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STEP Prep Thread 2017

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Original post by Zacken
You can go if you like, but in my experience, there's nothing that's going to help with STEP other than sitting down on your own and working through questions.


Noted. In my current experience, this seems to be the case, too. But I am going to test whether or not it would be beneficial for me. And then see if I should continue. The previous cohort did very, very well. So the chances are high. However, I place emphasis on the fact that they worked hard to get those results themselves, so it's mostly just about guidance from the teachers.
Original post by physicsmaths
Gna midnap him n take him to homerton


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lol lim jeck is a joke fam
Original post by Injective
lol lim jeck is a joke fam


I wonder what Lim Jeck thinks of all these jokes about him lol (if he's seeing them).
Original post by IrrationalRoot
I wonder what Lim Jeck thinks of all these jokes about him lol (if he's seeing them).


Lol I doubt he even visits TSR. I spend more time on AoPS than this site, so wouldn't be surprised.
Original post by IrrationalRoot
I wonder what Lim Jeck thinks of all these jokes about him lol (if he's seeing them).


Looked him up, he's just spent two years in national service... what a waste
Reply 265
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Maths.
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Hi guys,

Thanks for your help with the second question of the year 2000 STEP 2 paper. Attached is my solution to the first question, any thoughts on my answer would be appreciated.

Also enjoy your weekend :smile:
Reply 266
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Maths.
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Original post by Maths.
Hi guys,

Thanks for your help with the second question of the year 2000 STEP 2 paper. Attached is my solution to the first question, any thoughts on my answer would be appreciated.

Also enjoy your weekend :smile:


Here is the attachment......
STEP 2 Paper 2000 Q1.pdf421.0KB
Reply 267
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Zacken
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Original post by IrrationalRoot
I wonder what Lim Jeck thinks of all these jokes about him lol (if he's seeing them).


Nah he doesn't see them thankfully :lol:


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Reply 268
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Tau
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This isn't from a STEP paper (AFAIK) but can someone check my thinking for this question please and tell me any nicer ways to do this question.

The question is: ∫0π2log⁡sin⁡θ dθ \int^\frac{\pi}{2}_0 \log \sin \theta \ \mathrm{d}\theta

Using the substitution θ↦π2−θ \theta \mapsto \frac{\pi}{2} - \theta transforms the integral into ∫0π2log⁡(sin⁡(π2−θ)) dθ=∫0π2log⁡cos⁡θ dθ \int^\frac{\pi}{2}_0 \log (\sin (\frac{\pi}{2} - \theta) ) \ \mathrm{d}\theta = \int^\frac{\pi}{2}_0 \log \cos \theta \ \mathrm{d}\theta

If we call the integral I I , then 2I=∫0π2log⁡sin⁡θ dθ+∫0π2log⁡cos⁡θ dθ=∫0π2log⁡(sin⁡θcos⁡θ) dθ 2I = \int^\frac{\pi}{2}_0 \log \sin \theta \ \mathrm{d}\theta + \int^\frac{\pi}{2}_0 \log \cos \theta \ \mathrm{d}\theta = \int^\frac{\pi}{2}_0 \log( \sin \theta \cos \theta ) \ \mathrm{d}\theta

So 2I=∫0π2log⁡(12sin⁡2θ)dθ=∫0π2log⁡(12) dθ+∫0π2log⁡(sin⁡2θ) dθ 2I = \int^\frac{\pi}{2}_0 \log(\frac{1}{2} \sin 2\theta) \mathrm{d}\theta = \int^\frac{\pi}{2}_0 \log(\frac{1}{2}) \ \mathrm{d}\theta + \int^\frac{\pi}{2}_0 \log(\sin 2\theta) \ \mathrm{d}\theta

We can rearrange this to give ∫0π2log⁡(sin⁡2θ) dθ=2I+π2log2 \int^\frac{\pi}{2}_0 \log(\sin 2\theta) \ \mathrm{d}\theta = 2I + \frac{\pi}{2} log2

Using the substitution x=2θ x = 2\theta transforms this integral into
12∫0πlog⁡sin⁡x dx=∫0π2log⁡sin⁡x dx=I \frac{1}{2} \int^\pi_0 \log \sin x \ \mathrm{d}x =\int^\frac{\pi}{2}_0 \log \sin x \ \mathrm{d} x = I

So I=2I+π2log2 I = 2I + \frac{\pi}{2} log2 , which means I=−π2log2 I = -\frac{\pi}{2} log2

Do I need to justify the use of 12∫0πlog⁡sin⁡x dx=∫0π2log⁡sin⁡x dx \frac{1}{2} \int^\pi_0 \log \sin x \ \mathrm{d}x =\int^\frac{\pi}{2}_0 \log \sin x \ \mathrm{d} x ?

