The Student Room Group
Uni students - Complete our survey >>
Uni students - Complete our survey >>
Uni students - Complete our survey >>

STEP 2016 Solutions

Scroll to see replies

Original post by Zacken
STEP III: Question 4

Sum to N



Sum to infinity



Hyperbolic sum



how much for all that but not rewriting blabla+2sechy as 2cosechy?
Reply 261
Avatar for Zacken
Zacken
OP
22
Original post by gasfxekl
how much for all that but not rewriting blabla+2sechy as 2cosechy?


19
Original post by Zacken
19


Zacin. I used x=e^2y but then in last prt i subtracted r=1 term cause i was rushin. I still doubled and added r=0. How many lost


Posted from TSR Mobile
Reply 263
Avatar for Zacken
Zacken
OP
22
Original post by physicsmaths
Zacin. I used x=e^2y but then in last prt i subtracted r=1 term cause i was rushin. I still doubled and added r=0. How many lost


Posted from TSR Mobile


Probs lose 5 or 6 marks.
Original post by Zacken
Probs lose 5 or 6 marks.



Decent ttntntntnt
Reply 265
Avatar for fa991
fa991
3
Original post by sweeneyrod
Q 12 was beautiful, probably the easiest STEP III stats question ever.


Ikr! I began it right at the end so ran out time 😥😥 Could've been an easy 20

Posted from TSR Mobile
Reply 266
Avatar for fa991
fa991
3
Original post by Mathemagicien
I know, I felt bad doing it, it really feels like cheating


How do you think the marks willbe allocated between each part?

Posted from TSR Mobile
Reply 267
Avatar for Zacken
Zacken
OP
22
Original post by Number Nine
bagsy q1

Unparseable latex formula:

\displastyle[br]\begin{equation*} I_n = \int^{\infty}_{-\infty}\frac{1}{(x^2 + 2ax + b)^n} \mathrm{d}x\end{equation*}



Using the substitution x+a=b−a2tan⁡u x + a = \sqrt{b-a^2} \tan u we have dxdu=b−a2sec⁡2u\frac{\mathrm{d}x}{\mathrm{d}u} = \sqrt{b-a^2} \sec^2 u

So that:

Unparseable latex formula:

\displaystyle [br]\begin{align*}I_n = \int^{\infty}_{-\infty}\frac{1}{((x+a)^2 + b - a^2)^n} \mathrm{d}x & = \int^{\pi/2}_{-\pi/2}\frac{\sqrt{b-a^2}\sec ^2 u}{((b-a^2)\tan ^2 u + (b-a^2))^n } \mathrm{d}u \\ & = \int^{\pi/2}_{-\pi/2}\frac{\sqrt{b-a^2}\sec ^2 u}{((b-a^2)\sec ^2 u)^n } du \\ & = \int^{\pi/2}_{-\pi/2}\frac{1}{(b-a^2)^{n-\frac{1}{2}} \sec ^{2n-2} u } du \end{align*}



So:

i) Find I1I_1

Unparseable latex formula:

\displaystyle [br]\begin{align*}I_1 = \int^{\pi/2}_{-\pi/2}\frac{1}{(b-a^2)^{\frac{1}{2}}} \mathrm{d}u & = \left[ \frac{x}{(b-a^2)^\frac{1}{2}} \right]_{-\pi/2}^{\pi/2} \\& = \frac{\pi}{\sqrt{b-a^2}} \end{align*}



as required


Cleaning up the LaTeX a little for you.
..
Original post by zacken
cleaning up the latex a little for you.
..

yeah can
Anyone got the pdf? Willing to have a bash at any unpopular questions
wow everyone here is so clever. have fun at oxbridge you guys :smile:
Original post by fefssdf
wow everyone here is so clever. have fun at oxbridge you guys :smile:


We don't talk about Oxford round these parts


Posted from TSR Mobile
Original post by drandy76
We don't talk about Oxford round these parts


Posted from TSR Mobile


ok im backing off
Reply 273
Avatar for Zacken
Zacken
OP
22
Original post by Llewellyn
Anyone got the pdf? Willing to have a bash at any unpopular questions


http://www.thestudentroom.co.uk/showthread.php?t=4136101&page=116&p=66077019#post66077019
Can someone write an answer for Q3 please
Original post by Zacken
Cleaning up the LaTeX a little for you.
..


http://www.thestudentroom.co.uk/showthread.php?t=4166421&page=14&p=66074739#post66074739
Please can you make this of adequate size again? :smile:
Reply 276
Avatar for Zacken
Zacken
OP
22
Original post by student0042


..
Original post by Number Nine
bagsy q1

In=∫−∞∞1(x2+2ax+b)ndx[br]Usingthesubstitution: I_n = \int^{\infty}_{-\infty}\frac{1}{(x^2 + 2ax + b)^n} dx[br] Using the substitution:

x+a=b−a2tan⁡u x + a = \sqrt{b-a^2} \tan u

⇒dxdu=b−a2sec⁡2u \Rightarrow \frac{dx}{du} = \sqrt{b-a^2} \sec ^2 u

⇒In=∫−∞∞1((x+a)2+b−a2)ndx \Rightarrow I_n = \int^{\infty}_{-\infty}\frac{1}{((x+a)^2 + b - a^2)^n} dx

⇒In=∫−π/2π/2b−a2sec⁡2u((b−a2)tan⁡2u+(b−a2))ndu \Rightarrow I_n = \int^{\pi/2}_{-\pi/2}\frac{\sqrt{b-a^2}\sec ^2 u}{((b-a^2)\tan ^2 u + (b-a^2))^n } du

tan⁡2u+1=sec⁡2u \tan ^2 u + 1 = \sec ^2 u

In=∫−π/2π/2b−a2sec⁡2u((b−a2)sec⁡2u)ndu I_n = \int^{\pi/2}_{-\pi/2}\frac{\sqrt{b-a^2}\sec ^2 u}{((b-a^2)\sec ^2 u)^n } du

In=∫−π/2π/21(b−a2)n−12sec⁡2n−2udu I_n = \int^{\pi/2}_{-\pi/2}\frac{1}{(b-a^2)^{n-\frac{1}{2}} \sec ^{2n-2} u } du

i)Find I1 i) Find\ I_1

I1=∫−π/2π/21(b−a2)12du I_1 = \int^{\pi/2}_{-\pi/2}\frac{1}{(b-a^2)^{\frac{1}{2}}} du

I1=[x(b−a2)12]−π/2π/2 I_1 = \left[ \frac{x}{(b-a^2)^\frac{1}{2}} \right]_{-\pi/2}^{\pi/2}

I1=πb−a2 as required I_1 = \frac{\pi}{\sqrt{b-a^2}}\ as\ required

apparently my latex is ugly


How many marks do you reckon the first part will be? 5-6?
Original post by Shrek1234
How many marks do you reckon the first part will be? 5-6?


4-5 id say @Zacken opinion
Reply 279
Avatar for Zacken
Zacken
OP
22
Original post by Shrek1234
How many marks do you reckon the first part will be? 5-6?


Original post by Number Nine
4-5 id say @Zacken opinion


5 sounds about right, yeah.

Quick Reply

Related discussions

Latest

Trending

Trending

Articles for you