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# old a level questions watch

1. came across some old alevel papers (80s AEB ) i found my answers to most but a few im missing any one offer solutions to save an old person some brain ache ?

1 with O as origin the fixed point F is (a,b) a,b>0. the variable point P is (p,0) where p>a. the line PF when produced meets the y axis at the point Q.
Prove the area of OPQ is bp^2/(2p-2a)
as p varies find the least area of OPQ in terms of a and b

2 prove for all m, Y=mx-2m^2 is a tangent to 8y=x^2. find the value of m for which the line Y=mx-2m^2 is also a tangent y^2=x

3 the rhombus ABCD of side 17 is the horizontal base of a pyramid VABCD. the vertex V is vertically above the point M where the diagonals AC and BD intersect. GIVen that VA=VC=17 and that VB=VD=sqrt(128). find AC and BD and verify AC:BD=15:8. find the cosine of the angle between the planes VBA and ABCD.

4 a soild right circular cone of greatest volume is to be cut from a solid sphere of radius 3R. using calculus, show that the height of this cone is 4R and calculate its volume in treams of R.
5 the points p(acos r,bsin r) and q( -asin r, bcosr) lie on the ellipse x^2/a^2+y^2/b^2=1 and O is the origin. prove thats as r varies the values of Op^2+Oq^2 and the area Opq remain constant and state their values in terms of a and b.The tangents to the ellipse at p and q meet at T. prove that as r varies T lies on the ellipse x^2/a^2+y^2/b^2=2. prove OpTq is a parallelogram and state its area.
2. Hell. Our exams have got so much easier. I'd have loved to have done A levels back then.

1)

Point F(a,b) and P(p,0).
The line FP has equation (x-a)/(p-a) = (y-b)/(0-b)

Point Q lies on line FP and on y axis so has x coordinate of 0.
Plugging that into FP equation:
-a/(p-a) = (y-b)/-b
ab/(p-a) + b = y
[ab + b(p-a)]/(p-a) = y
bp/(p-a) = y

Area of triangle OPQ = base*height/2
=0.5 * p * bp/(p-a)
=bp^2/(2p-2a)

Next we differentiate the area A of the triangle with respect to p. It's safe to cancel b/2 as we'll be putting it equal to 0 in a bit where it would be cancelled out anyway.

d[p^2/(p-a)]/dp = 2p/(p-a) - p^2/(p-a)^2 (by the product rule)
= [2p(p-a) - p^2]/(p-a)^2
max or min area when dA/dp = 0 which can only happen when:
2p(p-a) - p^2 = 0
p^2 - 2ap = 0
p(p-2a) = 0
p = 0 which is not valid as p>a>0
p = 2a
So min area = b(2a)^2/[2(2a) - 2a]
= 4ba^2/2a
= 2ab
3. 4)

I started out by calculating its volume in terms of R because then the calculus part is easy.

The things that affect volume are the height and the radius. We want to link them using R as the variable. As these can be placed in a 2D plane we only need to consider them inside a circle. It's easier if we think of the equation of a circle with radius 3 and centre (3,0). That's (x-3)^2 + y^2 = 9. We want to find out how the positive height of the circle changes with x.
y^2 = 9 - (x-3)^2
y = sqrt[9 - (x-3)^2] (that's the positive square root)
let y=r=radius of cone and x=R=height of the cone
r = sqrt[9 - (R-3)^2]
Volume of cone = 1/3*pi*r^2*h
= 1/3*pi*[9-(R-3)^2]R
= (2pi)R^2 - (pi/3)R^3 (after rearranging)

Now to find the max volume.
dV/dR = (4pi)R - piR^2
max/min when:
(4pi)R - piR^2 = 0
4R - R^2 = 0
R(4-R) = 0
R=0 (min volume)
R=4 (max volume).
This corresponds to a height of 4R.

Note that I don't like my answer to this one bit. Hoping somebody can sort out the variables to make it a bit easier to follow.
4. For 2)

Substitute y = mx -2m^2 into 8y = x^2

8mx - 16m^2 = x^2
x^2 - 8mx + 16m^2 = 0
Now you have a quadratic eqn in x,
For it to be tangent, b^2 - 4ac should be 0, and sure enough, it is 64m^2 - 64m^2 = 0

For second bit, substitute y=mx - 2m^2 into y^2 = x
(mx - 2m^2)^2 = x
m^2x^2 -x(4m^3 +1) +4m^4 = 0
Again, b^2 - 4ac - 0
(4m^3+1)^2 - 4*m^2*4m^4 = 0
16m^6 +8m^3 +1 - 16m^6 = 0
8m^3 = -1
m^3 = -1/8
m = -1/2

(Edited due to a minus sign error - props to SsEe)
5. Jesus christ they did get easier
6. sephonline.

