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# STEP 2016 Solutions

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1. (Original post by jahovasfitness)
I stupidly marked the coordinates of the points at the beginning and ends of the horizontal lines. Given that I still got all of the shapes right, how many marks would I have lost do you think?
If you still got the coordinates right, I don't see why you should lose any marks?
2. (Original post by smartalan73)
If you still got the coordinates right, I don't see why you should lose any marks?
Agreed.
3. Courtesy of physicsmaths:

Question 6, STEP I:

Spoiler:
Show

4. Q5
Spoiler:
Show
Draw in radii to tangents and join centres giving you the following;

AC = a+c
AB = a+b
BC = b+c
AP = a
BQ = b
CR = c

WLOG assume a>c
draw in perpendiculars from c to AP
b to AP and CR

Now use pythagoras giving you the following relations

(a+b)^2 = (a-b)^2 + PQ^2
(a+c)^2 = (a-c)^2 + PR^2
(b+c)^2 = (c-b)^2 + QR^2

solving these we find that
2*sqrt(ab) = PQ and so forth

Obviously PQ + QR = PR giving us;
sqrt(ab) + sqrt(bc) = sqrt(ac)
divide by sqrt(abc) to get the desired result

For the second part simply square the result to find 1/b and square again to find 1/b^2
Substitute in a result follows easily

(ii)
square the RHS and cancel to get the following
1/a^2 + 1/b^2 + 1/c^2 = 2(1/ab + 1/bc + 1/ac)
multiply by (abc)^2
b^2(a^2 + c^2) + (ac)^2 = 2(a^2bc + ab^2c + abc^2)
b^2(a-c)^2 -2(ac)(a+c)b + (ac)^2 = 0
b = [2(ac)(a+c) +- 4(ac)^{3/2} ]/2(a-c)^2
b = [ac(sqrt(a) +- sqrt(c))^2]/(a-c)^2
sqrt(b) = [sqrt(ac)(sqrt(a) - sqrt(c))]/(a-c)
sqrt(b) = sqrt(ac)/(sqrt(a) + sqrt(c))
1/sqrt(b) = 1/sqrt(a) + 1/sqrt(c)
5. STEP 1 2016 Q11 solution. This is one I did in the paper... Thought it was a very long one actually, perhaps there was a shortcut I missed (would appreciate people looking for one). Ended in a show that which is always comforting.
Attached Images
6. STEP1 2016 Q11.compressed.pdf (324.1 KB, 110 views)
7. (Original post by rohanpritchard)
STEP 1 2016 Q11 solution. This is one I did in the paper... Thought it was a very long one actually, perhaps there was a shortcut I missed (would appreciate people looking for one). Ended in a show that which is always comforting.
Lol you just best me. I proceeded in a slightly different way, feel free to compare

550695[/ATTACH]ww.thestudentroom.co.uk/app]Posted from TSR Mobile
8. Q13
Spoiler:
Show
Disclaimer; my wording here is bad, please let me know how i can fix it. Also i use g = lamda ( don't know how to latex)
Spoiler:
Show
P(T_1=t) = ge^{-gt}
P(T_1<t) = 1-e^{-gt}
P(T_1>t) = e^{-gt}
All independent
P(T_1,T_2,...,T_n > t) = e^{-ngt}
P(not all T > t) = 1-e^{-ngt}
This means that the lowest time is < t
we want P(lowest time =t) so differentiate
P(lowest time = t) = nge^{-ngt}

E(t) = integral of ngte^{-ngt} from 0 to infinity
Use integration by parts and you get
E(t) = 1/gn
9. (Original post by KingRS)
Q13
Spoiler:
Show
Disclaimer; my wording here is bad, please let me know how i can fix it. Also i use g = lamda ( don't know how to latex)
Spoiler:
Show
P(T_1=t) = ge^{-gt}
P(T_1<t) = 1-e^{-gt}
P(T_1>t) = e^{-gt}
All independent
P(T_1,T_2,...,T_n > t) = e^{-ngt}
P(not all T > t) = 1-e^{-ngt}
This means that the lowest time is < t
we want P(lowest time =t) so differentiate
P(lowest time = t) = nge^{-ngt}

E(t) = integral of ngte^{-ngt} from 0 to infinity
Use integration by parts and you get
E(t) = 1/gn
This is only part 1?

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10. Q13

Will rotate at some point

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11. (Original post by Krollo)
..
12. (Original post by Zacken)
..
Ty bro

Forgot all my matrices dos they're not on step so I forgot how to rotate things

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13. (Original post by Zacken)
Agreed.
That's a relief:/
14. (Original post by student0042)
Q1. I feel like there was another part as it seems a bit small, but then again I only proved it for n=1 and said, do the same thing.
Attachment 550405 I'll let you blow it up again.
Note that p4(1) = 2^8 - 9(1)(3^3)=256-243=13
q4(1) = 2/2 = 1

So the two polynomials cannot be equal.

Much easier than wading through half a page of algebra!
15. Zacken

I'm obviously not going to upload any solutions until you give the go-ahead, but could I perhaps do the solutions for STEP II questions 6, 13 and that [redacted] one?
16. (Original post by Mathemagicien)
Zacken

I'm obviously not going to upload any solutions until you give the go-ahead, but could I perhaps do the solutions for STEP II questions 6, 13 and that one?
Believe it was question 7 or 8, but probably shouldn't mention what the topic is just in case

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17. (Original post by drandy76)
Believe it was question 7 or 8, but probably shouldn't mention what the topic is just in case

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You're right, my apologies!
18. (Original post by Mathemagicien)
You're right, my apologies!
Ith cool, only 21 hours to go!

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19. (Original post by MathHysteria)
Note that p4(1) = 2^8 - 9(1)(3^3)=256-243=13
q4(1) = 2/2 = 1

So the two polynomials cannot be equal.

Much easier than wading through half a page of algebra!
That probably could have saved me some time in the exam. It looks like I went a very long way around it. I was thinking, "surely there's an easier way to do this", but didn't think to do that.
20. can I do q5
21. Note the STEP Prep thread is now closed and will be re-opened at 9am (ish...) tomorrow.

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