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Date	May 2019	Marks available	2	Reference code	19M.1.SL.TZ2.S_9
Level	Standard Level	Paper	Paper 1	Time zone	Time zone 2
Command term	Find	Question number	S_9	Adapted from	N/A

Question

Let $θ$ $\theta$ be an obtuse angle such that $sin θ = \frac{3}{5}$ ${\text{sin}}\,\theta = \frac{3}{5}$ .

Let $f (x) = e^{x} sin x - \frac{3 x}{4}$ $f\left( x \right) = {{\text{e}}^x}\,{\text{sin}}\,x - \frac{{3x}}{4}$ .

Find the value of $tan θ$ ${\text{tan}}\,\theta$ .

[4]

a.

Line $L$ $L$ passes through the origin and has a gradient of $tan θ$ ${\text{tan}}\,\theta$ . Find the equation of $L$ $L$ .

[2]

b.

Find the derivative of $f$ $f$ .

[5]

c.

The following diagram shows the graph of $f$ $f$ for 0 ≤ $x$ $x$ ≤ 3. Line $M$ $M$ is a tangent to the graph of $f$ $f$ at point P.

Given that $M$ $M$ is parallel to $L$ $L$ , find the $x$ $x$ -coordinate of P.

[4]

d.

Markscheme

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

evidence of valid approach (M1)

eg sketch of triangle with sides 3 and 5, $co s^{2} θ = 1 - si n^{2} θ$ ${\text{co}}{{\text{s}}^2}\,\theta = 1 - {\text{si}}{{\text{n}}^2}\,\theta$

correct working (A1)

eg missing side is 4 (may be seen in sketch), $cos θ = \frac{4}{5}$ ${\text{cos}}\,\theta = \frac{4}{5}$ , $cos θ = - \frac{4}{5}$ ${\text{cos}}\,\theta = - \frac{4}{5}$

$tan θ = - \frac{3}{4}$ ${\text{tan}}\,\theta = - \frac{3}{4}$ A2 N4

[4 marks]

a.

correct substitution of either gradient or origin into equation of line (A1)

(do not accept $y = m x + b$ $y = mx + b$ )

eg $y = x tan θ$ $y = x\,{\text{tan}}\,\theta$ , $y - 0 = m (x - 0)$ $y - 0 = m\left( {x - 0} \right)$ , $y = m x$ $y = mx$

$y = - \frac{3}{4} x$ $y = - \frac{3}{4}x$ A2 N4

Note: Award A1A0 for $L = - \frac{3}{4} x$ $L = - \frac{3}{4}x$ .

[2 marks]

b.

$\frac{d}{d x} (\frac{- 3 x}{4}) = - \frac{3}{4}$ $\frac{{\text{d}}}{{{\text{d}}x}}\left( {\frac{{ - 3x}}{4}} \right) = - \frac{3}{4}$ (seen anywhere, including answer) A1

choosing product rule (M1)

eg $uv' + vu'$

correct derivatives (must be seen in a correct product rule) A1A1

eg ${\text{cos}}\,x$ , ${{\text{e}}^x}$

$f'\left( x \right) = {{\text{e}}^x}\,{\text{cos}}\,x + {{\text{e}}^x}\,{\text{sin}}\,x - \frac{3}{4}$ $\left( { = {{\text{e}}^x}\,\left( {{\text{cos}}\,x + {\text{sin}}\,x} \right) - \frac{3}{4}} \right)$ A1 N5

[5 marks]

c.

valid approach to equate their gradients (M1)

eg $f' = {\text{tan}}\,\theta$ , $f' = - \frac{3}{4}$ , ${{\text{e}}^x}\,{\text{cos}}\,x + {{\text{e}}^x}\,{\text{sin}}\,x - \frac{3}{4} = - \frac{3}{4}$ , ${{\text{e}}^x}\,\left( {{\text{cos}}\,x + {\text{sin}}\,x} \right) - \frac{3}{4} = - \frac{3}{4}$

correct equation without ${{\text{e}}^x}$ (A1)

eg ${\text{sin}}\,x = - {\text{cos}}\,x$ , ${\text{cos}}\,x + {\text{sin}}\,x = 0$ , $\frac{{ - {\text{sin}}\,x}}{{{\text{cos}}\,x}} = 1$

correct working (A1)

eg ${\text{tan}}\,\theta = - 1$ , $x = 135^\circ$

$x = \frac{{3\pi }}{4}$ (do not accept $135^\circ$ ) A1 N1

Note: Do not award the final A1 if additional answers are given.

[4 marks]

d.

Examiners report

[N/A]

a.

[N/A]

b.

[N/A]

c.

[N/A]

d.

Syllabus sections

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

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