Maths HL P 1 - aftermath

MollyJayne

Deleted User wrote: »

http://en.wikipedia.org/wiki/Line_%28geometry%29


It's a straight line. That being said, I didn't cop it until near the end of the exam even though it was my first question.

Ah, I see...damn it! =[
I agree that it is very misleading information! =[

MollyJayne

zam wrote: »

For alphaxbeta i got ±4
For alpha+beta i got 0 (which i think is wrong!)

alpha+beta = 2

from alpha^2+beta^2=(alpha+beta)^2-2(alphabeta)=12

shamoono

Making It Bad wrote: »

Nobody any answer to this?

I am just gonna put up the solution to the questions I did. It will be in the last post.

shi7zman

Ah Paper1 is over! forget abou it! i need paper2 predictions!! i know theorms will be up but what else??

zam

MollyJayne wrote: »

alpha+beta = 2

from alpha^2+beta^2=(alpha+beta)^2-2(alphabeta)=12

That's what I did! I put in 4 into it once and -4 into it the other time and put them equal to each other I'd say it was just a slip...hopefully....:P

Reillyman

MollyJayne wrote: »

Ah, I see...damn it! =[
I agree that it is very misleading information! =[

No it's not in fairness. I didn't spot that it was a line either, I said that the graph was [latex]increasing for f(x)[/latex]

But I'm not going to whinge that it was unfair or misleading, I mean when I look back at it it's so obvious!:P

NumberZeroNine

Question 7; (C)

Making It Bad wrote: »

Part 7(c) look really easy except I couldn't get it out! Got it to be "loge(1+x)" when differentiated except this didn't equal 0 when you substitute in for x it equalled "-1" where the hell did I go wrong?

when diff.ing it;
let u(x)=(1+X), v(x)= ln(1+X)
=> u'(x)=1, v'(x)=1/(1+X)

f'(x) = [v(x).u'(x)+u(x).v'(x)]
f'(x) = [1.ln(1+X)+(1+X)/(1+X)]
f'(x) = [ln(1+X)+1]

f''(x) = 1/(1+X)

At turning point; f'(x) = 0

[ln(1+X)+1] =0
ln(1+X) = -1
1+X = e^-1
X = e^-1 -1
X = 1/e - e/e
X = (1-e)/e

Sub into F(x)

y = f((1-e)/e)
y = ( (e+1-e)/e ).ln( (e+1-e)/e)
y = (1/e).ln(e^-1)
y = -1/e

At local Min f''(x) 0
At local Max f''(x) 0

F''((1-e)/e) = 1/((e+1-e)/e)
= 1/(1/e)
= e
> 0
Therefore Min.

blacklionboy

I just did induction. lol

MollyJayne

zam wrote: »

That's what I did! I put in 4 into it once and -4 into it the other time and put them equal to each other I'd say it was just a slip...hopefully....:P

Sorry, just realised it should be root 20! I forgot to change my signs when bringing over =[ What did you get?

MollyJayne

Reillyman wrote: »

No it's not in fairness. I didn't spot that it was a line either, I said that the graph was [latex]increasing for f(x)[/latex]

But I'm not going to whinge that it was unfair or misleading, I mean when I look back at it it's so obvious!:P

I wasn't whinging! I'm really happy with how I did. Maybe you should stop judging people?!

zam

MollyJayne wrote: »

Sorry, just realised it should be root 20! I forgot to change my signs when bringing over =[ What did you get?

0 in the end! I dunno if it's right or not but sure.... it's the work that counts....

blacklionboy

here chaps 6cii) was grand. just find slope using the 2 points given and they turned up to be same

MollyJayne

zam wrote: »

0 in the end! I dunno if it's right or not but sure.... it's the work that counts....

I hope so, 'cause I make a lot of adding mistakes! XD

shamoono

So here is my solutions to the questions I did .

Q1. (a) was pretty easy (b) use the quadratic formula (c) part iii was the hardest had no idea what was going on.

Q2. (a) easy again (b) just let a=a^2 and b=b^2 and solve for a+b and ab
(c) part one you just had to bring ab on the right hand side to the left and factorise and you would get (something)squared>0 which mean it was true.
Part ii you had to get common denominator and multiply across by a^2 x b^2 finally after some algebra you would end up with (a+b)(a-b)^2 > 0 which is true since (any number squared)>0 and a+b>0(which was given in the question).

