SPM.1.SL.TZ0.6

pestleMathematicsAASLPaper 1SPM· sl-3-6-pythagorean-identity-double-anglessource ↗

Show that ${\text{lo}}{{\text{g}}_9}\left( {{\text{cos}}\,2x + 2} \right) = {\text{lo}}{{\text{g}}_3}\sqrt {{\text{cos}}\,2x + 2}$ .

[3]

Hence or otherwise solve ${\text{lo}}{{\text{g}}_3}\left( {{\text{2}}\,{\text{sin}}\,x} \right) = {\text{lo}}{{\text{g}}_9}\left( {{\text{cos}}\,2x + 2} \right)$ for $0 < x < \frac{\pi }{2}$ .

[5]

Markscheme / solution

attempting to use the change of base rule M1

${\text{lo}}{{\text{g}}_9}\left( {{\text{cos}}\,2x + 2} \right) = \frac{{{\text{lo}}{{\text{g}}_3}\left( {{\text{cos}}\,2x + 2} \right)}}{{{\text{lo}}{{\text{g}}_3}9}}$ A1

$= \frac{1}{2}{\text{lo}}{{\text{g}}_3}\left( {{\text{cos}}\,2x + 2} \right)$ A1

$= {\text{lo}}{{\text{g}}_3}\sqrt {{\text{cos}}\,2x + 2}$ AG

[3 marks]

${\text{lo}}{{\text{g}}_3}\left( {{\text{2}}\,{\text{sin}}\,x} \right) = {\text{lo}}{{\text{g}}_3}\sqrt {{\text{cos}}\,2x + 2}$

${\text{2}}\,{\text{sin}}\,x = \sqrt {{\text{cos}}\,2x + 2}$ M1

${\text{4}}\,{\text{si}}{{\text{n}}^2}\,x = {\text{cos}}\,2x + 2$ (or equivalent) A1

use of ${\text{cos}}\,2x = 1 - 2\,{\text{si}}{{\text{n}}^2}\,x$ (M1)

$6\,{\text{si}}{{\text{n}}^2}\,x = 3$

${\text{sin}}\,x = \left( \pm \right)\frac{1}{{\sqrt 2 }}$ A1

$x = \frac{\pi }{4}$ A1

Note: Award A0 if solutions other than $x = \frac{\pi }{4}$ are included.

[5 marks]

Examiners’ report

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