EXM.1.AHL.TZ0.4

pestleMathematicsAAHLPaper 1EXM· ahl-2-13-rational-functionssource ↗

Let $f\left( x \right) = \frac{{2x + 6}}{{{x^2} + 6x + 10}}{\text{,}}\,\,x \in \mathbb{R}.$

Show that $f\left( x \right)$ has no vertical asymptotes.

[3]

Find the equation of the horizontal asymptote.

[2]

Find the exact value of $\int\limits_0^1 {f\left( x \right)} \,dx$ , giving the answer in the form ${\text{ln}}\,q{\text{,}}\,\,q \in \mathbb{Q}$ .

[3]

Markscheme / solution

${x^2} + 6x + 10 = {x^2} + 6x + 9 + 1 = {\left( {x + 3} \right)^2} + 1$ M1A1

So the denominator is never zero and thus there are no vertical asymptotes. (or use of discriminant is negative) R1

[3 marks]

$x \to \pm \infty {\text{,}}\,\,f\left( x \right) \to 0$ so the equation of the horizontal asymptote is $y =0$ M1A1

[2 marks]

$\int\limits_0^1 {\frac{{2x + 6}}{{{x^2} + 6x + 10}}} \,dx = \left[ {{\text{ln}}\left( {{x^2} + 6x + 10} \right)} \right]_0^1 = {\text{ln}}\,17 - {\text{ln}}\,10 = {\text{ln}}\,\frac{{17}}{{10}}$ M1A1A1

[3 marks]

Examiners’ report

[N/A]