STEP Maths I, II, III 1994 Solutions

Scroll to see replies

Reply 60

For the second integral.

make the substitution

x = tan(theta)

then you can make the same substitution from the first one and go from there.

Reply 61

Speleo

Are you sure? Shouldn't it be tan theta?

Reply 62

nota bene

STEP I Q 13

i) X=X_1+X_2+...+X_n since X_i=1 (and 0 otherwise) if a red is removed and X denotes the number of reds removed.

ii) $E(X_i)=\frac{m}{M}$

iii) $E(X)=\frac{nm}{M}$ This can be seen by thinking it is a mean when we draw n objects of which have a probability p to be red. This p is then determined by (total number of red)/(total number of objects) i.e. the answer in ii)
edit: As DFranklin points out the following should probably be said to make it formally valid:
$E(X) = \displaystyle\sum_1^n E(X_i) = \displaystyle\sum_1^n \frac{m}{N} = \frac{mn}{N}$
iv) The probability will be (number of way to get our selection)/(total number of selections). The total number of selections of n objects from M is $\begin{pmatrix} M \\ n \end{pmatrix}$ . Now we are concentrating on how we can choose k red from m red. Let k=number of red drawn. Then we can chose $\begin{pmatrix} m \\ k \end{pmatrix}$ and the ways non-reds can be chosen from our sample is naturally $\begin{pmatrix} M-m \\ n-k \end{pmatrix}$ (as M-m is the number of non-reds and n-k is the number of chosen non-reds).

Then follows $P(x=k)=\frac{\begin{pmatrix} m \\ k \end{pmatrix} \begin{pmatrix} M-m \\ n-k \end{pmatrix}}{\begin{pmatrix} M \\ n \end{pmatrix}}$

v) Deduce that $\displaystyle\sum_{k=1}^n k\begin{pmatrix} m \\ k \end{pmatrix}\begin{pmatrix} M-m \\ n-k \end{pmatrix}=m\begin{pmatrix} M-1 \\ n-1 \end{pmatrix}$
Uhm, I'm giving in for sleep I think, can't for my life realise how to deduce part v) right now :redface:

(Maybe iii) needs to be far more rigorously proven, if so I'd be happy if someone did it)

Reply 63

insparato

Speleo

Are you sure? Shouldn't it be tan theta?

Yes, damn people for being up at this ungodly hour. I was just laying in bed and thinking that i gave the wrong substitution. I did have the tan theta substitution written down on the paper :p:

.

Ah! Frustrating, i was thinking of the 2I thing but i didnt really know whether i could do that. Its just an angle isnt i guess.

Reply 64

DFranklin

insparato

Hmm i cant see where to go on this the change in variable doesnt really help at all.

$\displaystyle \int_0^{\frac{\pi}{4}} ln (1+tan\theta) \hspace5 d\theta$

$\displaystyle -\int_{\frac{\pi}{4}}^0 ln \left(\frac{2}{1+tan\phi}\right)$
$= \int_0^{\pi/4}\ln 2 \,d\phi - \int_0^{\pi/4} \ln(1+\tan \phi) \,d\phi$

The other well known integral where you can play a similar trick is

$\int_0^{\pi/2} \ln \sin \theta \,d\theta$

Spoiler

Reply 65

DFranklin

nota bene

iii) $E(X)=\frac{nm}{M}$ This can be seen by thinking it is a mean when we draw n objects of which have a probability p to be red. This p is then determined by (total number of red)/(total number of objects) i.e. the answer in ii)

(Maybe iii) needs to be far more rigorously proven, if so I'd be happy if someone did it)

I think you are supposed to use the fact the expectations can be summed, so $E(X) = \sum_1^n E(X_i) = \sum_1^n \frac{m}{N} = \frac{mn}{N}$

v) Deduce that $\displaystyle\sum_{k=1}^n k\begin{pmatrix} m \\ k \end{pmatrix}\begin{pmatrix} M-m \\ n-k \end{pmatrix}=m\begin{pmatrix} M-1 \\ n-1 \end{pmatrix}$
Uhm, I'm giving in for sleep I think, can't for my life realise how to deduce part v) right now :redface:

Use your expression for P(X=k) to find a different expression for E(X) and then compare with your answer to (iii).

Reply 66

DFranklin

zrancis

ii)

If the wagon is not to collide with the pulley, then as the ball is on the point of hitting the floor, it must have a velocity that is less than or equal to 0:I don't think this is right. I think the scenario you are supposed to be considering is "ball falls towards floor, accelerating wagon. Ball hits floor, wagon carries on under it's own momentum, (but now has gravity providing -ve acceleration). Does wagon have enough momentum to hit top?"

(Incidentally, the condition $M > m sin \theta$ is basically the condition that the ball is heavy enough to accelerate the wagon in the first place).

Reply 67

Kolya

Speleo

Thanks for that. :smile:

Reply 68

Speleo

No probs.

