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

STEP I:
1: Solution by nuodai
2: Solution by Unbounded
3: Solution by Unbounded
4: Solution by Unbounded
5: Solution by Unbounded
6: Solution by Unbounded
7: Solution by Unbounded
8: Solution by Unbounded
9: Solution by nuodai
10: Solution by Sk1lz
11: Solution by nuodai
12: Solution by darkness9999
13: Solution by brianeverit
14: Solution Farhan.Hanif93

STEP II:
1: Solution by SimonM
2: Solution by Daniel Freedman
3: Solution by SimonM
4: Solution by Glutamic Acid
5: Solution by SimonM
6: Solution by Daniel Freedman
7: Solution by SimonM
8: Solution by sonofdot
9: Solution by Farhan.Hanif93
10: Solution by Glutamic Acid
11: Solution by Farhan.Hanif93
12: Solution by nuodai
13: Solution by brianeverit
14: Solution by brianeverit

STEP III:
1: Solution by DeanK22
3: Solution by SimonM
4: Solution by SimonM
5: Solution by Daniel Freedman
6: Solution by sonofdot
7: Solution by sonofdot
8: Solution by SimonM
9: Solution by sonofdot
10: Solution by brianeverit
11: Solution by brianeverit
12: Solution by sweeneyrod
13: Solution by brianeverit
14: Solution by sonofdot

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
3
13 years ago
#2
STEP I, Question 3, 2005
(i)

So there are two distinct real roots if, and only if, a and b are both positive or both negative. Q.E.D.
(ii)

There is exactly one real solution when the discriminant of this quadratic in x is equal to zero.

Noting that

But also that

Let us simply prove that c is non-zero:

Suppose that c = 0, then the equation for (ii) becomes exactly the same as (i). Hence we require a and b to both be positive or negative. Then the condition of exactly one root, which is no longer holds, since the LHS is zero whereas the RHS is negative - contradiction. Hence c is non-zero.

And finally, we have proven that . QED
2
13 years ago
#3
Question 4, STEP I, 2005
(a)

(b)

or

Given the restriction on theta, we need

4
13 years ago
#4
I'm not going to type it up, but STEP III Question 7 is one of my favourite STEP questions
0
13 years ago
#5
STEP III 2005 Question 7

This is one of my favourites too

We have

Making a substitution

(i)

Consider:

By above result:

(The integral is undefined when n=1)
(ii)

Consider

Making the substitution gives:

Hence, by the above result:

Special case: when n=2, the integral becomes:

1
13 years ago
#6
STEP II 2005, Question 6

Spoiler:
Show

i)

Let

Let

ii)

as required.

Let

Let

1
13 years ago
#7
(Original post by sonofdot)
.
What happens when n = 2 in part ii)?
0
13 years ago
#8
STEP II 2005, Question 2

Spoiler:
Show

a) i)

ii)

But where . Therefore

which is a product of integers and is therefore an integer.

b)

i) Statement is false

Take m = n = 2 but

ii) Statement is true

iii) Statement is false

Let p = 3, q = 4

f(3) = 2, f(4) = 2, f(12) = 4

c)

Therefore p = 11, m = 5.
0
13 years ago
#9
(Original post by Daniel Freedman)
What happens when n = 2 in part ii)?
Good point, edited
0
13 years ago
#10
STEP I: Question 12
Part 1
Spoiler:
Show

Group 1:

Part 2
Spoiler:
Show

Group 1:

Group 2:

Unparseable or potentially dangerous latex formula. Error 4: no dvi output from LaTeX. It is likely that your formula contains syntax errors or worse.
\implies \lambda p=\frac{2}{5}\times({\lambda p +(1-\lambda)q -\frac{1}{4}[\lambda p +(1-\lambda)q] + \frac{1}{4}[\lambda\times(1-p) + (1- \lambda)\times(1-q)])

\Rightarrow 10\lambda p=4\lambda p + 4(1-\lambda)q - \lambda p -(1-\lambda)q +\lambda(1-p) + (1-\lambda)(1-q)

\Rightarrow 10\lambda p= 4\lambda p + 4q - 4\lambda q - \lambda p - q + \lambda q + \lambda - \lambda p + 1 - q - \lambda - \lambda q

\Rightarrow 10\lambda p = 2\lambda p + 2q - 2\lambda q + 1

\Rightarrow \boxed{\lambda= \frac{4q + 2}{5+6p + 4q}}

1
13 years ago
#11
STEP III 2005 Question 14

that was a long one...

