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
5: Solution by Dadeyemi
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:
1: Solution by Adje
2: Solution by SimonM
3: Solution by Adje
4: Solution by tommm
5: Solution by SCE1912
6: Solution by SimonM
7: Solution by Daniel Freedman
8: Solution by Adje
9: Solution by brianeverit
10: Solution by tommm
11: Solution by brianeverit
12: Solution by brianeverit
13: Solution by toastedlion
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
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 1
 17042009 21:43
Last edited by SimonM; 19052013 at 10:02. 
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 2
 17042009 22:40
Last edited by Adjective; 17042009 at 22:59. 
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 3
 17042009 23:06
STEP III, Question 14
Last edited by SimonM; 17042009 at 23:18. 
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 4
 17042009 23:19
Given that's what I've written down on my whiteboard, I'll assume that was a typo

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 5
 18042009 00:03
STEP I question 1
First part:
Spoiler:Show
Second part:
Last edited by Horizontal 8; 18042009 at 02:54. 
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 6
 18042009 00:09
III, 3
Denominator ≠ 0
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 (1x)⁵, so f(x) tends to infinity.
As x tends to minus infinity, so do (1+x)⁵ and (1x)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)
Last edited by Adjective; 18042009 at 01:21. 
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 7
 18042009 01:54
STEP I question 4
(Not 100% confident, since I always make mistakes with intervals)
Last edited by Horizontal 8; 18042009 at 02:58. 
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 8
 18042009 07:23

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 9
 18042009 13:33
III/8
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:
Last edited by Adjective; 18042009 at 22:25. 
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 10
 18042009 13:52
I would've never through of that
But does my solution check out?
PS: I think you missed post #5Last edited by Horizontal 8; 18042009 at 13:55. 
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 11
 18042009 14:30
STEP II Q3
First Part:
Spoiler:Show
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:
Last edited by Horizontal 8; 18042009 at 19:24. 
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 12
 18042009 15:31
I did III/Q4 yesterday, I'll type it up soon.

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 13
 18042009 17:03
Question 5, STEP I, 2003

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 14
 18042009 17:16
STEP I question 3
Let x= theta for notational purposes
(i)
(ii)
(iii)

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 15
 18042009 17:20
Question 7, STEP I, 2003
First PartSecond Partfor and
Now as 99a and c(c+1)(c1) are multiples of 3, therefore b(10b) 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)(c1) 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)(c1), with respect to our possible values of c:
We can also write down all the possible values of b(10b) with respect to our set of b:
b(10b) = {9, 21, 24}
Notice that, going back to the equation 99a = c(c+1)(c1)  b(10b), 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)(c1):
c(c+1)(c1) = 120. We see that b(10b) = 21 provides us with a solution, ie: c = 5, b = {7, 3} a = 1
c(c+1)(c1) = 210, and by checking the set of b(10b), subtracting them from 210, we do not get a multiple of 99, so there is no solution for c = 6
c(c+1)(c1) = 336, and once more no solution, with the possible values of b(10b), so no solution for c = 7
c(c+1)(c1) = 504, and by checking the set of b(10b), we see that b(10b) = 9 provides us with a solution, ie: c = 8, b = {1, 9}, a = 5
c(c+1)(c1) = 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. 
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 16
 18042009 17:26
Question 9, STEP I, 2003
First PartSecond PartLast edited by Unbounded; 18042009 at 17:28. 
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 17
 18042009 17:35
Question 8, STEP I, 2003
Spoiler:ShowLet 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 
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 18
 18042009 17:43
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 .
Last edited by qgujxj39; 18042009 at 17:46. 
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 19
 18042009 18:01
Question 10, STEP I, 2003
First PartLast edited by Unbounded; 04052009 at 23:32. 
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 20
 18042009 18:07
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 .
Last edited by qgujxj39; 04052009 at 12:44.
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