I'm a bit confused about how to solve part b. I've done the "solve directly" method shown on the markscheme but I don't understand how they've gone from line 1 to line 2 (where has the modulus sign gone to?) I usually use the method where you square both sides to remove the modulus and it didn't work. I don't know whether it's supposed to work and I made a mistake or it shouldn't work for this question. Please could someone explain this to me.
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C3 modulus help. watch
- Thread Starter
- 21-01-2017 21:51
- 21-01-2017 22:02
From the graph, they chose instead of because it's the left branch of the function, which has a negative gradient (of -1). Given that the intersection point lies on the right branch of (with a positive gradient), they replaced it with .
Squaring both sides would have worked. Expanding , you should get , meaning you'll have to verify at the end if in .
- 21-01-2017 22:11
Squaring only works if everything is in the modulus sign, so if you had mod(x)=mod(x-2+1), then squaring would work. Edit: didn't know squaring would work if the modulus sign was retained, interesting.
You'd have to do it by considering both moduli to be positive, and then one to be negative. But, as there's only one intersection you only get one solution (1.5), which you get from the left branch, as MartyO says, that would have gone down below the y axis without the modulus