STEP 2013 Solutions Watch

Alexn159
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#61
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(Original post by Mastermind2)
STEP I 2013 Question 7:
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7.i) y = ux ---> dy/dx = xdu/dx + u
Substitute into DE to give xdu/dx + u = u + 1/u ---> Int(u)du = Int(1/x)dx ---> (1/2)u^2 = lnx + c. Now substitute u = y/x back in, to give: (1/2)(y^2/x^2) = lnx + c. Substituting the initial conditions gives c = 2, and we get y = xroot(4 + 2lnx), as required.

ii) Using the same substitution, we get: xdu/dx + u = 1/u + 2u ---> xdu/dx = (u^2 + 1)/u ---> 1/2Int(2u/(u^2 + 1)) = Int(1/x)dx ---> (1/2)ln(u^2 + 1) = lnx + c. Substituting u = y/x back in gives: (1/2)ln((y^2/x^2)+1) = lnx + c. Substituting the initial conditions gives c = (1/2)ln5, and we get ln((y^2 + x^2)/x^4) = ln5 ---> y^2 + x^2 = 5x^4 ---> y = xroot(5x^2 -1)

iii) We need a different substitution for this part. Let y = (1/2)ux^2 ---> dy/dx = ux + (1/2)x^2du/dx. Substituting this into the DE gives: (1/2)x^2du/dx = 2/u ---> Int(u) = Int(4x^-2) ---> lnu = -4/x + c. Substituting the initial conditions gives: c = 2ln2 + 4. So we have: y = 2x^2e^(4 - 4/x)

Please let me know if I've made any mistakes.
Used the same substitution in part ii and think I might have got a different answer, don't remember my answer though.

Part iii I used a different substitution and got the same answer (though I took the 4 out of the brackets), took me so long to do that question part, like 40 minutes :/.



Can't really remember my answers but I did questions 1,2 4 and 7 iirc.

This was so much better than step II, only got 1 question out on that paper .

How many questions do you think will need to get out for each grade? Would like at least a 2, if it was 4 questions for a 1, might scrape a 1 but I am sure I must have messed up something somewhere (always do :rolleyes:).


The graphs question about n and number of solutions I am unsure of my answer.

I looked at one solution being between (1/2) < (n) \< (2/3) when n is 1 so for n solutions used something like (n/n+1) < N \< (n+1/n+2) cant remember the exact combination of letters it asked us for but I think that was it.

And the graph I put a weird curves thing bit like this. Don't remember what question this was (1? 2?)

Apologies for awful paint drawing XD. Line is y = 7/12
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MathsNerd1
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#62
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(Original post by Mastermind2)
I don't remember what I did in the exam tbh, my answer might have been different to what I put here.
I'm just very concerned about the boundaries as I can just see them being sky high! :-/
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GeneralOJB
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(Original post by Vaz_Tê)
I used the substitution y=(x^2)(u) and it worked, though I might have made a mistake. I'll have to see the paper again.
I did this too. Took me ages to think to do that, I just thought I'd try it and it worked. Was such a lovely moment, since I didn't even really understand why it worked.
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Vaz_Tê
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(Original post by Mastermind2)
Did you get the same final solution as me?
I definitely got something similar looking, though can't quite remember.
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GeneralOJB
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What did people get the for line of symmetry for question 5? y=-1/2x - 1 or something?
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GG.TintiN
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(Original post by Mastermind2)
I don't remember what I did in the exam tbh, my answer might have been different to what I put here.
You are completely right for part i and ii GJ...but iii I used y=u*x^2 and got a different answer...


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Mastermind2
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(Original post by GG.TintiN)
You are completely right =) GJ


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That's nice to know
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Slowbro93
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Getting ready to type up Q1.
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Alexn159
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^^that looks like Q7 not Q8.
I can't see anything wrong with your part iii), but I got the same answer as the last person that posted in, not this one.
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GeneralOJB
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(Original post by Alexn159)
Used the same substitution in part ii and think I might have got a different answer, don't remember my answer though.

Part iii I used a different substitution and got the same answer (though I took the 4 out of the brackets), took me so long to do that question part, like 40 minutes :/.



