Further Math A2s 4.1.1 Piecewise density

[Maximum number: 4]

Figure for Question CAIE A-Level Further Math A2 4.1.1 Piecewise Density Functions Questions question 1 — CAIE A-Level Further Math A2

As shown in the diagram, the continuous random variable X has probability density function f given by

f(x)= \begin{cases}m x & 0 \leqslant x \leqslant 2 \\ \frac{k}{x^{2}}+c & 2 \leqslant x \leqslant 6 \\ 0 & \text { otherwise }\end{cases}

where m, k and c are constants.

Given that $\mathrm{P}(X \leqslant 2)=\frac{1}{3}$ , show that $m=\frac{1}{6}$ and find the values of k and c.

$\begin{aligned} \mathrm{P}(X \leqslant 2)=\frac{1}{3}, \quad \text { so }\left[\frac{m x^{2}}{2}\right]_{0}^{2}=\frac{1}{3} m\left(\frac{4}{2}\right)=\frac{1}{3}, \quad m=\frac{1}{6} \end{aligned}$
AG
Reasoning required.

$\int_{2}^{6}\left(\frac{k}{x^{2}}+c\right) \mathrm{d} x=\frac{2}{3} \quad$ so $\quad\left[-\frac{k}{x}+c x\right]_{2}^{6}=4 c+\frac{1}{3} k=\frac{2}{3} \quad$ or $\quad k+12 c=2$
Attempt to integrate and correct use of correct limits to
form linear equation in k and c.

Also, $2 m=\frac{k}{4}+c$ or $\frac{1}{3}=\frac{k}{4}+c$ or 3 k+12 c=4
Matching at x=2.

Solve, $k=1, \quad c=\frac{1}{12}$
4

CAIE A-Level Further Math A2 4.1.1 Piecewise Density Functions Questions question 1