1 for correct answer only
Part c
Answer
Marks
Guidance
Answer: \(f(x)\) is an increasing function
Marks: 1 (A1)
Guidance: \(f'(x) = +ve + +ve > 0\), so \(f(x)\) is an increasing function
## Part a
**Answer:** $A = 2$, $B = -3$, $C = 1$ | **Marks:** 4 | **Guidance:** M1 for uses a correct identity; Either $-6x^2 + 24x - 9 \equiv A(x-2)(1-3x) + B(1-3x) + C(x-2)$ OR $10x - 5 \equiv B(1-3x) + C(x-2)$ in a complete method to find values for $B$ and $C$; B1 for $A = 2$; M1 for attempts to find the value of either $B$ or $C$ from their identity; A1 for $B = -3$ and $C = 1$
## Part b
**Answer:** $f'(x) = 3(x-2)^{-2} + 3(1-3x)^{-2}$ or $3(x-2)^{-2} + 3(1-3x)^{-2}$ | **Marks:** 2 | **Guidance:** M1 for differentiates $2 - 3(x-2)^{-1} + (1-3x)^{-1}$ to give $A(x-2)^{-1} + B(1-3x)^{-1}$; A1 for correct answer only
## Part c
**Answer:** $f(x)$ is an increasing function | **Marks:** 1 (A1) | **Guidance:** $f'(x) = +ve + +ve > 0$, so $f(x)$ is an increasing function
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11. $\frac { - 6 x ^ { 2 } + 24 x - 9 } { ( x - 2 ) ( 1 - 3 x ) } \equiv A + \frac { B } { x - 2 } + \frac { C } { 1 - 3 x }$\\
a. Find the values of the constants $A , B$ and $C$.\\
b. Using part (a), find $\mathrm { f } ^ { \prime } ( x )$.\\
c. Prove that $\mathrm { f } ( x )$ is an increasing function.\\
\hfill \mbox{\textit{Edexcel PMT Mocks Q11 [7]}}