Can't really remember my answers but I did questions 1,2 4 and 7 iirc.

This was so much better than step II, only got 1 question out on that paper .

How many questions do you think will need to get out for each grade? Would like at least a 2, if it was 4 questions for a 1, might scrape a 1 but I am sure I must have messed up something somewhere (always do :rolleyes:).


The graphs question about n and number of solutions I am unsure of my answer.

I looked at one solution being between (1/2) < (n) \< (2/3) when n is 1 so for n solutions used something like (n/n+1) < N \< (n+1/n+2) cant remember the exact combination of letters it asked us for but I think that was it.

And the graph I put a weird curves thing bit like this. Don't remember what question this was (1? 2?)

Apologies for awful paint drawing XD. Line is y = 7/12
The graph is wrong on the left...
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Alexn159
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Oh wow just saw that :/, big mess up there.
Should have the lines flipped the other way.


For mistakes like that how many marks will I lose, does not actually effect any of the rest of the question.
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GG.TintiN
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(Original post by Mastermind2)
That's nice to know
My guess is that they will accept both ways as there could be more than one solution that satisfy the equation...and I'm really sorry to tell you that you made a mistake of int(u)...it should be u^2/2...you wrote it as ln(u)...I am so careless =(
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FJacob
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(Original post by Mastermind2)
STEP I 2013 Question 7:
Spoiler:
Show
7.i) y = ux ---> dy/dx = xdu/dx + u
Substitute into DE to give xdu/dx + u = u + 1/u ---> Int(u)du = Int(1/x)dx ---> (1/2)u^2 = lnx + c. Now substitute u = y/x back in, to give: (1/2)(y^2/x^2) = lnx + c. Substituting the initial conditions gives c = 2, and we get y = xroot(4 + 2lnx), as required.

ii) Using the same substitution, we get: xdu/dx + u = 1/u + 2u ---> xdu/dx = (u^2 + 1)/u ---> 1/2Int(2u/(u^2 + 1))du = Int(1/x)dx ---> (1/2)ln(u^2 + 1) = lnx + c. Substituting u = y/x back in gives: (1/2)ln((y^2/x^2)+1) = lnx + c. Substituting the initial conditions gives c = (1/2)ln5, and we get ln((y^2 + x^2)/x^4) = ln5 ---> y^2 + x^2 = 5x^4 ---> y = xroot(5x^2 -1)

iii) We need a different substitution for this part. Let y = (1/2)ux^2 ---> dy/dx = ux + (1/2)x^2du/dx. Substituting this into the DE gives: (1/2)x^2du/dx = 2/u ---> Int(u)du = Int(4x^-2)dx ---> lnu = -4/x + c. Substituting the initial conditions gives: c = 2ln2 + 4. So we have: y = 2x^2e^(4 - 4/x)

Please let me know if I've made any mistakes.
I have a different answer for iii), and Wolfram seems to agree.

I used the same substitution y=ux, and the equation reduces to
x*(du/dx) = x/u + u
Then you do the same thing again and let u=vx and the final solution is something like x*sqrt(6x^2-2x).

EDIT: I just realized that amounts to the same thing as doing y=vx^2 as someone else already suggested.
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Slowbro93
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Question 1:

Part 1
You get given the equation y^2+3y-\frac{1}{2}=0 and get given the substitution y=\sqrt{x}

Using this, you then have:

y^2+3y-\frac{1}{2}=0



\Rightarrow \left(y+\frac{3}{2}\right)^2-\frac{11}{4}=0



\Rightarrow y=\frac{-3\pm \sqrt{11}}{2}

As y\geq0, \Rightarrow y=\frac{-3 + \sqrt{11}}{2}

Now solve for x:





\sqrt{x}=y=\frac{-3 + \sqrt{11}}{2}



\therefore x=\frac{10-3\sqrt{11}}{2}



part ii a


The next question states to solve x+10\sqrt{x+2} -22=0

This time, use a substitution of y=\sqrt{x+2}, \Rightarrow x=y^2-2

\Rightarrow (y^2-2)+10y-22=0



\Rightarrow y^2=10y-24=0



\Rightarrow (y+12)(y-2)=0



\therefore y=2, -12

Remember that y\geq0 by the definition of the square root. This means that you take the value of y=2.

