(a)     Let $a > 0$ . Draw the graph of $y = \left| {x - \frac{a}{2}} \right|$ for $- a \leqslant x \leqslant a$ on the grid below.

(b)     Find k such that $\int_{ - a}^0 {\left| {x - \frac{a}{2}} \right|{\text{d}}x = k\int_0^a {\left| {x - \frac{a}{2}} \right|{\text{d}}x} }$ .

(a)

    A1A1

Note: Award A1 for the correct x-intercept,

A1 for completely correct graph.

(b)     METHOD 1

the area under the graph of $y = \left| {x - \frac{a}{2}} \right|$ for $- a \leqslant x \leqslant a$ , can be divided into ten congruent triangles;     M1A1

the area of eight of these triangles is given by $\int_{ - a}^0 {\left| {x - \frac{a}{2}} \right|{\text{d}}x}$ and the areas of the other two by $\int_0^a {\left| {x - \frac{a}{2}} \right|{\text{d}}x}$     M1A1

so, $\int_{ - a}^0 {\left| {x - \frac{a}{2}} \right|{\text{d}}x}  = 4\int_0^a {\left| {x - \frac{a}{2}} \right|{\text{d}}x}  \Rightarrow k = 4$     A1     N0

METHOD 2

use area of trapezium to calculate     M1

$\int_{ - a}^0 {\left| {x - \frac{a}{2}} \right|{\text{d}}x}  = a \times \frac{1}{2}\left( {\frac{{3a}}{2} + \frac{a}{2}} \right) = {a^2}$     A1

and area of two triangles to obtain     M1

$\int_0^a {\left| {x - \frac{a}{2}} \right|{\text{d}}x}  = 2 \times \frac{1}{2}{\left( {\frac{a}{2}} \right)^2} = \frac{{{a^2}}}{4}$     A1

so, k = 4     A1     N0

METHOD 3

use integration to find the area under the curve

$\int_{ - a}^0 {\left| {x - \frac{a}{2}} \right|{\text{d}}x}  = \int_{ - a}^0 { - x + \frac{a}{2}{\text{d}}x}$     M1

$= \left[ { - \frac{{{x^2}}}{2} + \frac{a}{2}x} \right]_{ - a}^0 = \frac{{{a^2}}}{2} + \frac{{{a^2}}}{2} = {a^2}$     A1

and

$\int_0^a {\left| {x - \frac{a}{2}} \right|{\text{d}}x} = \int_0^{\frac{a}{2}} { - x + \frac{a}{2}{\text{d}}x + \int_{\frac{a}{2}}^a {x - \frac{a}{2}{\text{d}}x} }$     M1

$= \left[ { - \frac{{{x^2}}}{2} + \frac{a}{2}x} \right]_0^{\frac{a}{2}} + \left[ {\frac{{{x^2}}}{2} - \frac{a}{2}x} \right]_{\frac{a}{2}}^a = \frac{{{a^2}}}{8} + \frac{{{a^2}}}{4} + \frac{{{a^2}}}{2} - \frac{{{a^2}}}{2} - \frac{{{a^2}}}{8} + \frac{{{a^2}}}{4} = \frac{{{a^2}}}{4}$     A1

so, k = 4     A1     N0

[7 marks]

Most candidates attempted this question but very often produced sketches lacking labels on axes and intercepts or ignored the domain of the function. For part (b) many candidates attempted to use integration to find the areas but seldom considered the absolute value. A small number of candidates used geometrical methods to determine the areas, showing good understanding of the problem.