# STEP Maths I, II, III 1988 solutionsWatch

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10 years ago
#61
(Original post by kabbers)
III/1:

Sketch

I'm not going to bother posting most of the differentiating legwork, there arent too many tricks, its just an algebra bash.

Differentiating (quotient rule is your friend), we get

So we have turning points at x = 0, and x =

Differentiating again, we get

Substituting in values of x for the turning points, we find that x = 0 is a minimum, and the other two maximums.

Noticing the denominator of y, we find that there is an asymptote at x = -1.

And because of the exponential properties of e^-x, y tends to zero as

y tends to as since the numerator remains positive and given the exponenential properties of e^x much greater than the denominator, while the denominator becomes negative.

So the graph will look http://www.thestudentroom.co.uk/show...8&postcount=14

Prove

First note that

Hence we may split the integral into

Consider

Now consider

I posit the inequality for x > 0

So our inequality holds.

So:

Thus

And therefore,

Now notice that the graph of is greater than 0 for x > 0, and hence so is the infinite integral.

So,

I'm probably wrong but to show that the integral lies between 0 and 1 would it not be sufficient to say that

(x^2)(e^-x)/(x + 1) < (x^2)(e^-x)/x = x(e^-x) for 0 < x < infinity

then integrate x(e^-x), which equals 1. Seems a lot simpler?
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10 years ago
#62
Yeah, that's fine.

While we're nitpicking - taking the 2nd derivative to deduce the nature of the turning points is a lot more work than simply sketching x(2-x^2) to argue whether the derivative is increasing or decreasing.
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10 years ago
#63
1988 Paper 2 number 4
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10 years ago
#64
1988 Paper 2 number 8
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10 years ago
#65
1988 Paper 2 numbers 10,11 & 12
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10 years ago
#66
1988 Paper 2 numbers 13 & 14
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10 years ago
#67
1988 Paper 2 numbers 15 & 16
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#68
Question 8, STEP III

Spoiler:
Show

Therefore the equation of the gradient and normal respectively are:

and

For the triangle formed by the tangents, the vertices will be the intersection of the tangents. By symmetry we need only compute this for one vertex:

Therefore the area of the triangle of tangents is:

For the triangle formed by normals, the vertices will be the intersection of the normals. Again by symmetry we need only consider one pair:

This gives us an area

As required.

If they are congruent, the area of the triangle will be zero:
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9 years ago
#69
Can somebody please post the solution for STEP III Q3?
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8 years ago
#70
(Original post by SimonM)
Done
Am I missing something or is the solution, well, missing?
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8 years ago
#71
Where can i get STEP 2 solution 2
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8 years ago
#72
(Original post by SimonM)
....
STEP III Q10
Consider the positions of the dogs as time goes on. They will form a square of decreasing side length that rotates about the centre. If we take the position of one of the dogs to be from the centre, then we see by symmetry that the position of one the next dogs along is . We can also see that the tangent at any moment goes through the position of the next dog along, so:

let:

let be the angle between the x axis and the straight line to the point . Since and, by trigonometry, . So, substituting into the RHS and integrating we get:
.
By trigonometry, again, where r is the distance from the origin to the point
is constant so for some constant as required.
To generalise to n-dogs held at the vertices of an n-gon it must be noted that the next dog along will have the coordinates of the first dog bur rotated about the centre of the field i.e:

So, like the previous part,:

let:

let:

Integrating both sides we get:

And now, just like in the previous part:
for some constant . Letting n=4 we can check and see that it gives the same answer as the first part.
EDIT: Just realised that the final result simplified to the much nicer form that it is currently in.
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8 years ago
#73
Where can I find the step past papers before 1997? I have been looking for them for days, but all the links do not work now. Can you help me? thanks a lot!
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8 years ago
#74
(Original post by Dzwx777)
Where can I find the step past papers before 1997? I have been looking for them for days, but all the links do not work now. Can you help me? thanks a lot!
http://www.mathshelper.co.uk/oxb.htm
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7 years ago
#75
(Original post by insparato)
III/3

is the circle on the complex plane with radius 2 and centre (0,i)

so x^2 + (y - 1)^2 = 2

and therefore in paramatrised form

I've omitted sketches however I have described them in sufficient detail to make an easy sketch if one chooses.

(i)

Okay so its a circle of radius and the centre is (1,0) [/latex]

(ii)

If z = x + iy and is real then y must = 0

so

-u + vx - 1 must = 0 (this is important bit on)

so

Now from before

so

So in the W plane, when z is real, you get a circle centre(0,0) with radius 1

(iii)

if z = x + iy and z is imaginary then x = 0

Wahey! wtf ? I'm sure i got something different when jotting it down, something that made sense! hmm have i gone wrong somewhere? I think I'm doing the right method.
erm....I have something different....
1
7 years ago
#76
1988 STEP III number 3.
I'm afraid I disagree with the solutions published so far. Here is my solution.

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7 years ago
#77
1988 STEP I question 1 Alternative Solution

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7 years ago
#78
1988 STEP I question 2 Alternative solution

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7 years ago
#79
1988 STEP I question 7 Alternative solution

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7 years ago
#80
(Original post by nota bene)
I/9

i) Recall that , now we can use this to integrate by parts.

Putting in the limits gives So

ii) Let
So the integral becomes
Going back to x we have
Alternative solution for part (b)

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