IBDP Maths analysis and approaches Topic: AHL 2.16 :The graph of 1/f(x) given the graph of y=f(x) HL Paper 1

Question

Sketch on the same axes the curve $$y = \left| {\frac{7}{{x – 4}}} \right|$$ and the line $$y = x + 2$$, clearly indicating any axes intercepts and any asymptotes.[3]

a.

Find the exact solutions to the equation $$x + 2 = \left| {\frac{7}{{x – 4}}} \right|$$.[5]

b.

Markscheme

A1 for vertical asymptote and for the $$y$$-intercept $$\frac{7}{4}$$

A1 for general shape of $$y = \left| {\frac{7}{{x – 4}}} \right|$$ including the $$x$$-axis as asymptote

A1 for straight line with $$y$$-intercept 2 and $$x$$-intercept of $$– 2$$     A1A1A1

[3 marks]

a.

METHOD 1

for $$x > 4$$

$$(x + 2)(x – 4) = 7$$     (M1)

$${x^2} – 2x – 8 = 7 \Rightarrow {x^2} – 2x – 15 = 0$$

$$(x – 5)(x + 3) = 0$$

$$({\text{as }}x > 4{\text{ then}}){\text{ }}x = 5$$     A1

Note:     Award A0 if $$x = – 3$$ is also given as a solution.

for $$x < 4$$

$$(x + 2)(x – 4) = – 7$$     M1

$$\Rightarrow {x^2} – 2x – 1 = 0$$

$$x = \frac{{2 \pm \sqrt {4 + 4} }}{2} = 1 \pm \sqrt 2$$     (M1)A1

Note:     Second M1 is dependent on first M1.

[5 marks]

METHOD 2

$${(x + 2)^2} = \frac{{49}}{{{{(x – 4)}^2}}}$$     M1

$${x^4} – 4{x^3} – 12{x^2} + 32x + 15 = 0$$     A1

$$(x + 3)(x – 5)({x^2} – 2x – 1) = 0$$

$$x = 5$$     A1

Note:     Award A0 if $$x = – 3$$ is also given as a solution.

$$x = \frac{{2 \pm \sqrt {4 + 4} }}{2} = 1 \pm \sqrt 2$$     (M1)A1

[5 marks]

b.

Question

A rational function is defined by $$f(x) = a + \frac{b}{{x – c}}$$ where the parameters $$a,{\text{ }}b,{\text{ }}c \in \mathbb{Z}$$ and $$x \in \mathbb{R}\backslash \{ c\}$$. The following diagram represents the graph of $$y = f(x)$$.

Using the information on the graph,

state the value of $$a$$ and the value of $$c$$;[2]

a.

find the value of $$b$$.[2]

b.

Markscheme

$$a = 1$$    A1

$$c = 3$$    A1

[2 marks]

a.

use the coordinates of $$(1,{\text{ }}0)$$ on the graph     M1

$$f(1) = 0 \Rightarrow 1 + \frac{b}{{1 – 3}} = 0 \Rightarrow b = 2$$    A1

[2 marks]

b.

Question

Sketch the graphs of $$y = \frac{x}{2} + 1$$ and $$y = \left| {x – 2} \right|$$ on the following axes.

[3]

a.

Solve the equation $$\frac{x}{2} + 1 = \left| {x – 2} \right|$$.[4]

b.

Markscheme

straight line graph with correct axis intercepts      A1

modulus graph: V shape in upper half plane      A1

modulus graph having correct vertex and y-intercept      A1

[3 marks]

a.

METHOD 1

attempt to solve $$\frac{x}{2} + 1 = x – 2$$     (M1)

$$x = 6$$      A1

Note: Accept $$x = 6$$ using the graph.

attempt to solve (algebraically) $$\frac{x}{2} + 1 = 2 – x$$     M1

$$x = \frac{2}{3}$$     A1

[4 marks]

METHOD 2

$${\left( {\frac{x}{2} + 1} \right)^2} = {\left( {x – 2} \right)^2}$$      M1

$$\frac{{{x^2}}}{4} + x + 1 = {x^2} – 4x + 4$$

$$0 = \frac{{3{x^2}}}{4} – 5x + 3$$

$$3{x^2} – 20x + 12 = 0$$

attempt to factorise (or equivalent)       M1

$$\left( {3x – 2} \right)\left( {x – 6} \right) = 0$$

$$x = \frac{2}{3}$$     A1

$$x = 6$$      A1

[4 marks]

b.
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