Post your Maths Answers (higher)

SimpleLogic

No doubt I got lots wrong, so correct me please. Some answers arent done as they are proofs

Q1 A) 2/x-2


1/3, 3/2 ,3


Q2 A) [x+5]^2 + 7

i) 47
ii) 322 not sure about this sounds abit strange

Q3 A) -7 2
5 0

B)1-i

4 A) 3 nearly sure this is wrong

5 A) -2<x<5

ii) -5/2 +- 3^.5/2

6 A) -3/2 times square root X

C) -12

7 A) 2 + 2cos2x

-10x-6y/10y+6x

8 A) x^2 + sin3x/3 + C

B i) e-1
ii) 12 - loge5

C i) 6

Fuascailt

Burner- wrote: »

No doubt I got lots wrong, so correct me please. Some answers arent done as they are proofs

Q1 A) 2/x-2yes


1/3, 3/2 ,3yes


Q2 A) [x+5]^2 + 7yes

i) 47
ii) 322 not sure about this sounds abit strangeI got this as well! did think it was a bit weird

Q3 A) -7 2
5 0as far as i can remember, yes

B)1-i yup

4 A) 3 nearly sure this is wrong

5 A) -2<x<5

ii) -5/2 +- 3^.5/2

6 A) -3/2 times square root Xyup

C) -12i got -36, but i think -12 is right, hopefull just a slip

7 A) 2 + 2cos2xyes, thought that was suspiciously short

-10x-6y/10y+6xyes

8 A) x^2 + sin3x/3 + C yes

B i) e-1 yes
ii) 12 - loge5yes

C i) 6

This has cheered me up a little:)

Yuugib

how the hell are you getting 12 -log 5?
i have (1/4) log 5

tabouli

Q1 a) 6/(x+2)(x-2)

b)x= 3, 3/2, 1/3

Q2) a) (x+5)^2

b)i) 47
ii) 322

3) a) (7 -2)
(-5 0)

b)i) 1-i
ii) 8i

4)a) 3

c)ii) 2n^3 +4n^2 +7n +2^n -1

5)a) -2<x<5

b)i) x= -27/4 (? not a clue)

6)a) 3rootx/2

c)i) b=-12
ii) -20<c<7

7)a) 2 +2cos2x

b)i) -5x-3y/5y+3x

c) 1/1+x^2

8)a) x^2 +sin3x/3 (I may have left out the c... )

b)i) e-1

ii) 120 (think this is very wrong )

c) 16/3

LC90

Burner- wrote: »

No doubt I got lots wrong, so correct me please. Some answers arent done as they are proofs

Q1 A) 2/x-2 Same


1/3, 3/2 ,3 Same



Q2 A) [x+5]^2 + 7 Same

i) 47 Same
ii) 322 not sure about this sounds abit strange Same

Q3 A) -7 2
5 0

B)1-i

4 A) 3 nearly sure this is wrong

5 A) -2<x<5

ii) -5/2 +- 3^.5/2

6 A) -3/2 times square root X Don't think there was a minus

C) -12 Same

7 A) 2 + 2cos2x Same

-10x-6y/10y+6x Same

8 A) x^2 + sin3x/3 + C Ah,ffs I put the 3 in front of the Sin

B i) e-1 Same
ii) 12 - loge5 Went with a sneaky misreading

C i) 6

.

mahamageehad

tabouli wrote: »

Q1 a) 6/(x+2)(x-2) Yippee same!!:cool:

b)x= 3, 3/2, 1/3 i got x=3, got no more
Q2) a) (x+5)^2 same...tout it was 2 easy2 b tru!!

b)i) 47 Ya ur right, but i messed it up and put a + in formula:mad:
ii) 322

3) a) (7 -2) hmm..i got da same nos as ya, but all positive except 7
(-5 0)

b)i) 1-i same
ii) 8i same...would it matter much if i left out the i tho??

4)a) 3

c)ii) 2n^3 +4n^2 +7n +2^n -1

5)a) -2<x<5

b)i) x= -27/4 (? not a clue)

6)a) 3rootx/2 right again, but i messed that one up as well

c)i) b=-12 yes!!
ii) -20<c<7

7)a) 2 +2cos2x yippee!!!

b)i) -5x-3y/5y+3x same but i didnt cancel em

c) 1/1+x^2

8)a) x^2 +sin3x/3 (I may have left out the c... ) YAH!!!

b)i) e-1

ii) 120 (think this is very wrong )

c) 16/3

whew...still tink im a borderline fail. does any1 no how much attempt marks are worth????

eoin2nc

For Q3bii i stupidly wrote 8 instead of 8i. How many marks would I lose?

mahamageehad

eoin2nc wrote: »

For Q3bii i stupidly wrote 8 instead of 8i. How many marks would I lose?

me too!!!!! :mad::mad::mad: i am very angry at myself. im hopin dey wont mark down much for it coz i had the i the whole way down except in da last line!!!:(

Davidius

eoin2nc wrote: »

For Q3bii i stupidly wrote 8 instead of 8i. How many marks would I lose?

