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## Past Paper Problems Problem questions from PP's.

### sparky

Posted 30 April 2004 - 08:09 PM

Hmmm...it appears to be one of those horrible vector ones...

I know you go a.(b+c)
= a.b + a.c
but after that I am lost. Edited by sparky, 30 April 2004 - 08:10 PM.

### Steve

Posted 01 May 2004 - 04:04 PM

Here goes,

a is the line from P to Q and b+c is the line joining Q and S.

The angle at R is 120Â° (since the shape is a regular hexagon - look here for more details).

Looking at triangle QRS, the angles at Q and S are both 30Â° (angles in a triangle add up to 180).

So the angle between PQ and QS is 90Â° (120-30).

You can then use the geometric form of the scalar product formula (given). You don't need to know the magnitude of b+c since the angle between the vectors is 90Â° and cos90Â°=0.

Therefore a.(b+c)=0

### Ally

Posted 01 May 2004 - 09:59 PM

Thanks for that!

I'll print off your working & get my past paper book, because right now I'm so confused.

Are we meant to know about all those shapes in the link?
It's got me worried now Thanks again for the solution.

### sparky

Posted 02 May 2004 - 12:27 PM

Three vectors, p, q and r are such that p.(q+r) = q.(p+r).

Show that r must be perpendicular to p-q

3 marks

Any help would be appreciated.

### George

Posted 02 May 2004 - 01:33 PM

OK, expanding p.(q+r) = q.(p+r) gives

p.q + p.r = q.p + q.r

And since p.q = q.p they can be cancelled out giving

p.r = q.r
p.r - q.r = 0
r.(p - q) = 0

So r is perpendicular to p - q as required.

### sparky

Posted 02 May 2004 - 02:31 PM

Thanks, I did expand, but never thought about cancelling!! ### hssc11045

Posted 04 May 2004 - 05:09 PM

Hey......am new to these forums and was jus wonderin if any1 can provide some help with questions:

2c) ii)
6)
11b)

of the 2000 calculator paper.

Thanks

P.S a knw the working has already been posted 4 question 6 but a cnt open tha link ### Steve

Posted 04 May 2004 - 08:23 PM

I've fixed the link to the working for Question 6 from 2000 - here is a link to it.

### Ally

Posted 04 May 2004 - 08:42 PM

 QUOTE (hssc11045 @ May 4 2004, 05:09 PM) Hey......am new to these forums and was jus wonderin if any1 can provide some help with questions:2c) ii)

Well for:

2 © (iii) - I was stuck on this question aswell but i thought it through and I think this is the answer :

I think that you put y=9 (answer to previous question) into the equation x+2y=9 giving x+18=9 which gives x=-9 so the co-ordinates of T are (-9,9).

Also for this question, I think that you can do it the long way by finding the tangent at P and equating this to y=9 which also gives (-9,9), but this is probably not the right way to do it as its only worth 2 marks. Maybe someone could clear this up?

### Dave

Posted 04 May 2004 - 08:58 PM

i tried doin it the wrong way and it didn't work out i got (-13,9) which is wel lout i must have made a hell of a mistake somewhere.

Question 11)

You should ave worked the equation to be P = 0.6Q + 1.8

if you sub the logs into this you get:

log base e(p) = 0.6 log base e (q) + 1.8 log base e (e)

one of the log rules states log xy = log x + log y so:

ln (p) = ln (e^1.8) ln (q^0.6)

if you calculate this out you get

ln(p) = 6.05 ln q^0.6)

remove the ln and you get

P = 6.05 q ^ 0.6

a = 6.05
b = 0.6

### hssc11045

Posted 04 May 2004 - 09:02 PM

thanks guys.....all makes sense nw ### Ally

Posted 04 May 2004 - 09:06 PM

 QUOTE (Dave @ May 4 2004, 08:58 PM) i tried doin it the wrong way and it didn't work out i got (-13,9) which is wel lout i must have made a hell of a mistake somewhere.

Did you get the tangent at P to be 3y+4x=-9 using the gradient -4/3? This gives x to be -9.

### Dave

Posted 05 May 2004 - 06:34 PM

Not likily i can't find where i did the working but it doesn't look familer

### sparky

Posted 05 May 2004 - 07:58 PM

Find algebraically the exact value of
sinÃ¸ + sin(Ã¸ +120) +cos(Ã¸+150)

I know you multiply these out using the double angle rules, and then substitute in the values of the related angles, but I still dont get this is cancel down to O which is the correct answer!

Any help appreciated

### \$impl¥_®i¢h

Posted 05 May 2004 - 09:08 PM

Is this question from one of the pastpapers?

### AndyW

Posted 05 May 2004 - 11:06 PM

Using the shorthand c = cos theta and s = sin theta, it expands to:

s + s cos 120 + c sin 120 + c cos 150 - s sin 150

= s + s ( - 0.5 ) + c ( root3over2) + c ( - root3over2) - s ( 0.5)

= s + s ( - 0.5 ) - s ( 0.5)

= s - 0.5 s - 0.5 s

= 0

Quite tricky that one.

### hssc11045

Posted 06 May 2004 - 05:37 PM

alrite everyone its me again with a couple more questions:

5a from tha 2001 non cal..........i narrowed it down 2 tha expanision or tha wave function but i canot seem to get any to work

7b) ii) get all tha previous questions in 7 correct but cannot seem to do this one 8) the answer I get is 80 whereas tha answer at tha bk says 81

FINALLY

question 11c)

Much appreciated

Thanks

### Allan

Posted 06 May 2004 - 06:02 PM

 QUOTE (hssc11045 @ May 6 2004, 06:37 PM) 5a from tha 2001 non cal..........i narrowed it down 2 tha expanision or tha wave function but i canot seem to get any to work

You can't use the wave function here as you have an "x" and a "2x" value.

So you need to use the double angle forumla as given in the formula sheet.

Subsituting you get:

2sinxcosx - cosx = 0

Taking out the common factor cos x:

cos x(2sinx - 1)=0

Set both parts to 0:

cos x = 0 2sinx - 1 = 0

Then you can solve it from there like a normal trig equation to get x = 30,90,150

### Allan

Posted 06 May 2004 - 06:07 PM

 QUOTE (hssc11045 @ May 6 2004, 06:37 PM) 7b) ii) get all tha previous questions in 7 correct but cannot seem to do this one g(h(x)) = cosxcos(pi/4) - sinxsin(pi/4)
= 1/√2 cosx - 1/√2 sinx

Substitue these into the formula given:

(1/√2sinx + 1/√2 cosx) - (1/√2 cosx - 1/√2sinx)

Simplifying this gives:

2/√2 sin x = 1
sin x = √2/2
inverse sin (√2/2) = (pi/4)

so x = (pi/4) and (3pi/4) [1st and 2nd quadrants]

### Allan

Posted 06 May 2004 - 06:12 PM

 QUOTE (hssc11045 @ May 6 2004, 06:37 PM) 8) the answer I get is 80 whereas tha answer at tha bk says 81

4 log [x] 6 - 2log [x]4 = 1

Using log rules:

log [x] 6[power 4] - log [x] 4Â²

log [x] 1296 - log [x] 16 = 1

Using more log rules:

log [x] 1296/16 = 1

log [x] 81 = 1

More log rules

81 = x[power 1]

x = 81

[NB The log rules should be in your textbook, you've just got to apply them here]

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