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STEP 2003 Solutions Thread

(edited 11 years ago)

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Reply 1

III/1

Differentiating arcsin



Differentiating arcosh



First integral



Second integral

Reply 2

STEP III, Question 14

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Reply 3

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

Reply 4

STEP I question 1

First part:

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Second part:

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Reply 5

III, 3

0 Denominator ≠



f'(x)



Graphs

Reply 6

STEP I question 4

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

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Reply 7

You could show

1+sinxcosx=cot(π4x2)=tan(x2π4)\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

III/8

Differentiation



General form of derivative



Solution of Differential Equation

Reply 9

SimonM
You could show

1+sinxcosx=cot(π4x2)=tan(x2π4)\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

STEP II Q3

First Part:

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Second part:

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Last part:

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Reply 11

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

Reply 12

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

STEP III 2003, Q4

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Reply 14

STEP III 2003 Q10

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Reply 15

STEP III, Question 2

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Reply 16

STEP III, Question 6

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Reply 17

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

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 An=m+Un1 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 n \to \infty

Is this what you mean?

Reply 19

STEP II 2003 Q1

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