STEP I, II, III 2002 Solutions
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Re: STEP I, II, III 2002 SolutionsThank you for your solution！(Original post by Glutamic Acid)
II/7: (Scary scary vectors.)
Let the lines have direction vector
which I'll denote by r, and without loss of generality let |r| = 1 so a^2 + b^2 + c^2 = 1;
.
b = 1 - a; c = 1 - a so
The solutions are a =1, b = 0, c = 0 and a = 1/3, b = 2/3, c = 2/3.
m3 has direction
; m4 direction 
.
Considering
gives 
.
(i) A has position vector
; B 
.
Let P =, and Q =
AQ . BP = 0 so
. Showing that the latter is > 0; 1 - sqrt(6)/3 > 0 therefore 1 > sqrt(6)/3 therefore 1 > 6/9, which is true.
(ii)
So
The second equation gives
. Substituting into the third gives 
. Substituting these into the first:
; multiplying out gives 
, so no non-zero solutions.
However I think there is a mistake in the final part:
Firstly, after the line
The second equation gives.
When you substitute into the third it gives not but 
and if you substitute these values of alpha and beta into the first equation you will find the
terms don't cancel out. Finally you will get a root 
But this is not correct as makes meaningless and also if you substitute it into the second equation you get 0=2/3.
The correct "zero value" of should arise when you substitute into the third equation. Here you must have cancelled a from both side and this is exactly the "zero value" of required by last part of the question.
I don't know if I have made any stupid mistakes here and I hope someone can check my work. And if I am right, I need to say that I still don't know why the root arises here.
Last edited by klausw; 11-04-2012 at 21:38. -
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Re: STEP I, II, III 2002 SolutionsStep 2 question 2
the one where you sub
followed most of the soln on here but for the second part of the question, after dividing the whole thing by
i ended up with the same equation
and then i just guessed factors which worked to factorise it into
and put each of the values of w into
but i got 8 solutions (each value of w giving a +/- value from quadratic formula.. where have i gone wrong? -
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Re: STEP I, II, III 2002 SolutionsJust did this question. Instead of guessing factors I used(Original post by 8inchestall)
Step 2 question 2
the one where you sub
followed most of the soln on here but for the second part of the question, after dividing the whole thing by i ended up with the same equation
and then i just guessed factors which worked to factorise it into
and put each of the values of w into
but i got 8 solutions (each value of w giving a +/- value from quadratic formula.. where have i gone wrong?
And got
and then
and then got to the same solutions of w as you.
And yes i got 8 solutions, should you not get 8 if the polynomial is of order 8? I need someone to confirm this.
Answers i got:
Spoiler:Show
(four solutions)
(two solutions)
(two solutions)
Last edited by desijut; 12-05-2012 at 18:29. -
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Re: STEP I, II, III 2002 Solutionsi got the same solutions as you except for this one:(Original post by desijut)
Just did this question. Instead of guessing factors I used
And got
and then
and then got to the same solutions of w as you.
And yes i got 8 solutions, should you not get 8 if the polynomial is of order 8? I need someone to confirm this.
Answers i got:
Spoiler:Show
(four solutions)
(two solutions)
(two solutions)
Spoiler:Show
where i got
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Re: STEP I, II, III 2002 SolutionsYes, my bad, copied it down wrong, i got that too(Original post by 8inchestall)
i got the same solutions as you except for this one:
Spoiler:Show
where i got
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Re: STEP I, II, III 2002 SolutionsJust a typo(Original post by SimonM)
STEP III, Question 1
Spoiler:ShowThe area is![\displaystyle \int_1^a \frac{\ln x}{x} \, dx = \left [ \frac{(\ln x)^2}{2} \right ]_1^a = \frac{(\ln a)^2}{2}](http://www.thestudentroom.co.uk/latexrender/pictures/76/7673d89701f436f38777242898b43a4e.png "\displaystyle \int_1^a \frac{\ln x}{x} \, dx = \left [ \frac{(\ln x)^2}{2} \right ]_1^a = \frac{(\ln a)^2}{2}")
As
, 
so the area tends to infinity as well
Volume of the solid of revolution is
