22M.1.AHL.TZ2.8

let $m$ be the median

EITHER

attempts to find the area of the required triangle          M1

base is $\left(m-a\right)$          (A1)

and height is $\frac{2}{\left(b-a\right)\left(c-a\right)}\left(m-a\right)$

area $=\frac{1}{2}\left(m-a\right)\times \frac{2}{\left(b-a\right)\left(c-a\right)}\left(m-a\right) \left(=\frac{{\left(m-a\right)}^2}{\left(b-a\right)\left(c-a\right)}\right)$         A1

 

OR

attempts to integrate the correct function          M1

$\underset{a}{\overset{m}{\int }}\frac{2}{\left(b-a\right)\left(c-a\right)}\left(x-a\right) dx$

$=\frac{2}{\left(b-a\right)\left(c-a\right)}{\left[\frac{1}{2}{\left(x-a\right)}^2\right]}_a^m$  OR  $\frac{2}{\left(b-a\right)\left(c-a\right)}{\left[\frac{x^2}{2}-ax\right]}_a^m$         A1A1

 

Note: Award A1 for correct integration and A1 for correct limits.

 

THEN

sets up (their) $\underset{a}{\overset{m}{\int }}\frac{2}{\left(b-a\right)\left(c-a\right)}\left(x-a\right) dx$ or area $=\frac{1}{2}$         M1

 

Note: Award M0A0A0M1A0A0 if candidates conclude that $m>c$ and set up their area or sum of integrals $=\frac{1}{2}$.

 

$\frac{{\left(m-a\right)}^2}{\left(b-a\right)\left(c-a\right)}=\frac{1}{2}$

$m=a\pm \sqrt{\frac{\left(b-a\right)\left(c-a\right)}{2}}$         (A1)

 

as $m>a$, rejects $m=a-\sqrt{\frac{\left(b-a\right)\left(c-a\right)}{2}}$

so $m=a+\sqrt{\frac{\left(b-a\right)\left(c-a\right)}{2}}$         A1

  

[6 marks]

[N/A]