STEP 2003 Solutions Thread

OP

18

STEP III, Question 14

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Reply 3

OP

18

Given that's what I've written down on my whiteboard, I'll assume that was a typo

Reply 4

10

STEP I question 1

First part:

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Second part:

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Reply 5

Adjective

12

III, 3

0 Denominator ≠

f'(x)

Graphs

step.jpg8.9KB

steps.jpg10.1KB

Reply 6

10

STEP I question 4

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

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Reply 7

OP

18

You could show

$\displaystyle \frac{1+\sin x}{\cos x} = \cot \left ( \frac{\pi}{4} - \frac{x}{2} \right ) = \tan \left ( \frac{x}{2} - \frac{\pi}{4} \right )$

Reply 8

Adjective

12

III/8

Differentiation

General form of derivative

Solution of Differential Equation

Reply 9

10

SimonM

You could show

$\displaystyle \frac{1+\sin x}{\cos x} = \cot \left ( \frac{\pi}{4} - \frac{x}{2} \right ) = \tan \left ( \frac{x}{2} - \frac{\pi}{4} \right )$

I would've never through of that :o:

But does my solution check out?

PS: I think you missed post #5 :smile:

Reply 10

10

STEP II Q3

First Part:

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Second part:

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Last part:

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Reply 11

17

I did III/Q4 yesterday, I'll type it up soon.

Reply 12

10

STEP I question 3

Let x= theta for notational purposes
(i)

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(ii)

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(iii)

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Reply 13

17

STEP III 2003, Q4

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Reply 14

17

STEP III 2003 Q10

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Reply 15

OP

18

STEP III, Question 2

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Reply 16

OP

18

STEP III, Question 6

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Reply 17

DFranklin

18

8 Horizontal

STEP II Q3

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I grant you it's not 100% clear, but I would interpret this part as "given an integer m, give an example of a sequence of irrational numbers that converges to m". (In other words, you have to produce a sequence that works for every m).

Reply 18

10

DFranklin

I grant you it's not 100% clear, but I would interpret this part as "given an integer m, give an example of a sequence of irrational numbers that converges to m". (In other words, you have to produce a sequence that works for every m).

Define a sequence $A_n = m+U_n-1$ 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. Yet it converges to m as $n \to \infty$

Is this what you mean?

Reply 19