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# AEA Mathematics 25/06 2013 Thread watch

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1. (Original post by brittanna)
What was the last part of question 5?
I think... The differianting e^(tanx) to find V. I saw the 'long' solution but didn't wanna go there .
2. (Original post by raqueloliveira)
does anyone has the solutions?
I've got a solution for you, girl . Yeah.... that's right. I said it.
3. (Original post by Victor Victus)
I think... The differianting e^(tanx) to find V. I saw the 'long' solution but didn't wanna go there .
I differentiated and then said let V=Bsec^2xe^tanx, integral V=Be^tanx. I then just substituted this into the the expression for (integral V)/V, and showed that it worked .
4. (Original post by brittanna)
I differentiated and then said let V=Bsec^2xe^tanx, integral V=Be^tanx. I then just substituted this into the the expression for (integral V)/V, and showed that it worked .
Nice.... .
5. (Original post by brittanna)
I differentiated and then said let V=Bsec^2xe^tanx, integral V=Be^tanx. I then just substituted this into the the expression for (integral V)/V, and showed that it worked .
I tried to do this but I couldn't get it to work
6. (Original post by brittanna)
I differentiated and then said let V=Bsec^2xe^tanx, integral V=Be^tanx. I then just substituted this into the the expression for (integral V)/V, and showed that it worked .
I did exactly this However I couldn't do the show part of the question :-/
7. (Original post by MathsNerd1)
I did exactly this However I couldn't do the show part of the question :-/
I usually assume the show that's are marks that I can definitely get, although in this paper that was definitely not the case . There were also too many questions where I was just bluffing my way to the end of the question.
8. Does anyone have the question paper to upload?
9. (Original post by brittanna)
I usually assume the show that's are marks that I can definitely get, although in this paper that was definitely not the case . There were also too many questions where I was just bluffing my way to the end of the question.
Yeah same here and that inequality question with the integrals was just bizarre really! I looked at it and just wanted to walk out when first seeing it but I think I did all bar 1 part correctly
10. I think after talking with other people, I've probably got about 60ish on this paper. I really hope that's a merit lol. Was hoping for a distinction but the integration questions and series question were too much for me
11. (Original post by Victor Victus)
BTW, for the last question, M was less than 5/16 but greater than -5/16. Messed that up. I'm fuming.
yes, im pissed of about that as well, i got to the critical value of m being 5/16 but in the exam I blanked out, as soon as i stopped it hit me the modulus of m<5/16. SO I'VE been kicking myself for that mistake.Also i made a copying error which meant for my stationary point i got the wrong y-coordinate, i hope this error is carried forward to my sketch.
12. I did q1,q2,q3,q4 (series) I did up to proving the second to last inequality and I used the 2nd to last inequality to prove the RHS of the last inequality.q5) complete although I had to come back to part a), q6 most'y complete, i went wrong proving the integral=12root3/5. But the last part I did fine using this result.
q7) I found the stationary point although miscalculated the y-cordinates, made a logical i think guess to the value of m that had to be stated. x=m
skipped to the graph sketch, and blanked out on finding the m for mx+1. Got to 5/16 but blanked there, so hopefully ill get error carried forward onto the sketch and some method marks on the mx+1.
I think it was a hard paper,it wasn't anything like before especially the integrals and series question
13. (Original post by brittanna)
I differentiated and then said let V=Bsec^2xe^tanx, integral V=Be^tanx. I then just substituted this into the the expression for (integral V)/V, and showed that it worked .
I did this as well.
14. Ill post my methods but not solutions .
1) do the expansion using the formula book or whatever, equate the coefficients, take them to one side and factorise carefully as to not lose any solutions.
2)first part was easy
second part it was of the form 2sin(a)=cos(b)/cos(a)
goes to 2sin(a)cos(a)=cos(b)
sin(2a)=sin(90-b)
so 2a=90-b etc
3) it was equation the lines, equating line with point, dot product.
For finding L3, i think I remembered from another paper that it would be
r=oc+t(bc+ac)
4)the terms were easy to work out,
ar+1 relationship you had to sub r+1 into ar the you got 2r+1-1=2*(2r-1)+1=2ar+1
Sum of (ar). the previous part was no help to me.
the series. I wrote it out in full noticed it was a geometric series of first term 2 and common ratio 2, take away the sum of -1.
showing 1/ar+1 was easy. 1/ar+1= 1/(2ar+1)
2ar+1>2ar so it followed that, 1/ar+1<1/2ar

Second to last part, I noticed the gp of the 1/2s. but I couldn't prove each terms is greater than those on the rhs, **** thinking about it now, I just had to show the infinite sum of the 1/7*Infinity sum of the gp of common ratio 0.5 and first term 1 was greater than the infinite sum of 1/ar. Which I have no idea how to do, maybe that wasn't the way.Hopefully method marks.