Nice answer. Better than my pooey effort involving differentiating, getting an equation of a tangent at any point and comparing variables.
One tiny thing. Double check the end:

16m^6 +8m^3 +1 - 16m^6 = 0
8m^3 = 1
m^3 = 1/8
m = 1/2

There's a mistake.
7. (3) Take A as the origin, the angle BAD as 2a, the height of V above ABCD as h. Remember rhombus has side 17. If we take the x-axis along AB then B =(17,0,0)

Then C = (17 + 17 cos(2a), 17 sin(2a),0) = 34 cosa (cosa,sina,0)
and M = 17 cosa (cosa,sina,0) and V = (17(cosa)^2, 17 sina cosa,h)

As VA = 17 and VA^2 = MA^2 + VM^2 then

17^2 = (17 cosa)^2 + h^2

So h = 17 sina.

As VB^2 = 128 and

the vector BV = (-17+17(cosa)^2,17 sina cos a, h)
= 34sina(-sina,cosa,0) + (0,0,h)

then

128 = (34 sin a)^2 + h^2 = 2h^2

giving h = 8.

Then sina = 8/17 and cosa = 15/17.

So the length AC = 34 cosa = 30
and the length BD = 34 sina = 16

So AC:BD = 30:16 = 15:8

To find the angle project M= (225/17, 120/17) down to AB to the point

N=(17(cosa)^2,0,0) = (225/17,0,0)

We now have a right-angled triangle NMV and we want the angle VNM. Note

tan(VNM) = MV/MN = 8/(120/17) = 17/15.
8. (Original post by fishpaste)
Jesus christ they did get easier
Note though that although the questions are harder (probably due to the fact that they are not split up into smaller chunks), fewer people used to get the top grades in maths!
9. thanx for the replies to the questions,all numerical answers agree with the answers in the book of papers.when i done alevel in 1989 i used 1970s papers because they were harder and more challenging than the 1980s papers, this business of exams getting easier has always happened,also AEB was a tough exam board.just in case you are curious here are some of the paper 2 (out of 2 papers) questions.had 3 hours to do 7 of these type of questions out of 10

1) prove the number of combinations of n different things taken r at a time is nCr where nCr= n!/(n-r)!r!
show n+1Cr=nCr+nC(r-1) hence prove by induction
(1+x)^n=sum (r=0 to n) (nCr)x^r.
by considering the coefficient of x^n in the expansion of (1+x)^n(1+x)^n and (1+x)^(2n) prove that
sum (r=0 to n) (nCr)^2=(2n)!/((n!)^2)

2 show sum (r=1 to n) 1/r(r+1) =n/(n+1). use the series expansion of ln(1-x) to prove that sum(r=1 to n) 1/r((1/2)^r) tends to ln2 as n tends to infinity.hence by expressing 1/r(r+1) in partial fractions find the value of
sum(r=1 to n) 1/r(r+1)((1/2)^r) as n tends to infinty.

3)given that z=cosx+isinx prove z^n+z^-n=2cosnx.show that i+1/i=0 and hence deduce if z is a cube root of i z^3+z^-3=0.prove further if cosx+isinx is a cube root of i then 3x=2kpi+pi/2 for integral k.find the cube roots of i in the form a+ib.
express cosx in the form cos^8x(1-p(tan^2)x+qtan^4 x -ptan^6 x+tan^8 x)
for integers p q to be found.hence prove
tan^8 pi/16+ 70tan^4pi/16+1=28tan^2 pi/16(1+tan^4 pi/16).

4 2 lines are parallel.on the first line there are 5 dots and on the second 4.how many possible triangles can be formed by joining 3 of these dots? if 3 different dots are chosen at random from the 9 what is the probability they form a triangle.if the 3 dots chosen do not form a triangle what is the probability that all 3 lie on the first line?
one dot is chosen at random on the first line and labelled A. a second point B is also chosen at random from that line.Adot c is chosen at random from the second line and a second,different,dot d is chosen at random from that line. in how many ways can the dots a,b,c,d be chosen? what is the probability that the lines ac and bd will meet at a point between the 2 parallel lines
10. I suck at combinatorics and stuff. I need practice.

4)

For this I'm assuming we're allowed to join two dots on the same line that have dots between them. If we can only join two dots that are next to each other then it's only a few simple changes.