Q3. (a) Matrices just multiply out and get two simultaneous equations and solve.
(b) part one easy and I understand people found part ii hard. What I did was put Z2(the complex number) on an argand diagram. I let the angle=3pi/4…..From the diagram it’s easy to see that Tan3pi/4=8/t and solve for t from there.
(c) z^5=1+(0)i plot on argand diagram. let the angle be 2n(pi), i really don’t know why but that’s how it’s done. Put it in the form of (cos 2npi + i sin 2npi) ^1/5. Use De Moivre’s theorem and you get (cos 2npi/5 + i sin 2npi/5).
-Now here is the really weird part– for the first root let n=0, for the second root let n=1 and so on…..
Part ii was easy you just used one of the roots that you got from above like (cos 2pi/5 + i sin 2pi/5) and let it equal to W and solve for W^2 + W^3.

Question six was pretty easy … I see people found (b)ii hard, since dy/dx was not equal to zero the graph has no turning points and also d2y/dx2=0 it has a point of inflection, therefore the graph is a straight line. Part c was easy enough.

Q.7 I also ****ed up at first on (b) I went back over found that everything cancels out came down to something like -2/1-2cosxsinx ( some thing like that ) and same for part II. If you get dy/dx=1+y^2 your answer was right

Part (c) you had to change LOG TO LN because in the log tables it is down as ln. Then just use your product rule(look up the log tables for the derivative of ln(x) which would equal to 1/x). differentiating ln(1+x) = 1/1+x

part ii is just d2y/dx2 and sub in if it’s greater than or smaller than 0.

Q8. (a) and (b) very easy. I got C part i and ii out correctly but dont know about part iii only got to do a part of it.

(c) let u=1+sinx etc……

citizenerased1

in the long run guys discussing on this wont change anything!...:)

chins up

Making It Bad

NumberZeroNine wrote: »

Question 7; (C)

FUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUCK. I FORGOT THE +1 ARRRRGH

NumberZeroNine

Question 2 (B)

zam wrote: »

For alphaxbeta i got ±4
For alpha+beta i got 0 (which i think is wrong!)

it tells you alpha > 0, Beta > 0

so -4 isn't a possible solution, nearly didn't catch that one.

Full Solution;

A^2.B^2 = 16
A^2+B^2 = 12

(i)
(A.B)^2 = 16
A.B = rt(16) = 4 (-4 isn't a solution because of phrazing of Q)

(ii)
(A+B)^2 = A^2 +2.AB +B^2
(A+B)^2 = (A^2+B^2) +2.(AB)
(A+B)^2 = (12) + 2.(4)
(A+B) = rt(20)
(A+B) = 2.rt(5)

blacklionboy

for 6 b i got a constant for the slope and said in has no max/min ause a constant cant equal zero..

CantStandMeNow

NumberZeroNine wrote: »

Question 2 (B)



it tells you alpha > 0, Beta > 0

so -4 isn't a possible solution, nearly didn't catch that one.

Full Solution;

A^2.B^2 = 16
A^2+B^2 = 12

(i)
(A.B)^2 = 16
A.B = rt(16) = 4 (-4 isn't a solution because of phrazing of Q)

(ii)
(A+B)^2 = A^2 +2.AB +B^2
(A+B)^2 = (A^2+B^2) +2.(AB)
(A+B)^2 = (12) + 2.(4)
(A+B) = rt(20)
(A+B) = 2.rt(5)

Yup that's definitely right.. nearly didnt catch the minus 4 aswell!!

MollyJayne

mark.oc wrote: »

Q.1 (a) k = -3, t = 9

(c)(i) -3-b
(ii) -2-2b
(iii) b = -4, c = 1, d = 6

Q.2 (a) x = 3, y = -2, z = -1

(b)(i) 4
(ii) rt20 or 2rt5

Q.3 (a) x = 4, y = 2

(b)(i) ±6
(ii) -8

(c) Didn't take them down as they were all decimals.

Q.4 (a) 47/99

(b)(i) a = -42, d = 6
(ii) 0

(c)(iii) 27230

Q.5 (a) x = 2

(c)(ii) 47

Q.6 (a) x = 4/3

(b)(i) 2/5
(ii) line

(c)(i) [LATEX](-4-2xy^3)/(2+3x^2y^2)[/LATEX]
(ii) 2x + y - 6 is the tangent.

Q.7 (b)(i) 2/(1-2cosxsinx)

(c)(ii) minimum

Q.8 (a) -(cos2x)/2 + (e^4x)/4 + c

(b) 37 units

(c)(i) ln((1+sinb)/(1+sina))
(ii) ln((1+cosa)/(1+cosb))

You're a genius and I hate you

zam

Note to self: avoid this thread. It's depressing

MollyJayne

zam wrote: »

Note to self: avoid this thread. It's depressing

Don't worry! It sounds like you've done fine! Like you said, it's the method that counts!

wonderfulname

niamhocxox wrote: »

You're answer to 1 (C) part (iii) is wrong cause they were 3 consecutive numbers......eg.1,2,3.....I got my answers as fractions with a difference of 1 between them all......

hate to break it to you your the one who's wrong...

an easy paper, but too easy it threw me. i didn't do well at all. and that play on the cubed roots of unity, i genuinely thought that bit was just on the course to look pretty. never expected it!

