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Core 1 Inequality questions

It's question 8) c) iii)
Someone please help im confused the ms is not helpful either

QUESTION


http://filestore.aqa.org.uk/subjects/AQA-MPC1-W-QP-JAN06.PDF


MARKSCHEME


http://filestore.aqa.org.uk/subjects/AQA-MPC1-W-MS-JAN06.PDF
Original post by akereem100
It's question 8) c) iii)
Someone please help im confused the ms is not helpful either

QUESTION


http://filestore.aqa.org.uk/subjects/AQA-MPC1-W-QP-JAN06.PDF


MARKSCHEME


http://filestore.aqa.org.uk/subjects/AQA-MPC1-W-MS-JAN06.PDF


What have you tried? :smile:
Reply 2
Avatar for akereem100
akereem100
OP
10
Original post by Slowbro93
What have you tried? :smile:


I was trying to replace the x^2 - 2x in the dy/dx
In 8) c) i), you worked out dy/dx to be 6x-3x^2. This is the gradient of the graph, and y will be decreasing if it is negative, so you need to simplify:

6x-3x^2<0
take out a factor of 3 from the LHS
3(2x-x^2)<0
divide both sides by 3
2x-x^2<0
If you divide both sides by -1, then the inequality sign changes.
x^2-2x>0

Hope this helps
Original post by akereem100
I was trying to replace the x^2 - 2x in the dy/dx


What did you get when you differentiated your function? And what does dy/dx give you?

From this? How can you get the inequality as desired?

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