# STEP Maths I, II, III 1988 solutionsWatch

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11 years ago
#21
If you look at the two posts it took me 9 minutes therefore I win .

I suspect i just picked the two easiest questions out of the two papers. I feel guilty for doings maths .
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11 years ago
#22
ok doing II/1if no one else is
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11 years ago
#23
I've had a look at it, on first glance it seems like not a bad question! You go for it though.
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11 years ago
#24
cheers yep, a very straightforward one (all the more embarrasing any mistakes will be )

ok firstly, work out the sum to infinity (of a geometric sequence)

therefore, iff |r| < 1, as

Also note that

therefore, iff |r| < 1, as

part 1,

|x| < 1, so by (*),

as required.

part 2,

since 1<|x|<2, both 1/x and x/2 fall within the applicable range (|r|<1), so by (**) & (*)

(I just copied the second part from above)

as required.

part 3,

for |x| > 2, is ok, but is not.

so, we need something that will give us 1/(x-2)

I guessed , it seemed to fit the way the question progresssed

by (**),

hooray it works

Therefore,

(as above)

So

is the desired expansion
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11 years ago
#25
Right so I've been working on II/7 but I'm a little bit confused.

It looks to me as if there have been a few substitutions going on here but the questions has been using the same variables across.

For example the first part is fairly easy, transforming the integral I into the first equation is easily done by the substitution x=1+y, which gives the same answer but in terms of y, however the question gives it in terms of x, the original variable which makes me think I've perhaps missed something and I'm not supposed to use a substitution after all?
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11 years ago
#26
STEP III - Question 7

When n=0: .

We have:

(3).

And so, subsituting (1) and (3) into (*):

.

So for (**) to be zero for all x;

(4) and (5) .

(5) .

Hence is a postive constant, . (since ).

So we now have: . And since , as .

Also, .

Now, as x , . Hence, since , as .

For n=1,

as , so , as .

Also, .
Both terms tend to zero and x tends to , and so as .

So the conditions hold for n=1 and n=0, and , where is a positive constant.

Now, Which is the same for .

Hence: .

If , then .

And:

=

Now, we are given that as , and .

Hence: as

and

as .

and so:

, since omega, m and n are just constants.
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11 years ago
#27
(Original post by insparato)
If you look at the two posts it took me 9 minutes therefore I win .

I suspect i just picked the two easiest questions out of the two papers. I feel guilty for doings maths .
#zomg#
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11 years ago
#28
(Original post by Square)
Right so I've been working on II/7 but I'm a little bit confused.

It looks to me as if there have been a few substitutions going on here but the questions has been using the same variables across.

For example the first part is fairly easy, transforming the integral I into the first equation is easily done by the substitution x=1+y, which gives the same answer but in terms of y, however the question gives it in terms of x, the original variable which makes me think I've perhaps missed something and I'm not supposed to use a substitution after all?
You can just rename y as x, because it doesnt really matter what the dummy variable is called since the integral is going to eventually be evaluated at 0 and 1.
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11 years ago
#29
(Original post by squeezebox)
#zomg#
I don't do a maths related degree far from it, medicine actually. So when I do a sneaky STEP question or two i feel bad because i really should be doing something more humanie.
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11 years ago
#30
(Original post by squeezebox)
You can just rename y as x, because it doesnt really matter what the dummy variable is called since the integral is going to eventually be evaluated at 0 and 1.
Good, just confused me slightly.
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11 years ago
#31
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.
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11 years ago
#32
II/7 - under construction! - sorry my latex skills aren't very good.

First part:

substitution:

For the parts afterwards I get slightly confused, I've noticed that a substitution of y=tanx gets me close to the answer. However I only noticed that by exanding out the denominator of the second integral. Some advice on this would help.

iii) Take the third integral from the second part of the question:

Substitute:

Last part, first subtitute in y=x+1 to get one of the integrals, then into that substitute y=x^2 and then equate and cancel. I'll write up shortly.
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11 years ago
#33
STEP I question 5.

I'm sure there must be a more elegant way of doing this, but here's my 2AM attempt:

Let

(comparing constant coefficients)

(comparing x coefficients)

Let

WWWWW

Valid for
2
11 years ago
#34
I/7.

Therefore

If |f(x)| <= 1, then -1 <= f(x) <= 1.
If |f'(x)| <= 4, then -4 <= f(x) <= 4.

f'(x) = f(1)(x + 1/2) + f(-1)(x - 1/2) - 2f(0)x.

The largest values of f'(x) will be when x = -1 or 1 to maximize the values of (x + 1/2) and -(x-1/2). The largest value of f(1)(x + 1/2) can be 3/2 when f(1) = 1. The largest value of f(-1)(x - 1/2) can be 1/2 when f(-1) = 1. The largest value of -2f(0)x can be 2 when f(0) = -1. 3/2 + 1/2 + 2 = 4, so |f'(x)| <= is satisfied. When x = -1, the largest value can be (-1)(-1/2) + (-1)(-3/2) - (2)(1)(-1) = 4.