\therefore \sqrt{x+2} = 2



\Rightarrow x=2


part ii b


Final question asks to find the solutions of x^2-4x + \sqrt{2x^2-8x-3}-9=0

Again, make y=\sqrt{2x^2-8x-3} Also note that the first two terms within the square root are twice of those in the equation.

\Rightarrow y^2+3=2x^2-8x, \Rightarrow x^2-4x=\frac{1}{2}\left(y^2+3\right)



\therefore \frac{1}{2}\left(y^2+3\right) +y-9 =0



\Rightarrow y^2+2y-15=0



\Rightarrow y=-5, 3



y\geq 0



\therefore y=3

Finally, solve for x.

y=3, \Rightarrow y^2=9



\Rightarrow 2x^2-8x-3=9



\Rightarrow 2x^2-8x-12=0



\Rightarrow x^2-4x-6=0

And by completing the square, you get that:

x=2\pm \sqrt{10}
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GeneralOJB
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(Original post by cpdavis)
Question 1:

Part 1
You get given the equation y^2+3y-\frac{1}{2}=0 and get given the substitution y=\sqrt{x}

Using this, you then have:

y^2+3y-\frac{1}{2}=0



\Rightarrow \left(y+\frac{3}{2}\right)^2-\frac{11}{4}=0



\Rightarrow y=\frac{-3\pm \sqrt{11}}{2}

As y\geq0, \Rightarrow y=\frac{-3 + \sqrt{11}}{2}

Now solve for x:





\sqrt{x}=y=\frac{-3 + \sqrt{11}}{2}



\therefore x=\frac{10-3\sqrt{11}}{2}



part ii a


The next question states to solve x+10\sqrt{x+2} -22=0

This time, use a substitution of y=\sqrt{x+2}, \Rightarrow x=y^2-2

\Rightarrow (y^2-2)+10y-22=0



\Rightarrow y^2=10y-24=0



\Rightarrow (y+12)(y-2)=0



\therefore y=2, -12

Remember that y\geq0 by the definition of the square root. This means that you take the value of y=2.

\therefore \sqrt{x+2} = 2



\Rightarrow x=2


part ii b


Final question asks to find the solutions of x^2-4x + \sqrt{2x^2-8x-3}-9=0

Again, make y=\sqrt{2x^2-8x-3} Also note that the first two terms within the square root are twice of those in the equation.

\Rightarrow y^2+3=2x^2-8x, \Rightarrow x^2-4x=\frac{1}{2}\left(y^2+3\right)



\therefore \frac{1}{2}\left(y^2+3\right) +y-9 =0



\Rightarrow y^2+2y-21=0



\Rightarrow y=-1 \pm \sqrt{22}



y\geq 0



\therefore y=-1+\sqrt{22}

Finally, solve for x.

y=-1+\sqrt{22}, \Rightarrow y^2=23-2\sqrt{22}



\Rightarrow 2x^2-8x-3=23-2\sqrt{22}



\Rightarrow 2x^2-8x-3-23+2\sqrt{22}=0



\Rightarrow x^2-4x(-13+\sqrt{22})=0

And by completing the square, you get that:

x=2\pm \sqrt{17-\sqrt{22}}
Erm, for part iib I got y^2 +2y - 15
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atsruser
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(Original post by mikelbird)
Will this do for Q3??
As an alternative to your part (ii) argument, I did this:
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Let f(p),f(q) be the maximum and minimum values of f(x)=x^3+3ax^2+3bx+c. Then for 3 real roots, a necessary condition is that 0&lt;p&lt;q with f(p)&gt;0, f(q)&lt;0 (via a graphical demonstration for this question)

Now p,q are the roots of f'(x) = 3x^2+6ax+3b=0 \Rightarrow x^2+2ax+b = 0, with two real roots, from which we find that a^2&gt;b by considering the discriminant.

Also we have that 0&lt;p+q=-2a \Rightarrow a &lt; 0 and 0&lt;pq = b giving the required a^2&gt;b&gt;0.

