To do part (iii), first you will need to be able to draw the modulus graph in part (i).

If my explanations are too long for you, you can jump to the summarised results at the last part of this text.

The way to approach this question is by graphical sketching and imagination.

I have drawn several lines, in varying colours, for lines which pass through the point (0, -1).

I will now discuss ALL the possible cases for the varying number of intersections here. The solution to this question is in CASE 8.

The point (0, -1) itself lies below the x-axis, while the modulus graph does not appear anywhere below the x-axis and has a vertex to the right of the x-axis at (2, 0). As a result, any downsloping line will never cut the modulus graph at exactly two points.

CASE 1

If the downsloping line is steeper than the downsloping part of the modulus graph (with gradient -2), the line will cut the modulus graph at exactly 1 point, as demonstrated in the steeper light blue line.

Therefore, for m < -2 (i.e. steeper line), there will be 1 point of intersection

CASE 2

If the downsloping line is parallel to, or is less steep than the downsloping part of the modulus graph, the line and the modulus graph will never intersect, as demonstrated by the less steep light blue line.

Therefore, for -2 <= m < 0, there will not be any intersection.

CASE 3

A horizontal line passing through (0, -1) will also not cut the modulus graph.

Therefore, for m = 0, there will also not be any intersection,

Now let's consider the upsloping lines.

THE REMAINING CASES

The first line to consider is the line connecting (0, -1) and the vertex (2, 0) of the modulus graph as demonstrated by the green line. Do note that the upsloping part of the modulus graph has a gradient of 2.

CASE 4

This line has a gradient of 0.5 and cuts the modulus graph only at 1 point, the vertex itself. Beyond this point (2, 0), there will not be any more intersection since the line and the upsloping part of the graph has different gradients

Therefore, for m = 0.5, there will be one point of intersection.

CASE 5

Any upsloping line gentler than a gradient of 0.5 will not produce any intersection.

Therefore, for 0 < m < 0.5, there will not be any intersection.

THREE MORE CASES TO GO

A line steeper than gradient 0.5 should cut the curve at two points, but how far does it go?

CASE 6

The second line to consider is a line of gradient 2, parallel to the upsloping part of the modulus graph with gradient 2 as demonstrated by the red line. After the first intersection with the downsloping part of the graph, there will not be any more intersection since the line and the upsloping part of the graph is now parallel to each other.

Therefore, for m = 2, there will be 1 point of intersection.

CASE 7

Any upsloping line steeper than gradient 2 will also cut the modulus graph at one point, since this line is steeper than the more slowly rising upsloping part of the modulus graph.

Therefore, for m > 2, there will be 1 point of intersection.

CASE 8

This leaves us with the case 0.5 < m < 2 which will produce 2 points of intersection as demonstrated by the black line.

SUMMARY

For m < -2, there will be 1 point of intersection.
For -2 <= m < 0.5, there will be no intersection.
For m = 0.5, there will be 1 point of intersection.
For 0.5 < m < 2, there will be 2 points of intersection.
For m >= 2, there will be 1 point of intersection.