Reply 69

ad absurdum

Speleo

I = INT (0 to pi/4) ln(1+tanx) dx
let y = pi/4 - x
dy/dx = -1

-> - INT (pi/4 to 0) ln(1+tan(pi/4 - y)) dy
= INT (0 to pi/4) ln(1+ tan(pi/4 - y)) dy
= INT ln(1 + [1-tany/1+tany]) dy
= INT ln(2/1+tany) dy
I = INT ln(2/1+tanx) dx
Add original integral:
2I = INT (0 to pi/4) ln(1+tanx) + ln(2/1+tanx) dx
= INT ln2 dx
I = (1/2) INT (0 to pi/4) ln2 dx
= (1/2)(pi/4)ln2

= pi.ln2/8

Tell me if you want me to do the other parts too.

EDIT: glancing at the other parts, the second one looks like x = tanu transforms it to the original integral or very close, the last one is a y = pi/2 - x substitution, noting that sin and cos can be swapped over with the given limits, and applying a similar trick to get a simple integral (adding the original integral).

I suppose I might aswell type the last parts up fully, but I think I done the very last part differently.

$\int_0^1 \frac{ln(1+x)}{1+x^2} dx$
Let $x=\tan\theta$ , so $\frac{d\theta}{dx} = \frac{1}{1+x^2}$ , and the limits go from 0->0 and 1->pi/4, and we have:
$\int_0^1 \frac{ln(1+tan\theta)}{1+x^2} (1+x^2) d\theta$
$\int_0^1 ln(1+tan\theta) d\theta = \frac{\pi}{8} ln2$

$\int_0^{\frac{\pi}{2}} ln(\frac{1+\sin x}{1 + \cos x}) dx$
$=\int_0^{\frac{\pi}{2}} ln(1+sinx) dx - \int_0^{\frac{\pi}{2}} ln(1+cosx) dx$
I thought that here I could have said this was zero, but just to make absolutely sure, for the first interal, let $\sin x=\tan\theta$ , so $dx = \frac{\sec^2\theta d\theta}{\sqrt{1-\tan^2\theta}}$ and the limits go from 0->0 and pi/2->pi/4. For the second integral, let $\cos x=\tan\theta$ , so $dx = - \frac{\sec^2\theta d\theta}{\sqrt{1-\tan^2\theta}}$ , and the limits go from 0->pi/4 and 1->0, giving:
$\int_0^{\frac{\pi}{4}} \frac{ln(1+\tan\theta) \sec^2\theta}{\sqrt{1-\tan^2\theta}} d\theta + \int_{\frac{\pi}{4}} ^0 \frac{ln(1+\tan\theta) \sec^2\theta}{\sqrt{1-\tan^2\theta}} d\theta$
$=\int_0^{\frac{\pi}{4}} \frac{ln(1+\tan\theta) \sec^2\theta}{\sqrt{1-\tan^2\theta}} d\theta - \int_0 ^{\frac{\pi}{4}} \frac{ln(1+\tan\theta) \sec^2\theta}{\sqrt{1-\tan^2\theta}} d\theta = 0$

Reply 70

zrancis

DFranklin

I don't think this is right. I think the scenario you are supposed to be considering is "ball falls towards floor, accelerating wagon. Ball hits floor, wagon carries on under it's own momentum, (but now has gravity providing -ve acceleration). Does wagon have enough momentum to hit top?"

(Incidentally, the condition $M > m sin \theta$ is basically the condition that the ball is heavy enough to accelerate the wagon in the first place).

I did think that about part ii, I sort of stumbled through it! I'll try take another look at it today.

Reply 71

Speleo

Last part - easy way.
I = INT (0 to pi/2) ln(1+sinx) - ln(1+cosx) dx
Let y = pi/2 - x. sin(pi/2 - y) = cosy; cos(pi/2 - y) = siny.
dy/dx = -1 and limits swapped, therefore no change in limits overall.
I = INT (0 to pi/2) ln(1+cosy) - ln(1+siny) dy
Add:
2I = INT (0 to pi/2) 0 dx = 0; I = 0.

Reply 72

ad absurdum

Speleo

Yeah that is better :smile:

I think I was too keen to use the result shown in the first part.

