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# P+N 93-94, paper I

### #1

Posted 18 April 2005 - 08:30 PM

18) Find of the values of p and q if (x-1)^2 is a factor of x^4 + 2x^3 + px - 4x + q

i have subbed 1 into this equation and got that p+q=1

however i dont know how to get their individual values... if there is only one factor given??

also

19) Two curves, y=2cos2x and y=1-cosx where x is between 0 and 180 inclusive.

By solving an appropriate equation algebraically, find the x co-ordinate of the point A, correct to one decimal place.

i know that 2cos2x+cosx-1=0 (i think!), but i think where i keep making mistakes is with what trig formulae i use!

if anyone can help ill love u 4ever!xthanxx

### #2

Posted 18 April 2005 - 09:38 PM

if any1 can help b4 12 pm 2night that wud be smashing!

### #3

Posted 18 April 2005 - 09:45 PM

Remember, you tried x=1 or the factor (x-1).

The question stated that (x-1) was a factor. So that means that another (x-1) will be a factor of the quotient we got.

So do another synthetic division using the quotient and x=1 to get p+6=0, i.e. p=-6. Hence, q=7.

H tends 2 infinity

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Never argue with an idiot. They drag you down to their level then beat you with experience.

### #4

Posted 18 April 2005 - 10:04 PM

2cos2x = 1 - cosx

Then rearrange to equal zero:

2cos2x + cosx - 1 = 0

There are 3 choices to replace cos2x, for this question use cos2x = 2cos x - 1

This will give you a quadratic in terms of cosx which you can factorise and solve:

2(2cos x - 1) + cosx - 1 = 0

4cos x + cosx - 3 = 0

(4cosx - 3)(cosx + 1) = 0

Either 4cosx - 3 = 0, cosx = 3/4, x = 41.4, 318.6 (by CAST)

but discard 318.6 as > 180.

Or cosx + 1 = 0, cosx = -1, x = 270 (discard as > 180)

So solution is x = 41.4

H tends 2 infinity

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Never argue with an idiot. They drag you down to their level then beat you with experience.

### #5

Posted 18 April 2005 - 10:06 PM

H

H tends 2 infinity

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Never argue with an idiot. They drag you down to their level then beat you with experience.

### #6

Posted 18 April 2005 - 10:34 PM

### #7

Posted 19 April 2005 - 07:19 AM

I've sorted it now and replaced that solution above, which doesn't need Pascal's Triangle. I did think it was a bit odd at the time but it was the end of a long day!!

Let me know if you need more help with the new solution to Q18.

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Never argue with an idiot. They drag you down to their level then beat you with experience.

### #8

Posted 19 April 2005 - 06:36 PM

### #9

Posted 19 April 2005 - 08:27 PM

thanx a lot!

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