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#1
You can let someone else take control of the OP whenever.

STEP I:
1: Solution by Horizontal 8
2: Solution by Horizontal 8
3: Solution by Horizontal 8
4: Solution by Horizontal 8
5: Solution by Unbounded
6: Solution by Unbounded
7: Solution by Unbounded
8: Solution by Unbounded
9: Solution by Unbounded
10: Solution by Unbounded
11: Solution by Unbounded
12: Solution by Unbounded
13: Solution by AnonyMatt and Jkn
14: Solution by nota bene

STEP II:
1: Solution by tommm
2: Solution by tommm
3: Solution by Horizontal 8
4: Solution by tommm
6: Solution by Horizontal 8
7: Solution by SimonM
8: Solution by tommm
9: Solution by Elongar
10: Solution by TwoTwoOne
11: Solution by Rocious
12: Solution by SimonM
13: Solution by SimonM
14: Solution by TwoTwoOne

STEP III:
2: Solution by SimonM
4: Solution by tommm
5: Solution by SCE1912
6: Solution by SimonM
7: Solution by Daniel Freedman
9: Solution by brianeverit
10: Solution by tommm
11: Solution by brianeverit
12: Solution by brianeverit
13: Solution by toasted-lion
14: Solution by SimonM

Solutions written by TSR members:
1987 - 1988 - 1989 - 1990 - 1991 - 1992 - 1993 - 1994 - 1995 - 1996 - 1997 - 1998 - 1999 - 2000 - 2001 - 2002 - 2003 - 2004 - 2005 - 2006 - 2007
6
13 years ago
#2
III/1
Differentiating arcsin

Differentiating arcosh

First integral

Second integral

4
#3
STEP III, Question 14

Spoiler:
Show

The probability generating function for one standard die is

The probability generating function for the sum of two will be the product of two (since they are independent) which gives us what we wanted

as required

For our tetrahedral die, we have

So we can have die
0
#4
Given that's what I've written down on my whiteboard, I'll assume that was a typo
0
13 years ago
#5
STEP I question 1

First part:
Spoiler:
Show

Let

Adding the second and third equations

Subbing back into other equations and

solving simultaneously and

so,

Second part:

Spoiler:
Show

Let

Solving simultaneously gives

Substituting a and c into other equations gives:

Subtracting

0
13 years ago
#6
III, 3
Denominator ≠ 0

Considering the expansion of , which is:

All odd powers cancel, so we're left with a load of positive terms. This is clearly not equal to zero (>2).

f'(x)

Expanding the numerator, terms with like expressions in powers of m and m-1 cancel, leaving:

Graphs

Turning points can only be at x = ±1 since the rest of f'(x) is never 0.

At x = 1, f(x) = 1
At x = -1, f(x) = -1

At 0, f(x) = 0

f'(x) is positive everywhere, so the turning points must be points of inflexion.

As x tends to infinity, so do(1+x)⁵ and -(1-x)⁵, so f(x) tends to infinity.

As x tends to minus infinity, so do (1+x)⁵ and -(1-x)⁵, so f(x) tends to minus infinity.

After sketching f(x), 1/f(x) can be sketched by inspection.

f(x) and 1/f(x)
0
13 years ago
#7
STEP I question 4

(Not 100% confident, since I always make mistakes with intervals)

Spoiler:
Show

let theta = x for notational purposes

Numerator changes sign when

for

or

Numerator changes sign when or

Denominator changes sign when or

Therefore we must consider certain intervals:

Let

Note:

Therefore when considering the interval

or or

3
#8
You could show

0
13 years ago
#9
III/8

Differentiation

General form of derivative

Solution of Differential Equation

Trying to solve this using dy/dx as above proved fruitless. I divided by -(n+1) to get the y in the denominator as it appears in the question:

So we have three equations to solve for three unknowns:

So:

Dividing these gives

Multiplying by (b+3)(a+3) and expanding:

Obviously a contender solution here is (from ab = 10), but this doesn't work when trying to find n in our original equations. So we must press on, and subtract 12 from each side.

I think it's safe to divide by (a - b) here, since we've discounted a = b as a solution.

Taking gives so a general solution to the differential equation is:

Taking gives so another solution is:

3
13 years ago
#10
(Original post by SimonM)
You could show

I would've never through of that
But does my solution check out?

PS: I think you missed post #5
0
13 years ago
#11
STEP II Q3

First Part:
Spoiler:
Show

Let x be an irrational number.

Suppose (in hope of a contradiction) that where p,q are integers

But x is irrational so this is a contradiction therefore our original assumption was false. Hence the cube root of x is irrational if x is irrational.

Second part:

Spoiler:
Show

Suppose statement is true for n=k.

By the inductive hypothesis is irrational and it follow from our result in the first part of the question that the cube root of this must be irrational therefore it must be true for n=k+1.

The case where n=1 is true (from the question) therefore the result is true for all positive integral value of n.

Last part:

Spoiler:
Show

Define a sequence of which all the terms are clearly irrational since U_n is irrational for all n and a rational number + irrational number is always irrational.

As

As n increases without bound.

Therefore the sequence converges to m

4
13 years ago
#12
I did III/Q4 yesterday, I'll type it up soon.
0
13 years ago
#13
Question 5, STEP I, 2003
(i)

and we require

therefore, the term independent of x is given by:

(ii)

therefore the term independent of x is

and as the powers of x are increasing, beginning at , there will be no term independent of x.
4
13 years ago
#14
STEP I question 3

Let x= theta for notational purposes
(i)
Spoiler:
Show

or

Now:

Where n is an integer.

But

Therefore

(ii)

Spoiler:
Show

Let

Where n is an integer.