To make a triangle we need 1 point on one line and 2 on the other.
So the number of triangles formed from 1 point from the 5 point line and 2 points from the 4 point line = 5 * 4C2 = 5*6 = 30
Number of triangles formed from 1 point from the 4 point line and 2 from the 5 point line = 4 * 5C2 = 4*10 = 40
Total possible triangles = 30 + 40 = 70

9C3 = 84 possibilities for 3 points chosen at random. 70 of which form a triangle. So there's a 70/84 = 5/6 probability of making a triangle.

84-70 = 14 possible ways of not making a triangle. 5C3 = 10 of these will be on the 5 point line. So given it's not a triangle there's a 10/14 = 5/7 probability they lie on the first line.

There are 5*4 * 4*3 = 20 * 12 = 240 ways of placing ABC and D.

There are 20 ways of placing points A and B. 10 of these will have them in the order AB and the other 10 they'll be in the order BA.
There are 12 ways of placing C and D. 6 ways CD and 6 DC.
We want either the order AB DC or the order BA CD for the lines to cross in the middle.
There are 10*6 = 60 ways of having AB DC and 10*6 = 60 ways of having BA CD.
That's a total of 60+60 = 120 ways of having the lines cross.
120/240 = 1/2. So probability the lines cross is 1/2.

Somebody please correct the mistakes that are probably there.
11. SsEe your answers are the ones given in the answer booklet.good work. i hate prob questions even now i suck at them.the prob question i would always leave when doing these papers! as for your answer to the other question 4 while the 4R is correct the volume in terms of R is given as 32piR^3/3
12. Thanks. Normally I hate probability questions. Our class has only been simple rules but I've looked up lots of stuff on combinations and permutations in an old T1 book as they're very handy tools. But I'm still not fully comfortable using them.

Ahh d'oh. Didn't do that properly at all. Here's another attempt:

Let the height of the cone = aR

Picture the circle again. Radius 3, centre (3,0).
Equation (a-3)^2 + r^2 = 9 where r is the radius of the cone.
r^2 = 9 - (a-3)^2
= 9 - (a^2-6a+9)
= 6a - a^2

Volume of cone = 1/3 * pi * radius^2 * height
= (pi/3)(6a - a^2)a
= (2pi)a^2 - (pi/3)a^3

dV/da = (4pi)a - (pi)a^2 = 0 at max/min
a(4-a) = 0
a=0 (min)
a=4 (max)
height = aR = 4R

And my brain must be completely fried. I can't find a nice mathematically sound way of approaching the final part. I have the max volume in terms of R written down but it's messy. Any ideas?
13. (Original post by fishpaste)
Jesus christ they did get easier
LOL. And AEB was suposed to be the easiest board, you were put in for AEB papers if they didn't think you'd pass on O+C or JMB
14. (Original post by SsEe)
4)

I started out by calculating its volume in terms of R because then the calculus part is easy.

The things that affect volume are the height and the radius. We want to link them using R as the variable. As these can be placed in a 2D plane we only need to consider them inside a circle. It's easier if we think of the equation of a circle with radius 3 and centre (3,0). That's (x-3)^2 + y^2 = 9. We want to find out how the positive height of the circle changes with x.
y^2 = 9 - (x-3)^2
y = sqrt[9 - (x-3)^2] (that's the positive square root)
let y=r=radius of cone and x=R=height of the cone
r = sqrt[9 - (R-3)^2]
Volume of cone = 1/3*pi*r^2*h
= 1/3*pi*[9-(R-3)^2]R
= (2pi)R^2 - (pi/3)R^3 (after rearranging)

Now to find the max volume.
dV/dR = (4pi)R - piR^2
max/min when:
(4pi)R - piR^2 = 0
4R - R^2 = 0
R(4-R) = 0
R=0 (min volume)
R=4 (max volume).
This corresponds to a height of 4R.

Note that I don't like my answer to this one bit. Hoping somebody can sort out the variables to make it a bit easier to follow.

The problem can be solved using the single variable x, which represents the discrepancy between the radius of the sphere and the height of the cone. The working is then considerably reduced.
15. eqn of sphere z^2+x^2+y^2=9R^2 if centred at the origin.
h=6R-k for some k.
volume of cone V=pi/3r^2h
but r^2=x^2+y^2=9R^2-z^2=9R^2-(3R-k)^2=k(6R-k)
so V=(pi/3)k(6R-k)(6R-k)
dV/dk=(pi/3){-2k(6R-k)+(6R-k)^2}=(pi/3)(6R-k){-2k+6R-k}
so dv/dk=o when k=6R or k=2R so have max h=6R-2R=4R as required
at h=4R
V=(pi/3)(2R)(4R)^2=(32pi/3)R^3

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