21N.2.AHL.TZ0.10

pestleMathematicsAAHLPaper 221N· sl-1-7-laws-of-exponents-and-logssource ↗

Consider the function $f (x) = \frac{x^{2} - x - 12}{2 x - 15}, x \in ℝ, x \neq \frac{15}{2}$ .

Find the coordinates where the graph of $f$ crosses the

$x$ -axis.

[2]

a.i.

$y$ -axis.

[1]

a.ii.

Write down the equation of the vertical asymptote of the graph of $f$ .

[1]

The oblique asymptote of the graph of $f$ can be written as $y = a x + b$ where $a, b \in ℚ$ .

Find the value of $a$ and the value of $b$ .

[4]

Sketch the graph of $f$ for $- 30 \leq x \leq 30$ , clearly indicating the points of intersection with each axis and any asymptotes.

[3]

Express $\frac{1}{f (x)}$ in partial fractions.

[3]

e.i.

Hence find the exact value of $\int_{0}^{3} \frac{1}{f (x)} d x$ , expressing your answer as a single logarithm.

[4]

e.ii.

Markscheme / solution

Note: In part (a), penalise once only, if correct values are given instead of correct coordinates.

attempts to solve $x^{2} - x - 12 = 0$ (M1)

$(- 3, 0)$ and $(4, 0)$ A1

[2 marks]

a.i.

Note: In part (a), penalise once only, if correct values are given instead of correct coordinates.

$(0, \frac{4}{5})$ A1

[1 mark]

a.ii.

$x = \frac{15}{2}$ A1

Note: Award A0 for $x \neq \frac{15}{2}$ .
Award A1 in part (b), if $x = \frac{15}{2}$ is seen on their graph in part (d).

[1 mark]

METHOD 1

$(a x + b) (2 x - 15) \equiv x^{2} - x - 12$

attempts to expand $(a x + b) (2 x - 15)$ (M1)

$2 a x^{2} - 15 a x + 2 b x - 15 b \equiv x^{2} - x - 12$

$a = \frac{1}{2}$ A1

equates coefficients of $x$ (M1)

$- 1 = - \frac{15}{2} + 2 b$

$b = \frac{13}{4}$ A1

$(y = \frac{x}{2} + \frac{13}{4})$

METHOD 2

attempts division on $\frac{x^{2} - x - 12}{2 x - 15}$ M1

$\frac{x}{2} + \frac{13}{4} + \dots$ M1

$a = \frac{1}{2}$ A1

$b = \frac{13}{4}$ A1

$(y = \frac{x}{2} + \frac{13}{4})$

METHOD 3

$a = \frac{1}{2}$ A1

$\frac{x^{2} - x - 12}{2 x - 15} \equiv \frac{x}{2} + b + \frac{c}{2 x - 15}$ M1

$x^{2} - x - 12 \equiv \frac{(2 x - 15) x}{2} + (2 x - 15) b + c$

equates coefficients of $x$ : (M1)

$- 1 = - \frac{15}{2} + 2 b$

$b = \frac{13}{4}$ A1

$(y = \frac{x}{2} + \frac{13}{4})$

METHOD 4

attempts division on $\frac{x^{2} - x - 12}{2 x - 15}$ M1

$\frac{x^{2} - x - 12}{2 x - 15} = \frac{x}{2} + \frac{\frac{13 x}{2} - 12}{2 x - 15}$

$a = \frac{1}{2}$ A1

$\frac{\frac{13 x}{2} - 12}{2 x - 15} = \frac{13}{4} + \dots$ M1

$b = \frac{13}{4}$ A1

$(y = \frac{x}{2} + \frac{13}{4})$

[4 marks]

two branches with approximately correct shape (for $- 30 \leq x \leq 30$ ) A1

their vertical and oblique asymptotes in approximately correct positions with both branches showing correct asymptotic behaviour to these asymptotes A1

their axes intercepts in approximately the correct positions A1

Note: Points of intersection with the axes and the equations of asymptotes are not required to be labelled.

[3 marks]

attempts to split into partial fractions: (M1)

$\frac{2 x - 15}{(x + 3) (x - 4)} \equiv \frac{A}{x + 3} + \frac{B}{x - 4}$

$2 x - 15 \equiv A (x - 4) + B (x + 3)$

$A = 3$ A1

$B = - 1$ A1

$(\frac{3}{x + 3} - \frac{1}{x - 4})$

[3 marks]

e.i.

$\int_{0}^{3} (\frac{3}{x + 3} - \frac{1}{x - 4}) d x$

attempts to integrate and obtains two terms involving ‘ln’ (M1)

$= {[3 \ln |x + 3| - \ln |x - 4|]}_{0}^{3}$ A1

$= 3 \ln 6 - \ln 1 - 3 \ln 3 + \ln 4$ A1

$= 3 \ln 2 + \ln 4 (= \ln 8 + \ln 4)$

$= \ln 32 (= 5 \ln 2)$ A1

Note: The final A1 is dependent on the previous two A marks.

[4 marks]

e.ii.

Examiners’ report

[N/A]

a.i.

[N/A]

a.ii.

[N/A]

e.i.

[N/A]

e.ii.