The Edexcel AEA (26/06/12 - AM) Revision Thread

I did this exam yesterday, just "to keep my eye in" on maths at this level. Anyway, for the domain of $gf(x)$, I worked along the lines of if $x > 2$ for g(x), then for $g(x^2 - 2x + 6)$ we need $x^2 - 2x + 6 > 2$, which is always satisfied. So I went with x can be any real number as the domain of $gf(x)$, which led to the range being greater than or equal to 6.

As for 7(f), the angle the tangent of C1 made with the horizontal was $\beta$, and the angle the tangent of C2 made with the horizontal was $-\beta$, so the angle between the two tangents was $2\beta$. However, since $\sqrt{\frac{16 - \pi^2}{32}} < 1$ then $\beta < \frac{\pi}{4}$, so $2\beta < \frac{\pi}{2}$. So the obtuse angle between the two tangents is the difference between $\pi$ and $2\beta$, i.e. $\pi - 2\beta$.


They were my lines of thought, anybody agree/disagree?

I did this exam yesterday, just "to keep my eye in" on maths at this level. Anyway, for the domain of $gf(x)$, I worked along the lines of if $x > 2$ for g(x), then for $g(x^2 - 2x + 6)$ we need $x^2 - 2x + 6 > 2$, which is always satisfied. So I went with x can be any real number as the domain of $gf(x)$, which led to the range being greater than or equal to 6.

As for 7(f), the angle the tangent of C1 made with the horizontal was $\beta$, and the angle the tangent of C2 made with the horizontal was $-\beta$, so the angle between the two tangents was $2\beta$. However, since $\sqrt{\frac{16 - \pi^2}{32}} < 1$ then $\beta < \frac{\pi}{4}$, so $2\beta < \frac{\pi}{2}$. So the obtuse angle between the two tangents is the difference between $\pi$ and $2\beta$, i.e. $\pi - 2\beta$.

They were my lines of thought, anybody agree/disagree?

How?

It had a height of y = (0+a)(0-b)(0-b) and a width (a+b), right?

Show me some proofs, guys!

S was the maximum, not the y intercept!

I believe that the second-last question was one of the easiest on the paper.

In the part in which we had to show ((a+b)^4)/12 in that i had expanded the stuff while integrating, after integrating i knew it should be ((a+b)^4)/12 but i couldn't bother factorizing it so i just expanded ((a+b)^4)/12 to show that its expanded form matches with the one i got, will they give me full marks for it?

I believe that the second-last question was one of the easiest on the paper.

In the part in which we had to show ((a+b)^4)/12 in that i had expanded the stuff while integrating, after integrating i knew it should be ((a+b)^4)/12 but i couldn't bother factorizing it so i just expanded ((a+b)^4)/12 to show that its expanded form matches with the one i got, will they give me full marks for it?

Idk besides the 1-2 marks for giving the domain the first question was ridiculously easy.

How did you show they were the same without factorising though?

Similar to you, I expanded (a+b)^4 /12 at the bottom of the page so I could see what I was aiming for, and then factorised it until the expressions were the same.

I think if you've made it obvious that your expression and the expansion are the same, you should get the marks. It wasn't exactly difficult, just lot's of messy algebra.

I believe that the second-last question was one of the easiest on the paper.

In the part in which we had to show ((a+b)^4)/12 in that i had expanded the stuff while integrating, after integrating i knew it should be ((a+b)^4)/12 but i couldn't bother factorizing it so i just expanded ((a+b)^4)/12 to show that its expanded form matches with the one i got, will they give me full marks for it?

When I got 1 4 6 4 1 I briefly mentioned the binomial theorem and Pascal's triangle, and wrote that it was equal to (a+b)^4. Do you reckon that's alright? I don't know how else you can show the factorisation anyway... Surely you just do it by inspection.

If you just integrate (x+a)(x-b)^2 by parts you don't have to factorise.

Also guys what do we reckon overall, compared to previous papers about the difficulty of this one. Possibly 2010 grade boundaries?

I don't want to be pessimistic but perhaps even a few marks higher.

Seens like no ones given an answer for the third integral yet. I did do it (i believe) in the exam and i thought it used the same trick as the second integral. After simplifying you had to integrate something like

2sinx*(cosx)^(1/3) dx

which you can use the same "trick" for. As in, the dervative of cos is -sin and using the chain rule you drop the powers by one so just reverse the process giving

-3/2 (cosx)^(4/3) +C

At least i hope this is true!

If you just integrate (x+a)(x-b)^2 by parts you don't have to factorise.

When I got 1 4 6 4 1 I briefly mentioned the binomial theorem and Pascal's triangle, and wrote that it was equal to (a+b)^4. Do you reckon that's alright? I don't know how else you can show the factorisation anyway... Surely you just do it by inspection.

are you sure about the bolded part?

i got

2(\sin^2x \cosx)^{1/3}

have no idea how to turn that in latex just tried and failed about 10 edits lol