Good morning Irah! Here are my workings for Q8. I will do Q7 in the afternoon. Let me know if you also need me to do Q9 and Q10.

Ah my year’s paper. There’s a lot of logical reasoning/visualising behind a lot of questions in 2015 as I recall. Part (I) is rather simple just draw the graph without modulus and then flip the negative Y area to the positive Y area. Part (II) theres 2 methods to solve can either use algebra and compare or the draw in the line y=3x-1 and then you can see that it cuts the “reflected line” but not the original line since the gradient diverges from y=2x-4 (3>2 so Line with M3 will not meet line with M2 where M is gradient) so can just equate y=3x-1 to y=-2x+4 which is the eqn of the “reflected line”. Part (III) is based a lot on visualisation and logical reasoning; you need to understand how you can get 2 intersections and how there’ll only be 1 intersection. Only if the gradient “converges” with y=2x-4 will you get 2 intersection which means you need a gradient less than 2 so that it meets the graph. However if you have a gradient of 0.5, you will see that the y=0.5x-1 only cuts (2,0) and nothing else which is why M must be bigger than 0.5 but smaller than 2. Remember to write in set notation as that’s what the question asks.

Good afternoon Irah! You can play around with these questions at the Desmos graphing calculator online by typing what I wrote there and you can play around with the values of m to see how the lines change, for you to have a better visualisation of what the other tutor meant.

Good afternoon Irah! Here are my explanations for Q10, but I am not sure if you have learnt it yet. If you have yet to learn this, you can save it for the time you learn this topic.

Good afternoon Irah! Here are my workings for Q9 and Q7a. I will look at the rest of Q7, as well as Q10, as a later time of say 2 pm.