The last part, I could prove the sum of 1/ar<33/21 or something like that using the result from second to last part. Could prove the greater than
5) ]
You differentiate this by the product rule, then divide though by uv
second part was given sin^2(x). the other one for v was cos^2(x)

Expression for u
Here we just solved the differential equation, the were had one if you didn't realise they were in the formula book like me (I noticed after looking and noticing the differential of cot(x))

e^{tan(x)}
you had to differentiate this, then make the differential=v and verify that

6) let L = lambda.
Cant remember the exact expression, but you had to expand it. to get the quadratic for L.
ii) Then use the discriminant to get the inequality.
iii) sub f(x)=1 and g(x)=(1+x3)0.5
then solve that basically
iv) wasn't sure about this, I couldn't get 12root3/5 out. But I though it was the case of b^2-4ac=0
v) You had to use f(x)=x^2 and g(x)=(1+x3)0.25

and the result from iv) and the inequality and it came out with what was required.
7) i)first part was standard. finding f'(x) then equation=0. Got +-6, but I incorrect subbed that into f(x) so my y-coordinates are wrong. y=+-4 I believe
ii) I saw that delta(y)/delta(x) for the line ---> infinity. and k>0 so I went k=6. Not too sure

skipped to the sketch which was pretty simple, reflected it in the x axis. I don't think graph paper was need because it was a sketch, other than the asymptotes, it would be hard to have a graph paper worth sketch of it. I think they will mark it. Obviously my new stationary points are incorrect here, hope for error carried forward. the new ones are (6,4) and (-6,4), by the same logic.

the last part,
equate mx+1=f(x)
get a quadratic and do the discriminant.
critcal value at m=5/16
I blanked out here, so I didn't finish, didn't figure out what to do with the m>= 5/16 . But the answer I believe is modulus of m<5/16. Hopefully i got 2/4 marks here , but this allowed me to go back to get a complete on q5) so I lost 2 marks but gained 3 .

That's what I done, feel free to correct me etc, but please dont! I've made enough mistakes already
15. I did NOT like that paper at all...oh dear.
16. (Original post by jarasta)
Ill post my methods but not solutions .
1) do the expansion using the formula book or whatever, equate the coefficients, take them to one side and factorise carefully as to not lose any solutions.
2)first part was easy
second part it was of the form 2sin(a)=cos(b)/cos(a)
goes to 2sin(a)cos(a)=cos(b)
sin(2a)=sin(90-b)
so 2a=90-b etc
Can you remember what you got for these? Sure I screwed up (1)(b) as I ended up with 4 solutions! Then I rejected n=0 and n=1 as this makes the coefficient of x^2 and x^3 zero?

Also on (2) I got 4 solutions for theta, but I stupidly tried to work backwards to covert things into cos(blah) instead of sin so I probably made a silly mistake :/
17. (Original post by Math-Magician)
Can you remember what you got for these? Sure I screwed up (1)(b) as I ended up with 4 solutions! Then I rejected n=0 and n=1 as this makes the coefficient of x^2 and x^3 zero?

Also on (2) I got 4 solutions for theta, but I stupidly tried to work backwards to covert things into cos(blah) instead of sin so I probably made a silly mistake :/
i got n=0,1, 5/2 and -1/2 i think, but didn't reject, the expansion would still be correct for n=0 and 1?? So yeah four solutions to 1b as well

I also got 4 solutions for theta i think.
18. I got 7 solutions or something for theta...looks like it's a fail for me.
19. (Original post by yl95)
I got 7 solutions or something for theta...looks like it's a fail for me.
may have hard more actually now i think about it, i think i did.
20. (Original post by jarasta)
i got n=0,1, 5/2 and -1/2 i think, but didn't reject, the expansion would still be correct for n=0 and 1?? So yeah four solutions to 1b as well

I also got 4 solutions for theta i think.
The expansion would be valid, but (1+aX)^1 = 1+aX, and (1+aX)^0 = 1, so they both give no X^2 or X^3 terms and I'm sure the question said the coefficients of these were non-zero? The next part then asked you to reject one of the other two solutions (as they told you X=1/2 and this made mod(aX)>1)

I definitely got 5/2 and -1/2 though so that's good!

I'm pleased if you got four solutions for theta as I thought I'd included spurious solutions or something, it looked like there were too many

I must have dropped 20-30 marks or so on this paper, plus all the style marks no doubt

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