Spoiler:
Show
V has the density function for

Spoiler:
Show
Now we have

Making a substitution gives:

The median value of V, is such that . Clearly from above result:

(since all values of v are >0)
Spoiler:
Show
Spoiler:
Show

Making the substitution gives:

Hnece the density function of T, g(x), is given by:

Spoiler:
Show
Making a substitution in P(T<t) of gives:

Clearly, the median value of t, is such that:

(since all values of t are >0)

Product of median time and median speed:

Product of expected time and expected speed:

0
13 years ago
#12
Question 2, STEP I, 2005
First Part

At P, the gradient of the tangent is hence 1/p and similarly at Q, the gradient of the tangent is 1/q

The equation of the tangent at P is:

or

Similarly the equation of the tangent at Q is

Finding the point of intersection of these two tangents, to find the point R:

Therefore
Second Part
The line PQ will have a gradient of

So the equation of the line PQ is

As the point (1,0) is on this line, we can find a condition linking p and q.

We proceed to find the coordinates of the point S; the equation of the normal at P is:

or

Similarly the equation of the normal at Q is

Solving them to find S:

and dividing by q-p yields

And finding the y-coordinate

Therefore
Final Part
Neat proof by Hashashin:
Longer and messier proof by myself spoilerised
The gradient of PR is and the gradient of SQ is , and as it's clear that and therefore PR and SQ are parallel.

The same method works to show QR and PS are parallel, and as PR and PS are definition perpendicular, as are QR and QS, therefore PSQR is a rectangle.
Spoiler:
Show
We can show that the lines RP and PS are perpendicular:

To show that they are perpendicular, the product of their gradients must be -1:

By symmetry, the lines RQ and QS must be perpendicular, and by some simple geometry, we can see that QR and RP, and also QS and SP must also be perpendicular, hence we have a quadrilateral with four right-angles, and hence a rectangle. Q.E.D.
1
13 years ago
#13
Question 7, STEP I, 2005
(i)

After cancelling, we are left with

Note that
(ii)
Note that

And using the results we've just found

(iii)

Again, expanding this, and watching the terms drop away, we end up with

And this is justified as n is even, so all the denominators of each term in the product are not zero, as so is not a multiple of , meaning that

When I originally looked at this part of the question, I just thought WTF, but it turned out to be quite nice =)
0
13 years ago
#14
II/4:

1st part

Substituting a^2 = bc - 1, LHS = , as required. Note, since a, b and c are positive, 1/(a+b), 1/(a+c) and 1/a will all lie between 0 and +infinity, so "implies and is implied by" implication signs can be used.

2nd part

Let p + q = "a", s = "b", t = "c" , from (*)

Let p + r = "a", u = b, v = c , from (*)

Let p = "a", b = "q" and r = "c", so , from (*), and we're done.

3rd part

We can use the result in the second part by choosing p = 7, q =1, s = 5, t = 13, r = 50, u = 25 and v = 130, and everything falls out neatly.
3
13 years ago
#15
STEP III

Question 1

1st part

If cos(B) = Sin(A) this implies A = (4n+1)pi/2 (+/-) B

Let B = A + x for some x in R

It follows the problem is now Cos(A+x) = Sin(A) *

Epanding using the addition formulae we see that * is transformed to;

Cos(A)Cos(x) - Sin(A)Sin(x) = Sin(A)

<->
<->

Spoiler:
Show

It is relativel easy to show that if after searching for GA formula / derivation it has not been found, we use the identity sin(x) = cos(pi/2 - x) = cos(2(pi/4 - x/2)) and expand using the double angle formulae and the result follows fairly soon.

We also note that multiples of pi can be added to the tangent functions argument without altering its value to obtain;

<->

and the result

follows - although interestingly we do not have wip is obviously needed to obtain this.

