Science:Math Exam Resources/Courses/MATH105/April 2011/Question 03/Solution 1

Notice from the definition of $f(x)$ that if $k<0$ then $f(x)$ is always 0, which cannot be a probability density function.

The same is true if $k=0$. Therefore we only need to consider positive values of $k$ as candidates.

We require that

$\int _{0}^{k/2}f(x)dx=1$

Thus,

$\int _{0}^{k/2}{\frac {2k}{(k+x)(k-x)}}dx=1.$

To deal with the integral, we use partial fractions. We write

${\frac {2k}{(k+x)(k-x)}}={\frac {A}{k+x}}+{\frac {B}{k-x}}$.

Now we multiply the equation through by $(k+x)(k-x)$ to find $2k=A(k-x)+B(k+x)$.

To find $A$ and $B$ we can choose ``convenient" values for $x$.

Setting $x=k$ yields $2k=0+B(k+k)$ so $B=1$.

Setting $x=-k$ yields $2k=A(k-(-k))+0$ so $A=1$.

Therefore,

${\begin{aligned}\int _{0}^{k/2}f(x)dx&=\int _{0}^{k/2}\left({\frac {1}{k+x}}+{\frac {1}{k-x}}\right)dx\\&=(\ln |k+x|-\ln |k-x|)|_{0}^{k/2}\\&=(\ln(3k/2)-\ln(k/2))-(\ln k-\ln k)\\&=\ln(3)+\ln(k)-\ln(2)-\ln(k)+\ln(2)\\&=\ln(3)\end{aligned}}$

Note this is independent of $k$ and it will never be $1$. Thus no such $k$ exist!