(iii)

Spoiler:
Show

But
Hence no solutions arise from this.

Also,

So we use to find the other angles.

Cosx is -ve for when considering 0<x<2pi

Therefore the angles that we want are:

And this is equal to

As required.

1
13 years ago
#15
Question 7, STEP I, 2003
First Part
with equality if k = 5.

for k being an integer between 0 and 9 inclusive

therefore

is merely a product of three consecutive integers, and we know that exactly one integer will be a multiple of 3, so we know that this expression will have a factor of 3, and so is divisible by 3.
Second Part
for and

Now as 99a and c(c+1)(c-1) are multiples of 3, therefore b(10-b) must also be a multiple of 3, so we require b = {1, 3, 4, 6, 7, 9}.
We also can see than c(c+1)(c-1) must be greater than or equal to 99, which implies that must be greater than 99.
This implies that c = {5, 6, 7, 8, 9}.

By rewriting the equation:

We can write down all possible values of the LHS, as we just look for the multiples of 99, with respect to the range of a:
99a = {99, 198, 297, 396, 495, 594, 693, 792, 891}

We can also write down all possible values of c(c+1)(c-1), with respect to our possible values of c:

We can also write down all the possible values of b(10-b) with respect to our set of b:
b(10-b) = {9, 21, 24}

Notice that, going back to the equation 99a = c(c+1)(c-1) - b(10-b), the maximum of the RHS is 720 - 9 = 711, so we can reject the two higher values of 99a, ie: 99a = {99, 198, 297, 396, 495, 594, 693}

Now by looking at cases of c(c+1)(c-1):

c(c+1)(c-1) = 120. We see that b(10-b) = 21 provides us with a solution, ie: c = 5, b = {7, 3} a = 1

c(c+1)(c-1) = 210, and by checking the set of b(10-b), subtracting them from 210, we do not get a multiple of 99, so there is no solution for c = 6

c(c+1)(c-1) = 336, and once more no solution, with the possible values of b(10-b), so no solution for c = 7

c(c+1)(c-1) = 504, and by checking the set of b(10-b), we see that b(10-b) = 9 provides us with a solution, ie: c = 8, b = {1, 9}, a = 5

c(c+1)(c-1) = 720, and once more, no solution, so no solution for c = 9.

Therefore our solutions are 135, 175, 518, 598.

I have a feeling I've made a mistake somewhere, and I've missed a solution.
4
13 years ago
#16
Question 9, STEP I, 2003
First Part
Considering the situation vertically, and using , and denoting t to be the time when the particle reaches P.

we find that:

And considering it horizontally:

substituting this into the first equation gives:

Second Part
Considering as a quadratic in T, there exist two distinct solutions (and correspondingly two distinct angles) if and only if the discriminant is greater than zero.

And as

Then

Third Part
Looking back at the quadratic in T:

Let it factorise into two linear factors:

ie. and

By expanding the brackets, we see that and also that

Now looking at the equation

which is necessarily true.

So the statement is true.
1
13 years ago
#17
Question 8, STEP I, 2003
Spoiler:
Show
Let the B = yV. the rate at which B increases with respect to time is:

From the question, we see that:

where C is some constant

where

When A completely turns into B, y = 1.

and we have a contradiction and can conclude that A never completely turns into B.

I have a very strong feeling I've missed something, in the question
1
13 years ago
#18
STEP III 2003, Q4

Spoiler:
Show

, therefore

Using , we obtain the equation of the tangent:

This meets the curve again when . Substituting this in and rearranging to find a quadratic in , we obtain:

By the factor theorem, we find that is a root. We can factorise and divide by (because ) to obtain

.

Using the quadratic formula, we obtain

One of these gives , which we can disregard.

Therefore, as required.

The double angle formula for tangent is:

Therefore,

So, if , then

Therefore, if , then

We can also apply the formula similarly to the cotangent function, so

Using the identities and , we get

as required.

Another value of which satisfies the required properties is .

1
13 years ago
#19
Question 10, STEP I, 2003
First Part
Because the lamina is uniform, we can model masses by areas.

The area of the rectangle ABCD is pq. It's centre of mass is at the point

The ara of the triangle ABX is qr/2 and it's centre of mass is

Therefore we have:

Equating the coordinates, we get:

And also:

Second Part

If , then r would be greater than q, which causes X not to lie in the closed interval [B,C], so the area cut away from the lamina would not be a triangle, however we are told it is a triangle, therefore r < q.

1
13 years ago
#20
STEP III 2003 Q10

Spoiler:
Show

n.b. I don't know how to LaTex the dots denoting differentiation w.r.t to time, so I'm just using x for displacement, v for speed and a for acceleration.

Integrating gives

and using the initial conditions gives

Separating variables gives

Using the initial conditions gives

Therefore, upon rearranging a bit, we get

As , the argument of tangent and hence as required.

In this second case, our first integration combined with the initial conditions gives

Now, for simplicity, let .

Separating variables we obtain

which becomes

Rearranging a lot, we get the rather nasty answer:

Unparseable or potentially dangerous latex formula. Error 4: no dvi output from LaTeX. It is likely that your formula contains syntax errors or worse.
x = \sqrt{d^2 - \frac{2U}{k}} \displaystyle\frac{1 + \frac{d - \sqrt{d^2 - \frac{2U}{k}}}}{d + \sqrt{d^2 - \frac{2U}{k}}} e^{\sqrt{d^2 - \frac{2U}{k}}kt}{1 - \frac{d - \sqrt{d^2 - \frac{2U}{k}}}{d + \sqrt{d^2 - \frac{2U}{k}}} e^{\sqrt{d^2 - \frac{2U}{k}}kt}

As , the fraction

Therefore .

1
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