![\displaystyle \pi \int_1^a \left ( \frac{\ln x}{x} \right )^2 \, dx = \pi \left ( \left [ - \frac{(\ln x)^2}{x} \right ]_1^a + \int_1^a \frac{2 \ln x}{x^2} \, dx \right) =](http://www.thestudentroom.co.uk/latexrender/pictures/99/99f28ffef32a5bec0f7e47f1fedd40ee.png "\displaystyle \pi \int_1^a \left ( \frac{\ln x}{x} \right )^2 \, dx = \pi \left ( \left [ - \frac{(\ln x)^2}{x} \right ]_1^a + \int_1^a \frac{2 \ln x}{x^2} \, dx \right) =")
![\displaystyle \pi \left [ - \frac{(\ln x)^2}{x} \right ]_1^a + \pi \left [ - \frac{2 \ln x}{x} \right ]_1^a +\int_1^a \frac{2}{x^2} \, dx =](http://www.thestudentroom.co.uk/latexrender/pictures/0c/0c3eeb2c8ed1b3a74b54739fb4d2b6b4.png "\displaystyle \pi \left [ - \frac{(\ln x)^2}{x} \right ]_1^a + \pi \left [ - \frac{2 \ln x}{x} \right ]_1^a +\int_1^a \frac{2}{x^2} \, dx =")
![\displaystyle \pi \left [ - \frac{(\ln x)^2}{x} - \frac{\ln x}{x} - \frac{2}{x} \right ]_1^a =](http://www.thestudentroom.co.uk/latexrender/pictures/01/0124d89e87a7bdec4eab54cf1411ac13.png "\displaystyle \pi \left [ - \frac{(\ln x)^2}{x} - \frac{\ln x}{x} - \frac{2}{x} \right ]_1^a =")
As
, the volume tends to 2\pi
![\displaystyle \pi \left [ - \frac{(\ln x)^2}{x} - \frac{2 \ln x}{x} - \frac{2}{x} \right ]_1^a =](http://www.thestudentroom.co.uk/latexrender/pictures/6b/6b1fcd6816092a8a0ece8e30b28f3de1.png "\displaystyle \pi \left [ - \frac{(\ln x)^2}{x} - \frac{2 \ln x}{x} - \frac{2}{x} \right ]_1^a =")
As, the volume tends to 
Last edited by klausw; 19-05-2012 at 21:51. -
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Re: STEP I, II, III 2002 SolutionsAgree with most of the III Q4 solution here but for the final bit the only solutions I get is (0, 0), (-1, 0), and(0, 1), any one any comments?(Original post by Dadeyemi)
Some more;
Did these quite a while ago I'm afraid some may be partial solutions. -
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Re: STEP I, II, III 2002 SolutionsThey are correct (assuming you've written it as y^3 - x^3 = y^2 + x^2 and your solutions are of the format (x,y)). But surely you can see that (1, -1) (and vice versa) is also a solution. Even if you don't see where it comes from you can see that it works.(Original post by klausw)
Agree with most of the III Q4 solution here but for the final bit the only solutions I get is (0, 0), (-1, 0), and(0, 1), any one any comments?
You don't really need much confirmation from other people. Sub it in, if it works then it's a solution. -
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Re: STEP I, II, III 2002 SolutionsYou are right, I forget the (1, -1)solution. Thx!(Original post by hassi94)
They are correct (assuming you've written it as y^3 - x^3 = y^2 + x^2 and your solutions are of the format (x,y)). But surely you can see that (1, -1) (and vice versa) is also a solution. Even if you don't see where it comes from you can see that it works.
You don't really need much confirmation from other people. Sub it in, if it works then it's a solution. -
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Re: STEP I, II, III 2002 SolutionsJust wondering, isn't the answer 81/40?(Original post by tommm)
STEP II 2002 Q10
Spoiler:Showafter t hours, the competitor has
km left to run
the remainder is run at
, in a time (T-t) hours
using speed = distance/time:
which rearranges to
This is a quadratic in t. Clearly t is real, therefore it has real roots, so
which rearranges to give
for all values of t.
If T = 3, then
which has the (repeated) solution
.
Therefore competitor one runs at 13km/h for 3/4 hours
then (29/2)km/h for the remains 9/4 hours
Therefore distance run by competitor one after time x
(The 9/8 is obtained by observing that the two expressions must agree when x = 3/4.)
Now the speed of competitor 2 after time x
we can integrate to find the distance:
distance
from considering the distance travelled initially (0) and after 3 hours (42+3/8), we find
Therefore distance =
therefore the distance between the two
Via differentiation, we find that the first expression has no maxima in the required range, but the second expression has a maximum at
.
Substituting this in, we find that the maximum distance is
km.
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Re: STEP I, II, III 2002 SolutionsI can't see at all where I've got my result from, must've posted that years ago!(Original post by Extricated)
Just wondering, isn't the answer 81/40? -
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Re: STEP I, II, III 2002 Solutions(Original post by tommm)
I can't see at all where I've got my result from, must've posted that years ago!
I posted a thread about this (ages ago though(Original post by Extricated)
Just wondering, isn't the answer 81/40?
