19M.1.AHL.TZ2.H_9

pestleMathematicsAAHLPaper 119M· sl-3-7-circular-functions-graphs-composites-transformationssource ↗

Consider the functions $f$ and $g$ defined on the domain $0 < x < 2\pi$ by $f\left( x \right) = 3\,{\text{cos}}\,2x$ and $g\left( x \right) = 4 - 11\,{\text{cos}}\,x$ .

The following diagram shows the graphs of $y = f\left( x \right)$ and $y = g\left( x \right)$

Find the $x$ -coordinates of the points of intersection of the two graphs.

[6]

Find the exact area of the shaded region, giving your answer in the form $p\pi + q\sqrt 3$ , where $p$ , $q \in \mathbb{Q}$ .

[5]

At the points A and B on the diagram, the gradients of the two graphs are equal.

Determine the $y$ -coordinate of A on the graph of $g$ .

[6]

Markscheme / solution

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

$3\,{\text{cos}}\,2x = 4 - 11\,{\text{cos}}\,x$

attempt to form a quadratic in ${\text{cos}}\,x$ M1

$3\left( {2\,{\text{co}}{{\text{s}}^2}\,x - 1} \right) = 4 - 11\,{\text{cos}}\,x$ A1

$\left( {6\,{\text{co}}{{\text{s}}^2}\,x + 11\,{\text{cos}}\,x - 7 = 0} \right)$

valid attempt to solve their quadratic M1

$\left( {3\,{\text{cos}}\,x + 7} \right)\left( {2\,{\text{cos}}\,x - 1} \right) = 0$

${\text{cos}}\,x = \frac{1}{2}$ A1

$x = \frac{\pi }{3}{\text{,}}\,\,\frac{{5\pi }}{3}$ A1A1

Note: Ignore any “extra” solutions.

[6 marks]

consider (±) $\int\limits_{\frac{\pi }{3}}^{\frac{{5\pi }}{3}} {\left( {4 - 11\,{\text{cos}}\,x - 3\,{\text{cos}}\,2x} \right)} \,{\text{d}}x$ M1

$= \left( \pm \right)\left[ {4x - 11\,{\text{sin}}\,x - \frac{3}{2}{\text{sin}}\,2x} \right]_{\frac{\pi }{3}}^{\frac{{5\pi }}{3}}$ A1

Note: Ignore lack of or incorrect limits at this stage.

attempt to substitute their limits into their integral M1

$= \frac{{20\pi }}{3} - 11\,{\text{sin}}\frac{{5\pi }}{3} - \frac{3}{2}{\text{sin}}\frac{{10\pi }}{3} - \left( {\frac{{4\pi }}{3} - 11\,{\text{sin}}\frac{\pi }{3} - \frac{3}{2}{\text{sin}}\frac{{2\pi }}{3}} \right)$

$= \frac{{16\pi }}{3} + \frac{{11\sqrt 3 }}{2} + \frac{{3\sqrt 3 }}{4} + \frac{{11\sqrt 3 }}{2} + \frac{{3\sqrt 3 }}{4}$

$= \frac{{16\pi }}{3} + \frac{{25\sqrt 3 }}{2}$ A1A1

[5 marks]

attempt to differentiate both functions and equate M1

$- 6\,{\text{sin}}\,2x = 11\,{\text{sin}}\,x$ A1

attempt to solve for $x$ M1

$11\,{\text{sin}}\,x + 12\,{\text{sin}}\,x\,{\text{cos}}\,x = 0$

${\text{sin}}\,x\left( {11 + 12\,{\text{cos}}\,x} \right) = 0$

${\text{cos}}\,x = - \frac{{11}}{{12}}\,\,\left( {{\text{or}}\,\,{\text{sin}}\,x = 0\,} \right)$ A1

$\Rightarrow y = 4 - 11\left( { - \frac{{11}}{{12}}} \right)$ M1

$y = \frac{{169}}{{12}}\,\left( { = 14\frac{1}{{12}}} \right)$ A1

[6 marks]

Examiners’ report

[N/A]