18M.1.AHL.TZ2.H_11

pestleMathematicsAIHLPaper 118M· sl-1-5-intro-to-logssource ↗

It is given that ${\text{lo}}{{\text{g}}_2}\,y + {\text{lo}}{{\text{g}}_4}\,x + {\text{lo}}{{\text{g}}_4}\,2x = 0$ .

Show that ${\text{lo}}{{\text{g}}_{{r^2}}}x = \frac{1}{2}{\text{lo}}{{\text{g}}_r}\,x$ where $r,\,x \in {\mathbb{R}^ + }$ .

[2]

Express $y$ in terms of $x$ . Give your answer in the form $y = p{x^q}$ , where p , q are constants.

[5]

The region R, is bounded by the graph of the function found in part (b), the x-axis, and the lines $x = 1$ and $x = \alpha$ where $\alpha > 1$ . The area of R is $\sqrt 2$ .

Find the value of $\alpha$ .

[5]

Markscheme / solution

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

METHOD 1

${\text{lo}}{{\text{g}}_{{r^2}}}x = \frac{{{\text{lo}}{{\text{g}}_r}\,x}}{{{\text{lo}}{{\text{g}}_r}\,{r^2}}}\left( { = \frac{{{\text{lo}}{{\text{g}}_r}\,x}}{{{\text{2}}\,{\text{lo}}{{\text{g}}_r}\,r}}} \right)$ M1A1

$= \frac{{{\text{lo}}{{\text{g}}_r}\,x}}{2}$ AG

[2 marks]

METHOD 2

${\text{lo}}{{\text{g}}_{{r^2}}}x = \frac{1}{{{\text{lo}}{{\text{g}}_x}\,{r^2}}}$ M1

$= \frac{1}{{2\,{\text{lo}}{{\text{g}}_x}\,r}}$ A1

$= \frac{{{\text{lo}}{{\text{g}}_r}\,x}}{2}$ AG

[2 marks]

METHOD 1

${\text{lo}}{{\text{g}}_2}\,y + {\text{lo}}{{\text{g}}_4}\,x + {\text{lo}}{{\text{g}}_4}\,2x = 0$

${\text{lo}}{{\text{g}}_2}\,y + {\text{lo}}{{\text{g}}_4}\,2{x^2} = 0$ M1

${\text{lo}}{{\text{g}}_2}\,y + \frac{1}{2}{\text{lo}}{{\text{g}}_2}\,2{x^2} = 0$ M1

${\text{lo}}{{\text{g}}_2}\,y = - \frac{1}{2}{\text{lo}}{{\text{g}}_2}\,2{x^2}$

${\text{lo}}{{\text{g}}_2}\,y = {\text{lo}}{{\text{g}}_2}\left( {\frac{1}{{\sqrt {2x} }}} \right)$ M1A1

$y = \frac{1}{{\sqrt 2 }}{x^{ - 1}}$ A1

Note: For the final A mark, $y$ must be expressed in the form $p{x^q}$ .

[5 marks]

METHOD 2

${\text{lo}}{{\text{g}}_2}\,y + {\text{lo}}{{\text{g}}_4}\,x + {\text{lo}}{{\text{g}}_4}\,2x = 0$

${\text{lo}}{{\text{g}}_2}\,y + \frac{1}{2}{\text{lo}}{{\text{g}}_2}\,x + \frac{1}{2}{\text{lo}}{{\text{g}}_2}\,2x = 0$ M1

${\text{lo}}{{\text{g}}_2}\,y + {\text{lo}}{{\text{g}}_2}\,{x^{\frac{1}{2}}} + {\text{lo}}{{\text{g}}_2}\,{\left( {2x} \right)^{\frac{1}{2}}} = 0$ M1

${\text{lo}}{{\text{g}}_2}\,\left( {\sqrt 2 xy} \right) = 0$ M1

$\sqrt 2 xy = 1$ A1

$y = \frac{1}{{\sqrt 2 }}{x^{ - 1}}$ A1

Note: For the final A mark, $y$ must be expressed in the form $p{x^q}$ .

[5 marks]

the area of R is $\int\limits_1^\alpha {\frac{1}{{\sqrt 2 }}} {x^{ - 1}}{\text{d}}x$ M1

$= \left[ {\frac{1}{{\sqrt 2 }}{\text{ln}}\,x} \right]_1^\alpha$ A1

$= \frac{1}{{\sqrt 2 }}{\text{ln}}\,\alpha$ A1

$\frac{1}{{\sqrt 2 }}{\text{ln}}\,\alpha = \sqrt 2$ M1

$\alpha = {{\text{e}}^2}$ A1

Note: Only follow through from part (b) if $y$ is in the form $y = p{x^q}$

[5 marks]

Examiners’ report

[N/A]