MollyJayne

niamhocxox wrote: »

You're answer to 1 (C) part (iii) is wrong cause they were 3 consecutive numbers......eg.1,2,3.....I got my answers as fractions with a difference of 1 between them all......

Consecutive means the next in the sequence, the common difference doesn't have to be 1!

shi7zman

C'mon weres my predictions???:(

mind_master

Was a simple paper i must say! Think i got every bit right! Had a few slips here and there i think! So 98% at least

MollyJayne

Reillyman wrote: »

No it's not in fairness. I didn't spot that it was a line either, I said that the graph was [latex]increasing for f(x)[/latex]

But I'm not going to whinge that it was unfair or misleading, I mean when I look back at it it's so obvious!:P

Apparently the whole line/curve thing comes up a lot in parametric equations...really wished I'd looked at those past papers more closely now!

NumberZeroNine

Okay, so I'm going to sleep now but if anyone who feels that they did really well on a certain bit please copy this with that bit filled in.

It'd be nice to have a complete solution for people to look at.
When we're done we can post them in a seperate thread for clarity and easy-access.






Question 1 (A)
x^2 -6x+t = (x+k)^2
x^2 -6x+t = x^2+2kx+k^2

=> -6=2k , t=k^2
=> k+-3 , t=9

Question 1 (B)
Real Roots, => b^2-4ac > 0
(-4p-1)^2 -4.2p > 0
16p^2 +8p +1 -8p > 0
16p^2 +1 > 0
True, as perfect square plus positive number must > 0

Question 1 (C)

(x-2) factor f(x)
when you long divide you get
d = -2c-4b-8
(x+1) factor f(x)
when you long divide you get
d = c-b+1

let them equal
c-b+1 = -2c-4b-8
3c = -3b-9
c = -b-3
(ii)
d = -2c-4b-8
d = -2(-b-3)-4b-8
d = -2b-2
(iii)
c=b+n, d=b+2n
-b-3=b+n, -2b-2=b+2n
2b=-n-3, 3b=-2n-2
b=(-n-3)/2, 3.(-n-3)/2=-2n-2
-3n-9=-4n-4
n=5
b=-4,c=1,d=6

Question 2 (A)


Question 2 (B)
it tells you alpha > 0, Beta > 0

so -4 isn't a possible solution, nearly didn't catch that one.

Full Solution;

A^2.B^2 = 16
A^2+B^2 = 12

(i)
(A.B)^2 = 16
A.B = rt(16) = 4 (-4 isn't a solution because of phrasing of Q)

(ii)
(A+B)^2 = A^2 +2.AB +B^2
(A+B)^2 = (A^2+B^2) +2.(AB)
(A+B)^2 = (12) + 2.(4)
(A+B) = rt(20)
(A+B) = 2.rt(5)

Question 2 (C)

Question 3 (A)

Question 3 (B)

Question 3 (C)

Question 4 (A)

Question 4 (B)

Question 4 (C)

Question 5 (A)

Question 5 (B)

Question 5 (C)

Question 6 (A)

Question 6 (B)

Question 6 (C)

Question 7 (A)

Question 7 (B)




Question 7 (C)
when diff.ing it;
let u(x)=(1+X), v(x)= ln(1+X)
=> u'(x)=1, v'(x)=1/(1+X)

f'(x) = [v(x).u'(x)+u(x).v'(x)]
f'(x) = [1.ln(1+X)+(1+X)/(1+X)]
f'(x) = [ln(1+X)+1]

f''(x) = 1/(1+X)

At turning point; f'(x) = 0

[ln(1+X)+1] =0
ln(1+X) = -1
1+X = e^-1
X = e^-1 -1
X = 1/e - e/e
X = (1-e)/e

Sub into F(x)

y = f((1-e)/e)
y = ( (e+1-e)/e ).ln( (e+1-e)/e)
y = (1/e).ln(e^-1)
y = -1/e

At local Min f''(x) 0
At local Max f''(x) 0

F''((1-e)/e) = 1/((e+1-e)/e)
= 1/(1/e)
= e
> 0
Therefore Min.

Question 8 (A)


Question 8 (B)
Limits 0-1, 1-3
A = Iy.dx(0-1) + Iy.dx(1-3)
A = [3x4-16x3+18x2](0-1) +[3x4-16x3+18x2](1-3)
A = [(5) – (0)] + [(-27) – (5)]
A = 5 + |-32|
A = 37 square units

Question 8 (C)

EuropeanSon

Q1(c) I multiplied the two factors before dividing, it made it much much easier.

muffinz

I think i passed, but barely. Did well in integration for once, that graph was easy... ballsed up question 3 completely, the one i was banking on as my favourite