Thanks guys :smile:
Original post by Tau
This isn't from a STEP paper (AFAIK) but can someone check my thinking for this question please and tell me any nicer ways to do this question.

The question is: ∫0π2log⁡sin⁡θ dθ \int^\frac{\pi}{2}_0 \log \sin \theta \ \mathrm{d}\theta

Using the substitution θ↦π2−θ \theta \mapsto \frac{\pi}{2} - \theta transforms the integral into ∫0π2log⁡(sin⁡(π2−θ)) dθ=∫0π2log⁡cos⁡θ dθ \int^\frac{\pi}{2}_0 \log (\sin (\frac{\pi}{2} - \theta) ) \ \mathrm{d}\theta = \int^\frac{\pi}{2}_0 \log \cos \theta \ \mathrm{d}\theta

If we call the integral I I , then 2I=∫0π2log⁡sin⁡θ dθ+∫0π2log⁡cos⁡θ dθ=∫0π2log⁡(sin⁡θcos⁡θ) dθ 2I = \int^\frac{\pi}{2}_0 \log \sin \theta \ \mathrm{d}\theta + \int^\frac{\pi}{2}_0 \log \cos \theta \ \mathrm{d}\theta = \int^\frac{\pi}{2}_0 \log( \sin \theta \cos \theta ) \ \mathrm{d}\theta

So 2I=∫0π2log⁡(12sin⁡2θ)dθ=∫0π2log⁡(12) dθ+∫0π2log⁡(sin⁡2θ) dθ 2I = \int^\frac{\pi}{2}_0 \log(\frac{1}{2} \sin 2\theta) \mathrm{d}\theta = \int^\frac{\pi}{2}_0 \log(\frac{1}{2}) \ \mathrm{d}\theta + \int^\frac{\pi}{2}_0 \log(\sin 2\theta) \ \mathrm{d}\theta

We can rearrange this to give ∫0π2log⁡(sin⁡2θ) dθ=2I+π2log2 \int^\frac{\pi}{2}_0 \log(\sin 2\theta) \ \mathrm{d}\theta = 2I + \frac{\pi}{2} log2

Using the substitution x=2θ x = 2\theta transforms this integral into
12∫0πlog⁡sin⁡x dx=∫0π2log⁡sin⁡x dx=I \frac{1}{2} \int^\pi_0 \log \sin x \ \mathrm{d}x =\int^\frac{\pi}{2}_0 \log \sin x \ \mathrm{d} x = I

So I=2I+π2log2 I = 2I + \frac{\pi}{2} log2 , which means I=−π2log2 I = -\frac{\pi}{2} log2

Do I need to justify the use of 12∫0πlog⁡sin⁡x dx=∫0π2log⁡sin⁡x dx \frac{1}{2} \int^\pi_0 \log \sin x \ \mathrm{d}x =\int^\frac{\pi}{2}_0 \log \sin x \ \mathrm{d} x ?

Thanks guys :smile:


Yeh show that
If f(x)=f(2a-x) then f is symmetric about a



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Reply 270
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Tau
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Original post by physicsmaths
Yeh show that
If f(x)=f(2a-x) then f is symmetric about a



Posted from TSR Mobile


Ah alright, thanks.
Reply 271
Avatar for Zacken
Zacken
OP
22
Original post by Tau
This isn't from a STEP paper (AFAIK) but can someone check my thinking for this question please and tell me any nicer ways to do this question.

The question is: ∫0π2log⁡sin⁡θ dθ \int^\frac{\pi}{2}_0 \log \sin \theta \ \mathrm{d}\theta

Using the substitution θ↦π2−θ \theta \mapsto \frac{\pi}{2} - \theta transforms the integral into ∫0π2log⁡(sin⁡(π2−θ)) dθ=∫0π2log⁡cos⁡θ dθ \int^\frac{\pi}{2}_0 \log (\sin (\frac{\pi}{2} - \theta) ) \ \mathrm{d}\theta = \int^\frac{\pi}{2}_0 \log \cos \theta \ \mathrm{d}\theta

If we call the integral I I , then 2I=∫0π2log⁡sin⁡θ dθ+∫0π2log⁡cos⁡θ dθ=∫0π2log⁡(sin⁡θcos⁡θ) dθ 2I = \int^\frac{\pi}{2}_0 \log \sin \theta \ \mathrm{d}\theta + \int^\frac{\pi}{2}_0 \log \cos \theta \ \mathrm{d}\theta = \int^\frac{\pi}{2}_0 \log( \sin \theta \cos \theta ) \ \mathrm{d}\theta

So 2I=∫0π2log⁡(12sin⁡2θ)dθ=∫0π2log⁡(12) dθ+∫0π2log⁡(sin⁡2θ) dθ 2I = \int^\frac{\pi}{2}_0 \log(\frac{1}{2} \sin 2\theta) \mathrm{d}\theta = \int^\frac{\pi}{2}_0 \log(\frac{1}{2}) \ \mathrm{d}\theta + \int^\frac{\pi}{2}_0 \log(\sin 2\theta) \ \mathrm{d}\theta

We can rearrange this to give ∫0π2log⁡(sin⁡2θ) dθ=2I+π2log2 \int^\frac{\pi}{2}_0 \log(\sin 2\theta) \ \mathrm{d}\theta = 2I + \frac{\pi}{2} log2

Using the substitution x=2θ x = 2\theta transforms this integral into
12∫0πlog⁡sin⁡x dx=∫0π2log⁡sin⁡x dx=I \frac{1}{2} \int^\pi_0 \log \sin x \ \mathrm{d}x =\int^\frac{\pi}{2}_0 \log \sin x \ \mathrm{d} x = I

So I=2I+π2log2 I = 2I + \frac{\pi}{2} log2 , which means I=−π2log2 I = -\frac{\pi}{2} log2

Do I need to justify the use of 12∫0πlog⁡sin⁡x dx=∫0π2log⁡sin⁡x dx \frac{1}{2} \int^\pi_0 \log \sin x \ \mathrm{d}x =\int^\frac{\pi}{2}_0 \log \sin x \ \mathrm{d} x ?

Thanks guys :smile:


Depends on how pedantic you want to be; if this came up in STEP (which it sorta did this year in II) then you'd get away with stating that or at most a quick sketch.

If you want to go for a more formal way, then what PM said works best, but may be overly much detail in STEP (imo).

(BTW, that solution seems very familiar, would it happen to be mine/my question? Just curious)


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(edited 7 years ago)
Reply 272
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Tau
10
Original post by Zacken
Depends on how pedantic you want to be; if this came up in STEP (which it sorta did this year in II) then you'd get away with stating that or at most a quick sketch.

If you want to go for a more formal way, then what PM said works best, but may be overly much detail in STEP (imo).

(BTW, that solution seems very familiar, would it happen to be mine/my question? Just curious)


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Okay, thanks. I saw the question in a post of yours about trigonometric symmetry but I don't think you ever posted the answer. Thanks for the question though :wink:
Original post by physicsmaths
Yeh show that
If f(x)=f(2a-x) then f is symmetric about a



Posted from TSR Mobile

I gave an explanation of this here:

http://www.thestudentroom.co.uk/showpost.php?p=41358736&postcount=5

I expect that if they expected any justification at all however, a sketch of the sine graph with the appropriate line of symmetry would be enough.
Original post by Maths.
Here is the attachment......


I guess they are still too busy to help you :2euk48l:


@Zacken :tongue:
@jneill
(edited 7 years ago)
Original post by Tau
This isn't from a STEP paper (AFAIK) but can someone check my thinking for this question please and tell me any nicer ways to do this question.

The question is: ∫0π2log⁡sin⁡θ dθ \int^\frac{\pi}{2}_0 \log \sin \theta \ \mathrm{d}\theta

Using the substitution θ↦π2−θ \theta \mapsto \frac{\pi}{2} - \theta transforms the integral into ∫0π2log⁡(sin⁡(π2−θ)) dθ=∫0π2log⁡cos⁡θ dθ \int^\frac{\pi}{2}_0 \log (\sin (\frac{\pi}{2} - \theta) ) \ \mathrm{d}\theta = \int^\frac{\pi}{2}_0 \log \cos \theta \ \mathrm{d}\theta

If we call the integral I I , then 2I=∫0π2log⁡sin⁡θ dθ+∫0π2log⁡cos⁡θ dθ=∫0π2log⁡(sin⁡θcos⁡θ) dθ 2I = \int^\frac{\pi}{2}_0 \log \sin \theta \ \mathrm{d}\theta + \int^\frac{\pi}{2}_0 \log \cos \theta \ \mathrm{d}\theta = \int^\frac{\pi}{2}_0 \log( \sin \theta \cos \theta ) \ \mathrm{d}\theta

So 2I=∫0π2log⁡(12sin⁡2θ)dθ=∫0π2log⁡(12) dθ+∫0π2log⁡(sin⁡2θ) dθ 2I = \int^\frac{\pi}{2}_0 \log(\frac{1}{2} \sin 2\theta) \mathrm{d}\theta = \int^\frac{\pi}{2}_0 \log(\frac{1}{2}) \ \mathrm{d}\theta + \int^\frac{\pi}{2}_0 \log(\sin 2\theta) \ \mathrm{d}\theta

We can rearrange this to give ∫0π2log⁡(sin⁡2θ) dθ=2I+π2log2 \int^\frac{\pi}{2}_0 \log(\sin 2\theta) \ \mathrm{d}\theta = 2I + \frac{\pi}{2} log2

Using the substitution x=2θ x = 2\theta transforms this integral into
12∫0πlog⁡sin⁡x dx=∫0π2log⁡sin⁡x dx=I \frac{1}{2} \int^\pi_0 \log \sin x \ \mathrm{d}x =\int^\frac{\pi}{2}_0 \log \sin x \ \mathrm{d} x = I

So I=2I+π2log2 I = 2I + \frac{\pi}{2} log2 , which means I=−π2log2 I = -\frac{\pi}{2} log2

Do I need to justify the use of 12∫0πlog⁡sin⁡x dx=∫0π2log⁡sin⁡x dx \frac{1}{2} \int^\pi_0 \log \sin x \ \mathrm{d}x =\int^\frac{\pi}{2}_0 \log \sin x \ \mathrm{d} x ?

Thanks guys :smile:


As physicsmaths pointed out about symmetry - I would at that point say: "So 2I +pi*ln(2)/2 = integral of ln(sin(2x)). Now we make substitution 2x = t to get integral of ln(sin(x))/2 from 0 to pi.

However, since the graph of ln(sin(x)) is symmetric about x=pi/2, this is 2I/2=I so we get;

2I+pi*ln(2)/2=I so I=-pi*ln(2)/2"

Isn't needed, but worthwhile imo.
Original post by atsruser
I gave an explanation of this here:

http://www.thestudentroom.co.uk/showpost.php?p=41358736&postcount=5

I expect that if they expected any justification at all however, a sketch of the sine graph with the appropriate line of symmetry would be enough.


Actually, I note that the post in question dealt with reflections in a line x=ax=a, which is of course related but not precisely the same.

However if:

for all x,f(x)=f(2a−x)x, f(x)=f(2a-x)

then let x=a+u⇒2a−x=2a−a−u=a−ux=a+u \Rightarrow 2a-x = 2a-a-u=a-u

so that f(a+u)=f(a−u)f(a+u)=f(a-u), for all uu

and so we have symmetry about the line x=ax=a since the function takes the same value if we move a bit up the x-axis from aa i.e. to a+ua+u as when we move down the x-axis from aa i.e. to a−ua-u.

This uses the trick of creating a new u-axis whose origin is at x=ax=a
Original post by Zacken


(BTW, that solution seems very familiar, would it happen to be mine/my question? Just curious)


A similar question, ∫0π1+cos⁡x dx\int_0^\pi 1+\cos x \ dx, came up on the hard integral thread, at least once. Someone or other posted this solution, whose elegance, efficiency, and panache may have stuck in your mind:

http://www.thestudentroom.co.uk/showpost.php?p=61361339&postcount=609
Reply 278
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Maths.
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Original post by Melanie Leconte
I guess they are still too busy to help you :2euk48l:




:confused:

:banghead:
Reply 279
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Zacken
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Original post by atsruser
A similar question, ∫0π1+cos⁡x dx\int_0^\pi 1+\cos x \ dx, came up on the hard integral thread, at least once. Someone or other posted this solution, whose elegance, efficiency, and panache may have stuck in your mind.



Heh, some self-plugging is always good. Good to see you around, helping the young 'uns.

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