17M.1.AHL.TZ1.H_11

pestleMathematicsAAHLPaper 117M· sl-2-6-quadratic-function, sl-5-11-definite-integrals-areas-under-curve-onto-x-axis-and-areas-between-curvessource ↗

Consider the function $f(x) = \frac{1}{{{x^2} + 3x + 2}},{\text{ }}x \in \mathbb{R},{\text{ }}x \ne - 2,{\text{ }}x \ne - 1$ .

Express ${x^2} + 3x + 2$ in the form ${(x + h)^2} + k$ .

[1]

a.i.

Factorize ${x^2} + 3x + 2$ .

[1]

a.ii.

Sketch the graph of $f(x)$ , indicating on it the equations of the asymptotes, the coordinates of the $y$ -intercept and the local maximum.

[5]

Hence find the value of $p$ if $\int_0^1 {f(x){\text{d}}x = \ln (p)}$ .

[4]

Sketch the graph of $y = f\left( {\left| x \right|} \right)$ .

[2]

Determine the area of the region enclosed between the graph of $y = f\left( {\left| x \right|} \right)$ , the $x$ -axis and the lines with equations $x = - 1$ and $x = 1$ .

[3]

Markscheme / solution

${x^2} + 3x + 2 = {\left( {x + \frac{3}{2}} \right)^2} - \frac{1}{4}$ A1

[1 mark]

a.i.

${x^2} + 3x + 2 = (x + 2)(x + 1)$ A1

[1 mark]

a.ii.

M17/5/MATHL/HP1/ENG/TZ1/B11.b/M

A1 for the shape

A1 for the equation $y = 0$

A1 for asymptotes $x = - 2$ and $x = - 1$

A1 for coordinates $\left( { - \frac{3}{2},{\text{ }} - 4} \right)$

A1 $y$ -intercept $\left( {0,{\text{ }}\frac{1}{2}} \right)$

[5 marks]

$\int\limits_0^1 {\frac{1}{{x + 1}} - \frac{1}{{x + 2}}{\text{d}}x}$

$= \left[ {\ln (x + 1) - \ln (x + 2)} \right]_0^1$ A1

$= \ln 2 - \ln 3 - \ln 1 + \ln 2$ M1

$= \ln \left( {\frac{4}{3}} \right)$ M1A1

$\therefore p = \frac{4}{3}$

[4 marks]

M17/5/MATHL/HP1/ENG/TZ1/B11.e/M

symmetry about the $y$ -axis M1

correct shape A1

Note: Allow FT from part (b).

[2 marks]

$2\int_0^1 {f(x){\text{d}}x}$ (M1)(A1)

$= 2\ln \left( {\frac{4}{3}} \right)$ A1

Note: Do not award FT from part (e).

[3 marks]

Examiners’ report

[N/A]

a.i.

[N/A]

a.ii.

[N/A]