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Date	November 2019	Marks available	2	Reference code	19N.1.AHL.TZ0.H_10
Level	Additional Higher Level	Paper	Paper 1 (without calculator)	Time zone	Time zone 0
Command term	Show and Sketch	Question number	H_10	Adapted from	N/A

Question

Consider $f\left( x \right) = \frac{{2x - 4}}{{{x^2} - 1}}{\text{, }} - 1 < x < 1$ .

For the graph of $y = f\left( x \right)$ ,

Find $f'\left( x \right)$ .

[2]

a.i.

Show that, if $f'\left( x \right) = 0$ , then $x = 2 - \sqrt 3$ .

[3]

a.ii.

find the coordinates of the $y$ -intercept.

[1]

b.i.

show that there are no $x$ -intercepts.

[2]

b.ii.

sketch the graph, showing clearly any asymptotic behaviour.

[2]

b.iii.

Show that $\frac{3}{{x + 1}} - \frac{1}{{x - 1}} = \frac{{2x - 4}}{{{x^2} - 1}}$ .

[2]

c.

The area enclosed by the graph of $y = f\left( x \right)$ and the line $y = 4$ can be expressed as ${\text{ln}}\,v$ . Find the value of $v$ .

[7]

d.

Markscheme

attempt to use quotient rule (or equivalent) (M1)

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

$= \frac{{ - 2{x^2} + 8x - 2}}{{{{\left( {{x^2} - 1} \right)}^2}}}$

[2 marks]

a.i.

$f'\left( x \right) = 0$

simplifying numerator (may be seen in part (i)) (M1)

$\Rightarrow {x^2} - 4x + 1 = 0$ or equivalent quadratic equation A1

EITHER

use of quadratic formula

$\Rightarrow x = \frac{{4 \pm \sqrt {12} }}{2}$ A1

OR

use of completing the square

${\left( {x - 2} \right)^2} = 3$ A1

THEN

$x = 2 - \sqrt 3$ (since $2 + \sqrt 3$ is outside the domain) AG

Note: Do not condone verification that $x = 2 - \sqrt 3 \Rightarrow f'\left( x \right) = 0$ .

Do not award the final A1 as follow through from part (i).

[3 marks]

a.ii.

(0, 4) A1

[1 mark]

b.i.

$2x - 4 = 0 \Rightarrow x = 2$ A1

outside the domain R1

[2 marks]

b.ii.

A1A1

award A1 for concave up curve over correct domain with one minimum point in the first quadrant
award A1 for approaching $x = \pm 1$ asymptotically

[2 marks]

b.iii.

valid attempt to combine fractions (using common denominator) M1

$\frac{{3\left( {x - 1} \right) - \left( {x + 1} \right)}}{{\left( {x + 1} \right)\left( {x - 1} \right)}}$ A1

$= \frac{{3x - 3 - x - 1}}{{{x^2} - 1}}$

$= \frac{{2x - 4}}{{{x^2} - 1}}$ AG

[2 marks]

c.

$f\left( x \right) = 4 \Rightarrow 2x - 4 = 4{x^2} - 4$ M1

( $x = 0$ or) $x = \frac{1}{2}$ A1

area under the curve is $\int_0^{\frac{1}{2}} {f\left( x \right){\text{d}}x}$ M1

$= \int_0^{\frac{1}{2}} {\frac{3}{{x + 1}} - \frac{1}{{x - 1}}{\text{d}}x}$

Note: Ignore absence of, or incorrect limits up to this point.

$= \left[ {3\,{\text{ln}}\,\left| {x + 1} \right| - {\text{ln}}\,\left| {x - 1} \right|} \right]_0^{\frac{1}{2}}$ A1

$= 3\,{\text{ln}}\frac{3}{2} - {\text{ln}}\frac{1}{2}\left( { - 0} \right)$

$= {\text{ln}}\frac{{27}}{4}$ A1

area is $2 - \int_0^{\frac{1}{2}} {f\left( x \right){\text{d}}x}$ or $\int_0^{\frac{1}{2}} {4\,{\text{d}}x} - \int_0^{\frac{1}{2}} {f\left( x \right){\text{d}}x}$ M1

$= 2 - {\text{ln}}\frac{{27}}{4}$

$= {\text{ln}}\frac{{4\,{{\text{e}}^2}}}{{27}}$ A1

$\left( { \Rightarrow v = \frac{{4\,{{\text{e}}^2}}}{{27}}} \right)$

[7 marks]

d.

Examiners report

[N/A]

a.i.

[N/A]

a.ii.

[N/A]

b.i.

[N/A]

b.ii.

[N/A]

b.iii.

[N/A]

c.

[N/A]

d.

Syllabus sections

Topic 2—Functions » AHL 2.13—Rational functions

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