The smallest values of f'(x) will be when x = -1 or 1 likewise. When x = 1, be (-1)(3/2) + (-1)(1/2) - 2(1) with f(0) = 1, f(1) = -1 and f(-1) = -1. This gives -4. When x = -1, it will be (1)(-1/2) + (1)(-3/2) - 2(1) = -4, with f(1) = 1, f(-1) = 1 and f(0) = -1, giving -4. Therefore |f'(x) <= 4 is still satisfied.

Letting f(0) = -1, this gives c = -1. Letting f(1) = 1, this gives a + b - 1 = 1 hence a + b = 2. Letting f(-1) = 1, this gives a - b = 2. Hence 2a = 4, a = 2 and b = 0.
f(x) = 2x^2 - 1.

Sorry for any errors.
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11 years ago
#35
relatively straightforward q, pointing out any mistakes will be appreciated as always

Ok first note that at the end of day n, the firm will owe everything it owed the previous day, the interest on this plus everything it has borrowed that day.

This may be expressed as

By evaluating the first few terms of this sequence, it becomes easy to 'guess' and then prove a total value for Sn:

ok my guess is that (by geometric series)

this fits the questions, but we have to prove it (induction!)

Basis case: , therefore the basis case works.

Inductive step:

This is in the same form as the above, so it works.

for part 2, firstly we can assert that (this is the initial amount owed)

the firm must pay interest on what it owes then pay back what it has earned.

Again, let's try a few values:

hypothesis:

(again by geometric)

we need to prove this, again by induction.

basis case: when m = 0, basis works

inductive step:

Hence the sum works. However the question did not ask for us to express S_T, so we must substitute in (see our above def. for S_T)

for part 3, note that if the sequence is decreasing, the firm will pay off its debt. so the conditions are:

Letting x = k+1

taking everything on to one side

k+1 > 0, so both x^m and (x-1) are greater than zero. Therefore, for the inequality to be satisfied,

as required.
1
11 years ago
#36
Doing II/5 and 4 - but if anyone else beats me to it, feel free .
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11 years ago
#37
ok here's my attempt at III/6, i'm not completely sure if it's adequate for STEP.. any tips appreciated.

part 1:

rewrite f(x) as an addition of sines, using the identity

note that ,
,
,
,

and hence
<=> ,

also note that ,
,
,
,

and hence
<=> ,

Therefore, when m < 4,

1988/4 = 497; therefore n = 497, m = 0

so

part 2: firstly note that if y is small, (this we can tell from the maclaurin expansion)
also note that if y is small,

so, subbing in :

using the above approximations,

is small enough to ignore. subbing in the other values,

To prove that is the smallest value > 0: there are no roots of 3^1988 sin(3x) between 0 and pi/3, this is because the lowest x ( > 0) at which sin(x) = 0 is pi. Thus our root is going to be very near pi/3 (we can neglect the sin x over here as it is tiny compared to the sin 3x), in fact slightly greater to compensate for sin(x) being equal to sqrt(3)/2.

latex edits.
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11 years ago
#38
Has anyone else done III/2? I think I have a solution, but since I've never solved a difference equation or worked with them like this before today, I dunno If its correct or not..

EDIT: I'll post my solution when I have time.
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11 years ago
#39
II/3.

- Both k and 1/k will satisfy the equation.
.
So 1/c = c hence c^2 = 1 and c = +-1.
And b/c = b so b = cb. When c = 1, b can be any real value. When c = -1, b = 0.

- Since we know c = 1, the quadratic will be of the form , as k^2 - 1 = 0 does not satisfy the conditions (because -1 is a root, but 1--1 = 2 isn't a root).
Also will satisfy it. This gives

Comparing coefficients, -2-b = b ==> b = -1 This also satisfies 2 - 1 =1. So the quadratic is .

- Call the roots of the equation .

Therefore
.

Multiplying through last equation by alpha^3 and dividing by r, you get:
.

Comparing coefficients, 1/r = 1 so r = 1.
q/r = p p/r = q so p = q.

Replacing q by p and expanding the second equation we get.

Comparing coefficients, -3-p = p ==> 2p = -3 and p = -3/2. Since q = p, q = -3/2.
So the cubic equation is .
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11 years ago
#40
UNDER CONSTRUCTION

STEP II - Question 5

Let z = (Where for convenience, and .

Then:

Also, So equating the real and imaginary parts:

and .

. Where t = .

So if , then either , or .

If , then the values of theta for which (*) = 0 are given by , provided , where n is an integer.

these willl give distinct values of .

Hence the roots of (*) are : .

Product of the roots = -7. Hence: .

Now,

So:

. (Chosen positive root since there are an even number of negative tans.)

Repeat the whole process for n=9, and we end up with:

.

And for n=11, we end up with:

(negative root since odd number of negative tans).

Phew, my hands hurt.
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