EXM.1.AHL.TZ0.3

pestleMathematicsAAHLPaper 1EXM· ahl-1-11-partial-fractionssource ↗

Let $f\left( x \right) = \frac{{4x - 5}}{{{x^2} - 3x + 2}}$ $x \ne 1{\text{,}}\,x \ne 2$ .

Express $f(x)$ in partial fractions.

[6]

Use part (a) to show that $f(x)$ is always decreasing.

[3]

Use part (a) to find the exact value of $\int\limits_{ - 1}^0 {f(x)dx}$ , giving the answer in the form ${\text{ln}}\,q$ , $q \in \mathbb{Q}$ .

[4]

Markscheme / solution

$f\left( x \right) = \frac{{4x - 5}}{{\left( {x - 1} \right)\left( {x - 2} \right)}} \equiv \frac{A}{{x - 1}} + \frac{B}{{x - 2}}$ M1A1

$\Rightarrow 4x - 5 \equiv A\left( {x - 2} \right) + B\left( {x - 1} \right)$ M1A1

$x = 1 \Rightarrow A = 1$ $x = 2 \Rightarrow B = 3$ A1A1

$f\left( x \right) = \frac{1}{{x - 1}} + \frac{3}{{x - 2}}$

[6 marks]

$f'\left( x \right) = - {\left( {x - 1} \right)^{ - 2}} - 3{\left( {x - 2} \right)^{ - 2}}$ M1A1

This is always negative so function is always decreasing. R1AG

[3 marks]

$\int\limits_{ - 1}^0 {\frac{1}{{x - 1}} + \frac{3}{{x - 2}}{\text{ }}} dx = \left[ {{\text{ln}}\left| {x - 1} \right| + 3\,{\text{ln}}\left| {x - 2} \right|} \right]_{ - 1}^0$ M1A1

$= \left( {3\,{\text{ln}}\,2} \right) - \left( {\,{\text{ln}}\,2 + 3\,{\text{ln}}\,3} \right) = 2\,{\text{ln}}\,2 - 3\,{\text{ln}}\,3 = \,{\text{ln}}\frac{4}{{27}}$ A1A1

[4 marks]

Examiners’ report

[N/A]