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Date	May 2019	Marks available	5	Reference code	19M.2.SL.TZ1.S_9
Level	Standard Level	Paper	Paper 2	Time zone	Time zone 1
Command term	Find	Question number	S_9	Adapted from	N/A

Question

Let $f\left( x \right) = \frac{{16}}{x}$ . The line $L$ is tangent to the graph of $f$ at $x = 8$ .

$L$ can be expressed in the form r $= \left( {\begin{array}{*{20}{c}} 8 \\ 2 \end{array}} \right) + t$ u.

The direction vector of $y = x$ is $\left( {\begin{array}{*{20}{c}} 1 \\ 1 \end{array}} \right)$ .

Find the gradient of $L$ .

[2]

a.

Find u.

[2]

b.

Find the acute angle between $y = x$ and $L$ .

[5]

c.

Find $\left( {f \circ f} \right)\left( x \right)$ .

[3]

d.i.

Hence, write down ${f^{ - 1}}\left( x \right)$ .

[1]

d.ii.

Hence or otherwise, find the obtuse angle formed by the tangent line to $f$ at $x = 8$ and the tangent line to $f$ at $x = 2$ .

[3]

d.iii.

Markscheme

* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.

attempt to find $f'\left( 8 \right)$ (M1)

eg $f'\left( x \right)$ , ${y'}$ , $- 16{x^{ - 2}}$

−0.25 (exact) A1 N2

[2 marks]

a.

u $= \left( {\begin{array}{*{20}{c}} 4 \\ { - 1} \end{array}} \right)$ or any scalar multiple A2 N2

[2 marks]

b.

correct scalar product and magnitudes (A1)(A1)(A1)

scalar product $= 1 \times 4 + 1 \times - 1\,\,\,\left( { = 3} \right)$

magnitudes $= \sqrt {{1^2} + {1^2}}$ , $\sqrt {{4^2} + {{\left( { - 1} \right)}^2}}$ $\left( { = \sqrt 2 {\text{,}}\,\,\sqrt {17} } \right)$

substitution of their values into correct formula (M1)

eg $\frac{{4 - 1}}{{\sqrt {{1^2} + {1^2}} \sqrt {{4^2} + {{\left( { - 1} \right)}^2}} }}$ , $\frac{{ - 3}}{{\sqrt 2 \sqrt {17} }}$ , 2.1112, 120.96°

1.03037 , 59.0362°

angle = 1.03 , 59.0° A1 N4

[5 marks]

c.

attempt to form composite $\left( {f \circ f} \right)\left( x \right)$ (M1)

eg $f\left( {f\left( x \right)} \right)$ , $f\left( {\frac{{16}}{x}} \right)$ , $\frac{{16}}{{f\left( x \right)}}$

correct working (A1)

eg $\frac{{16}}{{\frac{{16}}{x}}}$ , $16 \times \frac{x}{{16}}$

$\left( {f \circ f} \right)\left( x \right) = x$ A1 N2

[3 marks]

d.i.

${f^{ - 1}}\left( x \right) = \frac{{16}}{x}$ (accept $y = \frac{{16}}{x}$ , $\frac{{16}}{x}$ ) A1 N1

Note: Award A0 in part (ii) if part (i) is incorrect.
Award A0 in part (ii) if the candidate has found ${f^{ - 1}}\left( x \right) = \frac{{16}}{x}$ by interchanging $x$ and $y$ .

[1 mark]

d.ii.

METHOD 1

recognition of symmetry about $y = x$ (M1)

eg (2, 8) ⇔ (8, 2)

evidence of doubling their angle (M1)

eg $2 \times 1.03$ , $2 \times 59.0$

2.06075, 118.072°

2.06 (radians) (118 degrees) A1 N2

METHOD 2

finding direction vector for tangent line at $x = 2$ (A1)

eg $\left( {\begin{array}{*{20}{c}} { - 1} \\ 4 \end{array}} \right)$ , $\left( {\begin{array}{*{20}{c}} 1 \\ { - 4} \end{array}} \right)$

substitution of their values into correct formula (must be from vectors) (M1)

eg $\frac{{ - 4 - 4}}{{\sqrt {{1^2} + {4^2}} \sqrt {{4^2} + {{\left( { - 1} \right)}^2}} }}$ , $\frac{8}{{\sqrt {17} \sqrt {17} }}$

2.06075, 118.072°

2.06 (radians) (118 degrees) A1 N2

METHOD 3

using trigonometry to find an angle with the horizontal (M1)

eg ${\text{tan}}\,\theta = - \frac{1}{4}$ , ${\text{tan}}\,\theta = - 4$

finding both angles of rotation (A1)

eg ${\theta _1} = 0.244978{\text{, 14}}{\text{.0362}}^\circ$ , ${\theta _1} = 1.81577{\text{, 104}}{\text{.036}}^\circ$

2.06075, 118.072°

2.06 (radians) (118 degrees) A1 N2

[3 marks]

d.iii.

Examiners report

[N/A]

a.

[N/A]

b.

[N/A]

c.

[N/A]

d.i.

[N/A]

d.ii.

[N/A]

d.iii.

Syllabus sections

Topic 4—Statistics and probability » SL 4.1—Concepts, reliability and sampling techniques

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16N.2.SL.TZ0.S_10b:
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17M.2.SL.TZ2.S_8d:
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17N.1.SL.TZ0.S_5b:
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19M.2.SL.TZ1.S_9a:
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19M.2.SL.TZ1.S_3b:
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19M.2.SL.TZ1.S_9d.ii:
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21M.2.SL.TZ1.1a:
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18N.2.SL.TZ0.S_10b.i:
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16N.2.SL.TZ0.S_10c:
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21M.3.AHL.TZ1.2b:
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21M.2.SL.TZ2.1a:
State the sampling method being used.
16N.2.AHL.TZ0.H_11c:
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21M.3.AHL.TZ2.1b:
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EXN.1.SL.TZ0.1a:
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17M.2.SL.TZ2.S_8c.ii:
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18M.1.SL.TZ1.S_7:
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