)
![= \dfrac{pR(1-q^r)}{1-q}+p(1+2q+3q^2+\dots+rq^{r-1})+rq^r= \dfrac{pR(1-q^r)}{1-q}+p\left[ \displaystyle \sum_{t=1}^rtq^{t-1}\right]+rq^r](http://www.thestudentroom.co.uk/latexrender/pictures/eb/eb9d70df07930e1a6f6396bd9ed540c6.png "= \dfrac{pR(1-q^r)}{1-q}+p(1+2q+3q^2+\dots+rq^{r-1})+rq^r= \dfrac{pR(1-q^r)}{1-q}+p\left[ \displaystyle \sum_{t=1}^rtq^{t-1}\right]+rq^r")
![=\dfrac{pR(1-q^r)}{1-q}+p\left[\dfrac{d}{dq} \displaystyle \sum_{t=1}^rq^t\right]+rq^r = \dfrac{pR(1-q^r)}{1-q}+p\dfrac{d}{dq}\left(\dfrac{q( 1-q^r)}{1-q}\right)+rq^r](http://www.thestudentroom.co.uk/latexrender/pictures/b7/b7ba5afaa96821c8b5f7f46a22210e04.png "=\dfrac{pR(1-q^r)}{1-q}+p\left[\dfrac{d}{dq} \displaystyle \sum_{t=1}^rq^t\right]+rq^r = \dfrac{pR(1-q^r)}{1-q}+p\dfrac{d}{dq}\left(\dfrac{q( 1-q^r)}{1-q}\right)+rq^r")
![=R-Rq^r+\dfrac{p(1-q)[1-(r+1)^r]+pq(1-q^r)}{(1-q)^2}+rq^r \Rightarrow R= \dfrac{(1-q)[1-(r+1)q^r]+q(1-q^r)+rq^r(1-q)}{q^r(1-q)}](http://www.thestudentroom.co.uk/latexrender/pictures/ab/ab545545c437c7192761b70781f220d4.png "=R-Rq^r+\dfrac{p(1-q)[1-(r+1)^r]+pq(1-q^r)}{(1-q)^2}+rq^r \Rightarrow R= \dfrac{(1-q)[1-(r+1)q^r]+q(1-q^r)+rq^r(1-q)}{q^r(1-q)}")
![Pr(X_1 \geq X_2)=1-\displaystyle \int_0^x \lambda_1\text{e}^{-\lambda_1t}dt=1-\left[-\text{e}^{-\lambda_1t}\right]_0^x=1+\text{e}^{-\lambda_1x}-1=\text{e}^{-\lambda_1x}](http://www.thestudentroom.co.uk/latexrender/pictures/62/626f60ab5f4b4547aa4483a8835f8e83.png "Pr(X_1 \geq X_2)=1-\displaystyle \int_0^x \lambda_1\text{e}^{-\lambda_1t}dt=1-\left[-\text{e}^{-\lambda_1t}\right]_0^x=1+\text{e}^{-\lambda_1x}-1=\text{e}^{-\lambda_1x}")
![\text{if }X\sim \text{e}(\lambda) \text{ then E}(X)=\displaystyle \int_0^{\infty}x\lambda\text{e}^ {-\lambda x}dx=\left[-x\text{e}^{-\lambda x}\right]_0^\infty +\displaystyle \int_0^{\infty}\text{e}^{-\lambda x}dx=\left[-\dfrac{1}{\lambda}\text{e}^{-\lambda x}\right]_0^{\infre=ty}=\dfrac{1}{\lambda }](http://www.thestudentroom.co.uk/latexrender/pictures/21/21678f8dbacfe7aacab8ef6866085627.png "\text{if }X\sim \text{e}(\lambda) \text{ then E}(X)=\displaystyle \int_0^{\infty}x\lambda\text{e}^ {-\lambda x}dx=\left[-x\text{e}^{-\lambda x}\right]_0^\infty +\displaystyle \int_0^{\infty}\text{e}^{-\lambda x}dx=\left[-\dfrac{1}{\lambda}\text{e}^{-\lambda x}\right]_0^{\infre=ty}=\dfrac{1}{\lambda }")
![\displaystyle \int_0^m \dfrac{1}{10}\text{e}^{-0.1t}dt=\dfrac{1}{2} \Rightarrow \left[-\text{e}^{-0.1t}\right]_0^m=0.5\Rightarrow1-\text{e}^{-0.1m}=0.5](http://www.thestudentroom.co.uk/latexrender/pictures/40/4025fde7f6c818c9df07c6d81e33af9b.png "\displaystyle \int_0^m \dfrac{1}{10}\text{e}^{-0.1t}dt=\dfrac{1}{2} \Rightarrow \left[-\text{e}^{-0.1t}\right]_0^m=0.5\Rightarrow1-\text{e}^{-0.1m}=0.5")
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