| Date | May 2011 | Marks available | 5 | Reference code | 11M.1.hl.TZ2.3 |
| Level | HL only | Paper | 1 | Time zone | TZ2 |
| Command term | Find | Question number | 3 | Adapted from | N/A |
Question
The random variable X has probability density function f where
\[f(x) = \left\{ {\begin{array}{*{20}{c}}
{kx(x + 1)(2 - x),}&{0 \leqslant x \leqslant 2} \\
{0,}&{{\text{otherwise }}{\text{.}}}
\end{array}} \right.\]
Sketch the graph of the function. You are not required to find the coordinates of the maximum.
Find the value of k .
Markscheme
A1
Note: Award A1 for intercepts of 0 and 2 and a concave down curve in the given domain .
Note: Award A0 if the cubic graph is extended outside the domain [0, 2] .
[1 mark]
\(\int_0^2 {kx(x + 1)(2 - x){\text{d}}x = 1} \) (M1)
Note: The correct limits and =1 must be seen but may be seen later.
\(k\int_0^2 {( - {x^3} + {x^2} + 2x){\text{d}}x = 1} \) A1
\(k\left[ { - \frac{1}{4}{x^4} + \frac{1}{3}{x^3} + {x^2}} \right]_0^2 = 1\) M1
\(k\left( { - 4 + \frac{8}{3} + 4} \right) = 1\) (A1)
\(k = \frac{3}{8}\) A1
[5 marks]
Examiners report
Most candidates completed this question well. A number extended the graph beyond the given domain.
Most candidates completed this question well. A number extended the graph beyond the given domain.
