June 2018 Paper 2 Q11

Notes

(a) Way 2

M1: Way 1: Uses a correct identity \(1 + 11x - 6x^2 \equiv A(1 - 2x)(x - 3) + B(1 - 2x) + C(x - 3)\) in a complete method to find values for \(B\) and \(C\). Note: Allow one slip in copying \(1 + 11x - 6x^2\)
Way 2: Uses a correct identity \(-10x + 10 \equiv B(1 - 2x) + C(x - 3)\) (which has been found from long division) in a complete method to find values for \(B\) and \(C\)

B1: \(A = 3\)

M1: Attempts to find the value of either \(B\) or \(C\) from their identity
This can be achieved by either substituting values into their identity or by comparing coefficients and solving the resulting equations simultaneously

A1: See scheme

Note: Way 1: Comparing terms: \(x^2: -6 = -2A;\ \ x: 11 = 7A - 2B + C;\ \ \text{constant}: 1 = -3A + B - 3C\)
Way 1: Substituting: \(x = 3: -20 = -5B \Rightarrow B = 4;\ x = \dfrac{1}{2}: 5 = -\dfrac{5}{2}C \Rightarrow C = -2\)

Note: Way 2: Comparing terms: \(x: -10 = -2B + C;\ \ \text{constant}: 10 = B - 3C\)
Way 2: Substituting: \(x = 3: -20 = -5B \Rightarrow B = 4;\ x = \dfrac{1}{2}: 5 = -\dfrac{5}{2}C \Rightarrow C = -2\)

Note: \(A = 3, B = 4, C = -2\) from no working scores M1B1M1A1

Note: The final A1 mark is effectively dependent upon both M marks

Note: Writing \(1 + 11x - 6x^2 \equiv B(1 - 2x) + C(x - 3) \Rightarrow B = 4, C = -2\) will get 1^st M0, 2^nd M1, 1^st A0

Note: Way 1: You can imply a correct identity \(1 + 11x - 6x^2 \equiv A(1 - 2x)(x - 3) + B(1 - 2x) + C(x - 3)\) from seeing \(\dfrac{1 + 11x - 6x^2}{(x - 3)(1 - 2x)} \equiv \dfrac{A(1 - 2x)(x - 3) + B(1 - 2x) + C(x - 3)}{(x - 3)(1 - 2x)}\)

Note: Way 2: You can imply a correct identity \(-10x + 10 \equiv B(1 - 2x) + C(x - 3)\) from seeing \(\dfrac{-10x + 10}{(x - 3)(1 - 2x)} \equiv \dfrac{B(1 - 2x) + C(x - 3)}{(x - 3)(1 - 2x)}\)

Note: Be aware of the following alternative solutions, by initially dividing by "\((x - 3)\)" or "\((1 - 2x)\)"

• \(\dfrac{1 + 11x - 6x^2}{\text{``}(x - 3)\text{''}(1 - 2x)} \equiv \dfrac{-6x - 7}{(1 - 2x)} - \dfrac{20}{(x - 3)(1 - 2x)} \equiv 3 - \dfrac{10}{(1 - 2x)} - \dfrac{20}{(x - 3)(1 - 2x)}\)
  \(\dfrac{20}{(x - 3)(1 - 2x)} \equiv \dfrac{D}{(x - 3)} + \dfrac{E}{(1 - 2x)} \Rightarrow 20 \equiv D(1 - 2x) + E(x - 3) \Rightarrow D = -4, E = -8\)
  \(\Rightarrow 3 - \dfrac{10}{(1 - 2x)} - \left(\dfrac{-4}{(x - 3)} + \dfrac{-8}{(1 - 2x)}\right) \equiv 3 + \dfrac{4}{(x - 3)} - \dfrac{2}{(1 - 2x)};\ A = 3, B = 4, C = -2\)
• \(\dfrac{1 + 11x - 6x^2}{(x - 3)\text{``}(1 - 2x)\text{''}} \equiv \dfrac{3x - 4}{(x - 3)} + \dfrac{5}{(x - 3)(1 - 2x)} \equiv 3 + \dfrac{5}{(x - 3)} + \dfrac{5}{(x - 3)(1 - 2x)}\)
  \(\dfrac{5}{(x - 3)(1 - 2x)} \equiv \dfrac{D}{(x - 3)} + \dfrac{E}{(1 - 2x)} \Rightarrow 5 \equiv D(1 - 2x) + E(x - 3) \Rightarrow D = -1, E = -2\)
  \(\Rightarrow 3 + \dfrac{5}{(x - 3)} + \left(\dfrac{-1}{(x - 3)} + \dfrac{-2}{(1 - 2x)}\right) \equiv 3 + \dfrac{4}{(x - 3)} - \dfrac{2}{(1 - 2x)};\ A = 3, B = 4, C = -2\)

Notes

M1: Differentiates to give \(\{\mathrm{f}^{\prime}(x) =\}\ \pm\lambda(x - 3)^{-2} \pm \mu(1 - 2x)^{-2};\ \lambda, \mu \neq 0\)

A1ft: \(\mathrm{f}^{\prime}(x) = -4(x - 3)^{-2} - 4(1 - 2x)^{-2}\), which can be simplified or un-simplified

Note: Allow A1ft for \(\mathrm{f}^{\prime}(x) = -(\text{their } B)(x - 3)^{-2} + (2)(\text{their } C)(1 - 2x)^{-2};\ (\text{their } B), (\text{their } C) \neq 0\)

A1: \(\mathrm{f}^{\prime}(x) = -4(x - 3)^{-2} - 4(1 - 2x)^{-2}\) or \(\mathrm{f}^{\prime}(x) = -\dfrac{4}{(x - 3)^2} - \dfrac{4}{(1 - 2x)^2}\) and a correct explanation
e.g. \(\mathrm{f}^{\prime}(x) = -(+\text{ ve}) - (+\text{ ve}) \lt 0\), so \(\mathrm{f}(x)\) is a decreasing {function}

Note: The final A mark can be scored in part (b) from an incorrect \(A = \ldots\) or from \(A = 0\) or no value of \(A\) found in part (a)

(b) Alternative Method 1

(b) Alternative Method 2

Allow M1A1A1 for the following solution:
Given \(\mathrm{f}(x) = 3 + \dfrac{4}{(x - 3)} - \dfrac{2}{(1 - 2x)} = 3 + \dfrac{4}{(x - 3)} + \dfrac{2}{(2x - 1)}\)
as \(\dfrac{4}{(x - 3)}\) decreases when \(x \gt 3\) and \(\dfrac{2}{(2x - 1)}\) decreases when \(x \gt 3\)
then \(\mathrm{f}(x)\) is a decreasing {function}