STEP I question 10

Let e be the extension of the string
Resolve vertically: $\frac{\lambda e}{l}\cos\theta = mg$
Resolve horizontally: $\frac{\lambda e}{l}\sin\theta = mr\omega^2$
But, $\sin\theta = \frac{r}{l+e}$
so, $\frac{\lambda e r}{l(l+e)} = mr\omega^2$
Rearranging, $e = \frac{ml^2\omega^2}{\lambda - ml\omega^2}$
Subsitute into vertical equation and rearrange to get $\cos\theta = \frac{gl(\lambda-ml\omega^2)}{\lambda l \omega^2}$

Now, the motion is possible only if $0<\cos\theta<1$ (if $\cos\theta = 1$ , then the string is vertical and so no angular motion is actually taking place, if $\cos\theta \leq 0$ , then the string is horizontal or above, which is impossible due to the weight of the particle).
So, $\cos\theta > 0 \Rightarrow \lambda > m l \omega^2$ , so $\omega^2 < \frac{\lambda}{ml}$
And, $\cos\theta < 1 \Rightarrow gl(\lambda - m l \omega^2) < \lambda l \omega^2$ , $\omega^2(gml^2 + \lambda l) > gl\lambda$
$\omega^2 > \frac{g\lambda}{l(\lambda + mg)}$

Reply 73

Speleo

Does anyone mind if I start a 1993 thread?
This one is pretty much dead, and none of the remaining ~5 questions look all that interesting. We're doing the pre-1996 ones pretty much just for our own benefit, anyway.

Reply 74

Rabite

NO! You most certainly shall not! My gosh, the youth these days, trying to take everything in their own hands!
:p:

(I'm kidding, just in case you didn't pick that up)

I mean uh.
Sure that'd be great :biggrin:

Let's try to go all the way to 1990. Although the papers get increasingly weird (education changes, so that's to be expected I suppose) as we go back.

I mean, it's not like starting the 1993 means we are absolutely forbidden to add stuff to this thread, right?
Actually, I was going to ask if I could start one for 1990 (since I've been at it recently), but I guess it's more rational to go for 1993, since you've brought that up.
:smile:

Team maths! Forward!

Reply 75

DFranklin

Speleo

Go for it.

On a slight aside, do you know how to do the $\int_0^{\pi/2} \ln \sin x \,dx$ integral I mentioned? I dare say it won't come up (I've seen it on a STEP paper before but I don't recall which one), but it's a "classic" question that just might pop up again. (And if it does pop up again, it's very easy marks if you've seen it before).

Reply 76

Speleo

I'll make the thread.

DFranklin, does it involve swapping the sin for cos and then combining the two forms of the integral somehow?

Reply 77

DFranklin

Speleo

I'll make the thread.

DFranklin, does it involve swapping the sin for cos and then combining the two forms of the integral somehow?

Yes, but there's another "trick" as well. I've seen this question in at least 3 exams (S-level, STEP, some uni past paper), and I don't think I've ever seen it without some level of hint, usually asking you to show that

$\int_0^{\pi/2} \ln \sin x \,dx = \int_0^{\pi/2} \ln \cos x \, dx = \int_0^{\pi/2} \ln \sin 2x \, dx$

Reply 78

Speleo

I can do it from there, I just need to prove the lnsin2x bit. I'll look at it in a bit :smile:

Reply 79

insparato

I had alittle go at this myself

$\displaystyle \int_0^{\frac{\pi}{2}} ln (sin\theta) \hspace5 d\theta$

$\theta = \frac{\pi}{2} - \phi$

$d\theta = -d\phi$

$\displaystyle -\int_{\frac{\pi}{2}}^0 ln (sin \frac{\pi}{2} - \phi ) \hspace5 d\phi$

$\displaystyle = \int_0^{\frac{\pi}{2}} ln (sin \frac{\pi}{2}cos\phi - cos\frac{\pi}{2}cos\phi ) \hspace5 d\phi$

$\displaystyle \int_0^{\frac{\pi}{2}} ln (cos \phi) d\phi = \int_0^{\frac{\pi}{2}} ln (cos\theta) \hspace5 d\theta$

Second Integral

$\displaystyle \int_0^{\frac{\pi}{2}} ln (cos\theta) d\theta$

$\theta = \frac{\pi}{2} - 2\phi$

$d\theta = -2d\phi$

$\displaystyle -2\int_{\frac{\pi}{4}}^{0}} ln (cos (\frac{\pi}{2}-2\phi)) \hspace5 d\phi$

$\displaystyle 2\int_0^{\frac{\pi}{4}} ln (cos \frac{\pi}{2} cos 2\phi + sin \frac{\pi}{2} sin 2\phi ) \hspace5 d\phi$

$\displaystyle 2\int_0^{\frac{\pi}{4}} ln (sin2\phi) \hspace5 d\phi$

then im guessing that

$\displaystyle = \int_0^{\frac{\pi}{2}} ln (sin2\theta) \hspace5 d\theta$

Total guess but :smile:

Latest

Trending

Articles for you

Mathematics with economics degree personal statement example (1d) Imperial, UCL offers

Mathematics degree personal statement example (1m) Cambridge offer

Mathematics with economics degree personal statement example (1e) Cambridge, UCL offers

Mathematics degree personal statement example (2b) five offers including Cambridge, Warwick

How The Student Room is moderated

To keep The Student Room safe for everyone, we moderate posts that are added to the site.

Try out the app

Continue on web

STEP Maths I, II, III 1994 Solutions

Related discussions

Articles for you

How The Student Room is moderated