# A2 Maths modulus question

#1
Hi,
In the question attached, for part b, why must I reject the x=-37 solution?
Thank you
Last edited by Rhys_M; 4 weeks ago
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4 weeks ago
#2
Its not a solution?
0
#3
Forget I even posted that :/ Thanks.

But also for part c I know that a>2 for y=ax to intersect the positive section of the graph, but how do I arrive at the other condition of a<=1.25.
Thank you
0
4 weeks ago
#4
Can you not just sketch it in the original figure?
You know the line passes through the origin. So what range of gains are acceptable and how does the line intersect with the plotted mod function.
0
#5
(Original post by mqb2766)
Can you not just sketch it in the original figure?
You know the line passes through the origin. So what range of gains are acceptable and how does the line intersect with the plotted mod function.
Well I assumed that for the negative part of the mod function,
Y=2(-x-4)-5
Y=-2x-13, so the gradient is -2 hence a must be less than -2.
I don't know where to get -1.25
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4 weeks ago
#6
Literally get your pen or ruler and rotate it on the origin. When does it (and what is the condtion) intersect at least once with the mod function and what is the corresponding (range of) gradient?

Think about the sign of the gradient. Tbh, if youre making typo errors like this, its always worth sketching/verifying the values youre getting out of these type of mod questions.
Last edited by mqb2766; 4 weeks ago
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