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# Further Mats Help watch

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1. Please could someone help me with this P4 question, I think I did the first bit and ended up with x is more than 1, which doesn't seem to work out for part (b). Please could someone help? Thanks
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2. I get for part 1, x > 1 or x < -1/2

So for part 2, sin t is x, and so you have 2 inequalities;
sin t > 1
sin t < -1/2

The first is impossible, so persue the second.
3. Thanks a lot . I must have forgotten the minus half part, thanks for pointing it out, thanks .

Would people mind if I used this thread to post any further questions, as I'm working through a few papers at the moment?

And, would people mind if I used their answers along with mine (once I get the papers marked of course) for a website which could have markschemes for all the papers Edexcel are too stingy to provide (I can scan the papers too, at a much higher quality also).
4. I did that and got -1/2 < x, x > 1 for part a, how can you call that P4
5. (Original post by mik1a)
I did that and got -1/2 < x, x > 1 for part a, how can you call that P4
how can u judge p4 just by doing that qu,i'll give u another 1 p4 qu
6. I'm having a couple of troubles on the M3 also, if anyone would be kind enough to lend a hand, I'd me mucho gratefuls . Thanks in advance .
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7. And this one.

It was P4, I know that they're easy, inequalities, I just made a lousy error.
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8. second part I/m not sure about trig inequalities but I had a go and got either

pi/6 < t < pi/2
-5pi/6 < t < -pi/2
9. M3 one

I don't understand "with a force 5(AB) Newtons"

What does this mean with AB in brackets? is it notation or 5*AB N?
10. (Original post by mik1a)
M3 one

I don't understand "with a force 5(AB) Newtons"

What does this mean with AB in brackets? is it notation or 5*AB N?
AB is the distance of B from A
11. (a) 2x > (3x+1)/(x+1)

2x(x+1) > 3x+1 , x>-1 (assumption)
2x^2 - x - 1 > 0
(2x+1)(x-1) > 0

2x+1 > 0 and x-1 < 0 (1)
or,
2x+1 < 0 and x-1 > 0 (2)

(1) gives
x>-1/2 and x<1 which satisfies the condition (x < -1)

(2) gives

So, solution is
x>-1/2 and x<1
i.e.
-1/2 < x < 1
========

(b) the above solution range satisfies -pi < x < pi
so from part (a) we have
-1/2 < sint < 1
giving
-pi/6 < t < pi/2
==========
12. (Original post by Fermat)
(a) 2x > (3x+1)/(x+1)

2x(x+1) > 3x+1 , x>-1 (assumption)
2x^2 - x - 1 > 0
(2x+1)(x-1) > 0

2x+1 > 0 and x-1 < 0 (1)
or,
2x+1 < 0 and x-1 > 0 (2)

(1) gives
x>-1/2 and x<1 which satisfies the condition (x < -1)

(2) gives

So, solution is
x>-1/2 and x<1
i.e.
-1/2 < x < 1
========

(b) the above solution range satisfies -pi < x < pi
so from part (a) we have
-1/2 < sint < 1
giving
-pi/6 < t < pi/2
==========
i disagree because in that interval the rectangular hyperbolic curve is greater than y=2x and besides since x+1 can be negative or positive the jump from 2x>(3x+1)/(x+1) to 2x(x+1)>3x+1 is flawed and only gives u 1 solution
13. hope dis helps the confusion!,best thing to do with these qu is to draw the graphs
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14. I think IntegralAnomaly is right, multiplying both sides by the RHS denominator does remove a solution.
15. yea so by lookin at the graph the line is 'higher' than the curve in the intervals x>1 and -1<x<-1/2.
and since it has been said by the james than sint cannot be >1,
so -1<sint<-1/2.
16. for a full solution, you have to solve 2 quadratic inequalities:

2x(x+1) > 3x+1 for x>-1

2x(x+1) < 3x+1 for x<-1
17. (Original post by IntegralAnomaly)
i disagree because in that interval the rectangular hyperbolic curve is greater than y=2x and besides since x+1 can be negative or positive the jump from 2x>(3x+1)/(x+1) to 2x(x+1)>3x+1 is flawed and only gives u 1 solution
You're right.
And there's another error in my post
I guess I've been away from this stuff for too long!
18. (Original post by Fermat)
You're right.
And there's another error in my post
I guess I've been away from this stuff for too long!
aahh every1 makes mistakes....
19. Thanks guys you've really helped
20. Thanks guys, you've really helped .

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