Finally, for x&lt;p, f(x) is increasing (look at the graph for hand-waving proof), so if we let \alpha with 0 &lt; \alpha &lt; p be the smallest positive root, then we have c=f(0) &lt; f(\alpha) = 0 giving the required c &lt; 0

This differs slightly in that it does everything via consideration of the turning points, and also derives c &lt; 0, rather than reusing the result from part (i) (which I think is a good enough argument, in fact).
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FJacob
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(Original post by cpdavis)
Question 1:

Part 1
You get given the equation y^2+3y-\frac{1}{2}=0 and get given the substitution y=\sqrt{x}

Using this, you then have:

y^2+3y-\frac{1}{2}=0



\Rightarrow \left(y+\frac{3}{2}\right)^2-\frac{11}{4}=0



\Rightarrow y=(-3\pm \sqrt{11}){2}

As y\geq0, \Rightarrow y=\frac{-3 + \sqrt{11}}{2}

Now solve for x:





\sqrt{x}=y=\frac{-3 + \sqrt{11}}{2}



\therefore x=\frac{10-3\sqrt{11}}{2}



part ii a


The next question states to solve x+10\sqrt{x+2} -22=0

This time, use a substitution of y=\sqrt{x+2}, \Rightarrow x=y^2-2

\Rightarrow (y^2-2)+10y-22=0



\Rightarrow y^2=10y-24=0



\Rightarrow (y+12)(y-2)=0



\therefore y=2, -12

Remember that y\geq0 by the definition of the square root. This means that you take the value of y=2.

\therefore \sqrt{x+2} = 2



\Rightarrow x=2


part ii b


Final question asks to find the solutions of x^2-4x + \sqrt{2x^2-8x-3}-9=0

Again, make y=\sqrt{2x^2-8x-3} Also note that the first two terms within the square root are twice of those in the equation.

\Rightarrow y^2+3=2x^2-8x, \Rightarrow x^2-4x=\frac{1}{2}\left(y^2+3\right)



\therefore \frac{1}{2}\left(y^2+3\right) +y-9 =0



\Rightarrow y^2+2y-21=0



\Rightarrow y=-1 \pm \sqrt{22}



y\geq 0



\therefore y=-1+\sqrt{22}

Finally, solve for x.

y=-1+\sqrt{22}, \Rightarrow y^2=23-2\sqrt{22}



\Rightarrow 2x^2-8x-3=23-2\sqrt{22}



\Rightarrow 2x^2-8x-3-23+2\sqrt{22}=0



\Rightarrow x^2-4x(-13+\sqrt{22})=0

And by completing the square, you get that:

x=2\pm \sqrt{17-\sqrt{22}}
I think there might be a mistake on ii b
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I let:
y = x^2-4x, which gives
y + sqrt(2y-3) - 9 = 0

Then let z = sqrt(2y-3)
y = (z^2 + 3)/2

(z^2 + 3)/2 + z - 9 = 0
z^2 + 2z - 15 = 0
Delta = 64
z = (-2 +- 8)/2 >= 0
z = 3
y = 12/2 = 6

x^2-4x = 6
Delta = 16 + 24 = 40
x = 2 +- sqrt(10)

Which is verified by wolfram.
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Vaz_Tê
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#78
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I might be wrong but I think your part ii) b of question one is wrong. I got 2 +/- sqrt(10) and have checked this on my calculator. Your solutions give me 3.
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tinto99
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(Original post by GG.TintiN)
My guess is that they will accept both ways as there could be more than one solution that satisfy the equation...and I'm really sorry to tell you that you made a mistake of int(u)...it should be u^2/2...you wrote it as ln(u)...I am so careless =(
Seeing that solution for iii scared the hell out of me! Surely there is no e or ln term in the solution??? My answer to part iii was sort of the same format as my other ones
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MathsNerd1
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(Original post by tinto99)
Seeing that solution for iii scared the hell out of me! Surely there is no e or ln term in the solution??? My answer to part iii was sort of the same format as my other ones
I was thinking the exact same thing really, thought I screwed it up completely but hopefully I haven't
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