Step 1: Understand the equation of the line.
The equation of the line is \( x + y - 1 = 0 \), which can be rewritten as:
\[
x + y = 1.
\]
This is a straight line with slope \( -1 \) passing through the point \( (1, 0) \) on the \( x \)-axis and the point \( (0, 1) \) on the \( y \)-axis. The points \( (3, 2) \) and \( (\cos \theta, \sin \theta) \) must lie on the same side of this line.
Step 2: Check the side of the line for point \( (3, 2) \).
To determine which side of the line the point \( (3, 2) \) lies on, substitute \( x = 3 \) and \( y = 2 \) into the equation of the line:
\[
3 + 2 = 5.
\]
Since \( 5 > 1 \), the point \( (3, 2) \) lies on the side of the line where \( x + y > 1 \).
Step 3: Check the side of the line for the point \( (\cos \theta, \sin \theta) \).
For the point \( (\cos \theta, \sin \theta) \), substitute \( x = \cos \theta \) and \( y = \sin \theta \) into the equation of the line:
\[
\cos \theta + \sin \theta.
\]
For the points to lie on the same side of the line, we need \( \cos \theta + \sin \theta > 1 \). Now, we will find the range of \( \theta \) for which this inequality holds.
Step 4: Solve the inequality.
To solve \( \cos \theta + \sin \theta > 1 \), square both sides:
\[
(\cos \theta + \sin \theta)^2 > 1^2,
\]
\[
\cos^2 \theta + 2\cos \theta \sin \theta + \sin^2 \theta > 1,
\]
\[
1 + 2\cos \theta \sin \theta > 1.
\]
This simplifies to:
\[
2\cos \theta \sin \theta > 0.
\]
Since \( \cos \theta \sin \theta = \frac{1}{2} \sin(2\theta) \), the inequality becomes:
\[
\sin(2\theta) > 0.
\]
The sine function is positive in the interval \( \left( 0, \frac{\pi}{2} \right) \), so the condition holds for \( \theta \in \left( 0, \frac{\pi}{2} \right) \).
Step 5: Conclusion.
Thus, the range of \( \theta \) is \( \left( 0, \frac{\pi}{2} \right) \), and the correct answer is (b).
Let the line $\frac{x}{4} + \frac{y}{2} = 1$ meet the x-axis and y-axis at A and B, respectively. M is the midpoint of side AB, and M' is the image of the point M across the line $x + y = 1$. Let the point P lie on the line $x + y = 1$ such that $\Delta ABP$ is an isosceles triangle with $AP = BP$. Then the distance between M' and P is: