# STEP 2003 Solutions Thread

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Thread starter 12 years ago
#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
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 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
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Thread starter 12 years ago
#3
STEP III, Question 14

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Thread starter 12 years ago
#4
Given that's what I've written down on my whiteboard, I'll assume that was a typo
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12 years ago
#6
III, 3

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)
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12 years ago
#7
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Thread starter 12 years ago
#8
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12 years ago
#9
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12 years ago
#10
I would've never through of that But does my solution check out?

PS: I think you missed post #5 0
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12 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:

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12 years ago
#12
I did III/Q4 yesterday, I'll type it up soon.
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12 years ago
#13
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12 years ago
#14
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12 years ago
#15
Question 7, STEP I, 2003
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.
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12 years ago
#16
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12 years ago
#17
Question 8, STEP I, 2003
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12 years ago
#18
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12 years ago
#19
Question 10, STEP I, 2003
First Part
Second Part
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12 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 .

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