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Date	May 2019	Marks available	9	Reference code	19M.1.AHL.TZ1.H_9
Level	Additional Higher Level	Paper	Paper 1 (without calculator)	Time zone	Time zone 1
Command term	Find and Hence or otherwise	Question number	H_9	Adapted from	N/A

Question

Show that ${\left( {{\text{sin}}\,x + {\text{cos}}\,x} \right)^2} = 1 + {\text{sin}}\,2x$ .

[2]

Show that ${\text{sec}}\,2x + {\text{tan}}\,2x = \frac{{{\text{cos}}\,x + {\text{sin}}\,x}}{{{\text{cos}}\,x - {\text{sin}}\,x}}$ .

[4]

Hence or otherwise find $\int_0^{\frac{\pi }{6}} {\left( {{\text{sec}}\,2x + {\text{tan}}\,2x} \right)} {\text{d}}x$ in the form ${\text{ln}}\left( {a + \sqrt b } \right)$ where $a$ , $b \in \mathbb{Z}$ .

[9]

Markscheme

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

${\left( {{\text{sin}}\,x + {\text{cos}}\,x} \right)^2} = {\text{si}}{{\text{n}}^2}\,x + 2{\text{sin}}\,x\,{\text{cos}}\,x + {\text{co}}{{\text{s}}^2}\,x$ M1A1

Note: Do not award the M1 for just ${\text{si}}{{\text{n}}^2}\,x + {\text{co}}{{\text{s}}^2}\,x$ .

Note: Do not award A1 if correct expression is followed by incorrect working.

$= 1 + {\text{sin}}\,2x$ AG

[2 marks]

${\text{sec}}\,2x + {\text{tan}}\,2x = \frac{1}{{{\text{cos}}\,2x}} + \frac{{{\text{sin}}\,2x}}{{{\text{cos}}\,2x}}$ M1

Note: M1 is for an attempt to change both terms into sine and cosine forms (with the same argument) or both terms into functions of ${\text{tan}}\,x$ .

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

$= \frac{{{{\left( {{\text{sin}}\,x + {\text{cos}}\,x} \right)}^2}}}{{{\text{co}}{{\text{s}}^2}\,x - {\text{si}}{{\text{n}}^2}\,x}}$ A1A1

Note: Award A1 for numerator, A1 for denominator.

$= \frac{{{{\left( {{\text{sin}}\,x + {\text{cos}}\,x} \right)}^2}}}{{\left( {{\text{cos}}\,x - {\text{sin}}\,x} \right)\left( {{\text{cos}}\,x + {\text{sin}}\,x} \right)}}$ M1

$= \frac{{{\text{cos}}\,x + {\text{sin}}\,x}}{{{\text{cos}}\,x - {\text{sin}}\,x}}$ AG

Note: Apply MS in reverse if candidates have worked from RHS to LHS.

Note: Alternative method using ${\text{tan}}\,2x$ and ${\text{sec}}\,2x$ in terms of ${\text{tan}}\,x$ .

[4 marks]

METHOD 1

$\int_0^{\frac{\pi }{6}} {\left( {\frac{{{\text{cos}}\,x + {\text{sin}}\,x}}{{{\text{cos}}\,x - {\text{sin}}\,x}}} \right)} {\text{d}}x$ A1

Note: Award A1 for correct expression with or without limits.

EITHER

$= \left[ { - {\text{ln}}\left( {{\text{cos}}\,x - {\text{sin}}\,x} \right)} \right]_0^{\frac{\pi }{6}}$ or $\left[ {{\text{ln}}\left( {{\text{cos}}\,x - {\text{sin}}\,x} \right)} \right]_{\frac{\pi }{6}}^0$ (M1)A1A1

Note: Award M1 for integration by inspection or substitution, A1 for ${{\text{ln}}\left( {{\text{cos}}\,x - {\text{sin}}\,x} \right)}$ , A1 for completely correct expression including limits.

$= - {\text{ln}}\left( {{\text{cos}}\,\frac{\pi }{6} - {\text{sin}}\,\frac{\pi }{6}} \right) + {\text{ln}}\left( {{\text{cos}}\,0 - {\text{sin}}\,0} \right)$ M1

Note: Award M1 for substitution of limits into their integral and subtraction.

$= - {\text{ln}}\left( {\frac{{\sqrt 3 }}{2} - \frac{1}{2}} \right)$ (A1)

let $u = {\text{cos}}\,x - {\text{sin}}\,x$ M1

$\frac{{{\text{d}}u}}{{{\text{d}}x}} = - {\text{sin}}\,x - {\text{cos}}\,x = - \left( {{\text{sin}}\,x + {\text{cos}}\,x} \right)$

$- \int_1^{\frac{{\sqrt 3 }}{2} - \frac{1}{2}} {\left( {\frac{1}{u}} \right)} {\text{d}}u$ A1A1

Note: Award A1 for correct limits even if seen later, A1 for integral.

$= \left[ { - {\text{ln}}\,u} \right]_1^{\frac{{\sqrt 3 }}{2} - \frac{1}{2}}$ or $\left[ {{\text{ln}}\,u} \right]_{\frac{{\sqrt 3 }}{2} - \frac{1}{2}}^1$ A1

$= - {\text{ln}}\left( {\frac{{\sqrt 3 }}{2} - \frac{1}{2}} \right)\left( {{\text{ + ln}}\,1} \right)$ M1

THEN

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

Note: Award M1 for both putting the expression over a common denominator and for correct use of law of logarithms.

$= {\text{ln}}\left( {1 + \sqrt 3 } \right)$ (M1)A1

METHOD 2

$\left[ {\frac{1}{2}{\text{ln}}\left( {{\text{tan}}\,2x + {\text{sec}}\,2x} \right) - \frac{1}{2}{\text{ln}}\left( {{\text{cos}}\,2x} \right)} \right]_0^{\frac{\pi }{6}}$ A1A1