Answer
See the detailed answer below.
Work Step by Step
$$\color{blue}{\bf [a]}$$
We are told that the wave function must be continuous at $ x = 0 $. This means that the probability density function $ P(x)=|\psi(x)|^2 $ must also be continuous at $ x = 0 $.
For $ x = 0 $, the value of $ P(x) $ from the left side (where $ -1 \leq x < 0 $) must equal the value from the right side (where $ 0 \leq x \leq 1 $).
For $ -1 \leq x < 0 $, the probability density is given by:
$$ P(x) = \frac{a}{1 - x} $$
For $ 0 \leq x \leq 1 $, the probability density is given by:
$$ P(x) = b(1 - x) $$
At $ x = 0 $, we must have:
$$ \frac{a}{1 - 0} = b(1 - 0) $$
Thus,
$$ \boxed{a = b} $$
Therefore, $ a $ and $ b $ must be equal for the wave function to be continuous at $ x = 0 $.
$$\color{blue}{\bf [b]}$$
To determine the values of $ a $ and $ b $, we use the fact that the total probability of finding the particle must be 1.
$$ \int_{-1}^{1} P(x) \; dx = 1 $$
Thus, from the given formula,
$$ \int_{-1}^{0} \frac{a}{1 - x} \; dx + \int_{0}^{1} b(1 - x) \; dx = 1 $$
And since $ a = b $, we can simplify by setting both constants equal to $ a $.
$$ a \int_{-1}^{0} \frac{1}{1 - x} \; dx + a \int_{0}^{1} (1 - x) \; dx = 1 \tag 1$$
The first integral:
$$ \int_{-1}^{0} \frac{1}{1 - x} \; dx =\ln(1 - (-1)) - \ln(1 - 0) \\
= \ln(2) - \ln(1) = \ln(2) $$
The second integral:
$$ \int_{0}^{1} (1 - x) \; dx= \left[ x - \frac{x^2}{2} \right]_{0}^{1} \\
= 1 - \frac{1}{2} = \frac{1}{2} $$
Substituting the results into (1)
$$ a \ln(2) + a \cdot \frac{1}{2} = 1 $$
Therefore,
$$ a=b = \frac{1}{\ln(2) + \frac{1}{2}} \approx \color{red}{\bf 0.838} $$
$$\color{blue}{\bf [c]}$$
The probability density $ P(x) $ is given by:
$$ P(x) = \begin{cases}
\dfrac{0.838}{1 - x} & \text{for} \; -1 \leq x < 0 \\\\
0.838 (1 - x) & \text{for} \; 0 \leq x \leq 1
\end{cases} $$
$$\color{blue}{\bf [d]}$$
The probability of finding the particle to the left of the origin is the integral of $ P(x) $ from $ -1 \; \text{mm} $ to $ 0 \; \text{mm} $:
$$ P(-1\lt x < 0) = \int_{-1}^{0} \frac{a}{1 - x} \; dx $$
$$ P(-1\lt x < 0) =a\ln(1 - (-1)) - \ln(1 - 0) =a \ln(2) - a\ln(1) \\
=a\ln(2)= 0.838 \ln(2) $$
$$ P(-1\lt x < 0) =\bf{0.581}=\color{red}{\bf 58\% }$$