2nd part

Rcos(x-a) = Rcos(a)cos(x) + Rsin(a)sin(x)

It follws that if

psin(x) + qcos(x) = Rcos(x-a) we have;

p = Rsin(a) and q = Rcos(a)

Adding p^2 + q^2 will reveal that this is equal to R^2 so ;

R^2 = p^2 + q^2

Divding reveals that (p/q) = tan(a)

Consider;

cos(x) + sin(x)

p=q=1 and it follows cos(x) + sin(x) = root(2)cos(x-(pi/4))

consider

sin(x) - cos(x) = root(2)cos(x+(pi/4))

We now note that |cos(x)| <= 1 and it follows that

Using these results we see that for sin(sin(x)) = cos(cos(x)) we must have

Noting that and that |(4n+1)pi/2| => pi/2 for n in Z we need to determine whether root(2) => pi/2 which even crude approimations will quickly reveal to not be the case so it follows that there is no value of x that cos(cos(x)) = sin(sin(x))

Cheating with the graphs.
2
13 years ago
#16
STEP III 2005 Question 6

Spoiler:
Show
We have the equation and have to show it is satisfied by . Substituting gives:

Spoiler:
Show
Now we have with

We can let for some a, since

We can also let for some T, since cosh T>1 so as required.

(if we say T>0)

From first part, one of the roots is

Spoiler:
Show
Let

By comparing coefficients, and

So other roots are satisfied by:

This gives roots and , where

Spoiler:
Show
We now have

From above, if we let b=2 and c=3, the roots are:

Therefore the roots are:

1
13 years ago
#17
III/2 - with the proviso that I think I may have missed something somewhere.

Solution of Differential Equation

Separating variables and integrating:

with

with

d/dx and d²/dx² of (x² + y²)

Graph
This is a picture of . In general, the graph crosses the y-axis at ± c/a.

Point of minimum distance to origin with 0 < c < a^2

The points of minimum distance to the origin will be when is a minimum. Which is a minimum when is a minimum. Which is a minimum when:

and

.

So... there is a stationary value of x² + y² when x = 0. When this is the case,

, which is greater than 0 since c < a² . Subbing x = 0 into yields .

So our minimum distances from the origin are at .

If c > a²

Looking at the other case in which

i.e.

Taking x² with positive c, when subbed into this gives:

which is greater than 0 since c > a². Taking x² with negative c leads to a minimum too, but also to an imaginary x, which we don't want to bother with.

When :

.

So our new minima are .
1
13 years ago
#18
Question 8, STEP I, 2005

(i)
Let the solution be of the form

We see, and are told, that k = 1

So the solution is:

Subbing y = 2, x = 1 gives C = 2

which is a pair of lines perpendicular to each other, which form a cross centred at (-1,0).
(ii)

Let the solution be of the form ie. k = -1

Subbing y = 1, x = 1 yields C = 1

0
13 years ago
#19
(Original post by GHOSH-5)
Question 2, STEP I, 2005

Final Part
We can show that the lines RP and PS are perpendicular:

To show that they are perpendicular, the product of their gradients must be -1:

By symmetry, the lines RQ and QS must be perpendicular, and by some simple geometry, we can see that QR and RP, and also QS and SP must also be perpendicular, hence we have a quadrilateral with four right-angles, and hence a rectangle. Q.E.D.
For this part, is it not far simpler to note that as the gradient of PR is and the gradient of SQ is , and as it's clear that and therefore PR and SQ are parallel?

The same method works to show QR and PS are parallel, and as PR and PS are definition perpendicular, as are QR and QS, therefore PSQR is a rectangle.

The saves a fair bit of algebra, and as far as I can see there's no flawed reasoning.
1
13 years ago
#20
(Original post by Hashshashin)
For this part, is it not far simpler to note that as the gradient of PR is and the gradient of SQ is , and as it's clear that and therefore PR and SQ are parallel?

The same method works to show QR and PS are parallel, and as PR and PS are definition perpendicular, as are QR and QS, therefore PSQR is a rectangle.

The saves a fair bit of algebra, and as far as I can see there's no flawed reasoning.
That is a lot neater. I'll edit it
0
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