3 I think. Could be 1 though.

Peleus

eoin2nc wrote: »

For Q3bii i stupidly wrote 8 instead of 8i. How many marks would I lose?

if the exmainer thinks you knew it should be 8i but accidently left out the i, its -1 mark. if he thinks you thought it should be 8, he'll probably mark you down 3 (blunder)

depends how you wrote down your answer.

TheBigRedDog

For question 2c

Is (a^2 + b^2) = (a + b)^2 - 2ab

giving you 7^2 - 2(1) = 45???

florencexxx

49-2 = 47

Am I the only one that got 2-i for Q3(b)? Eep.
Other than that they all look pretty familiar, so here's hoping...=]

ShogunWarrior

Hey just wondering about Q6 (a), I keep doing it and by the chain rule it makes sense to me as 3x^2 / (sqrt(x^3))

Attached solution if someone would take a look.

Peslo

What was the story with there being no part (i) or (ii) in b or c of Q1!!!!!!! That was bang out of order! I mean, they've ALWAYS done that and they have done with EVERY other question on the paper!
If they had of broken up 1c into to parts, i think more people could have managed it.

less is more

1.(b) I guessed the root of 3. How were you supposed to get it? By inspection?

2.(c) was a weird question, but I think I wrote a passable answer. There was hardly any maths involved, mostly using common sense - I wrote something like:

(a^2 + b^2)/(ab) is always >2 when the signs are the same. (try it yourself with some values)

and a/-b + -b/a = -(a^2 + b^2)/(ab) is always <-2 for the same reason, but the outer minus changes the sign

Therefore, value never lies between -2 and 2

3.(c) Did no one else include the 1/11? I remember we had to premultipy by A^-1 [ie 1/11(forgot the matrix no.s )] to both sides

5. was ridiculously easy, but made up for it in part (c)! If I had more time I probably could have figured it out, but I did all 8 questions so I ran out.

6. I found fairly ok, the diff of two squares bit threw me in (b), but got the correct answer in the end

7. (a) I went a bit further and ended up with 4cos^2 x, using the theorems book. Hopefully I won't lose marks...

8. (b)(ii) I could not for the life of me figure this out! It's a quotient with no theorem on p41, substitution didn't work, integrating straight away didn't work!

(c) I got weird fractions for the various sections of the shaded area (like 47/24!!), but all added up to 19, which greatly surprised me! Hopefully that's correct and not just a cruel coincidence

NanoZoom555

less is more wrote: »

1.(b) I guessed the root of 3. How were you supposed to get it? By inspection?


3.(c) Did no one else include the 1/11? I remember we had to premultipy by A^-1 [ie 1/11(forgot the matrix no.s )] to both sides


8. (b)(ii) I could not for the life of me figure this out! It's a quotient with no theorem on p41, substitution didn't work, integrating straight away didn't work!


for q.1 b) you sub in values untill f(x) =0

for q.3c) where are you getting 1/11 - its induction (f(1),f(k),f(k+1))

for q.8 b)ii) let u = x^2 - 1

rewrite 2x^3 as 2x(x^2) and get du and just sub in so that its in terms of u and integrate and solve

hope that helps!

ShogunWarrior

Thankfully Q1 was actually my best I think, unless I made a slip somewhere in there.

I've attached (PDF) my solution to Q1. (c)

I can do out other solutions if anyone wants for discussion or otherwise.

Peslo

NanoZoom555 wrote: »

hope that helps!

Bit late now! lol

ShogunWarrior

Here's attached (PDF) Q1. (b) for anyone who wants (what I hope is) the solution.

less is more

NanoZoom555 wrote: »

less is more wrote: »

1.(b) I guessed the root of 3. How were you supposed to get it? By inspection?


3.(c) Did no one else include the 1/11? I remember we had to premultipy by A^-1 [ie 1/11(forgot the matrix no.s )] to both sides


8. (b)(ii) I could not for the life of me figure this out! It's a quotient with no theorem on p41, substitution didn't work, integrating straight away didn't work!


for q.1 b) you sub in values untill f(x) =0

for q.3c) where are you getting 1/11 - its induction (f(1),f(k),f(k+1))

for q.8 b)ii) let u = x^2 - 1

rewrite 2x^3 as 2x(x^2) and get du and just sub in so that its in terms of u and integrate and solve

hope that helps!

Sorry I meant 3. (a)
Thanks for your help with the other questions though, I CANNOT believe I didn't think of subbing values in for 1.(b)! Just shows what stress will do to ya

Peslo

What was the story with there being no part (i) or (ii) in b or c of Q1!!!!!!! That was bang out of order! I mean, they've ALWAYS done that and they have done with EVERY other question on the paper!
If they had of broken up 1c into to parts, i think more people could have managed it.

Timans

Paper really pissed me off.

No binomial or induction in the binomial/induction question? Taking the piss.

jennyq

Induction did come up in q.3 & you did have to know the rules for (n r) as regards binomial in fairness.

Did aaaanybody else at all do 5(c) :rolleyes: Here's my solution from the other thread, I'd like to know if anybody else did something different or the same though, doesn't seem like many did it.

http://www.boards.ie/vbulletin/showpost.php?p=56160914&postcount=81

Adventure

For question 8 (b)(ii) any one get

16+ln5 ??

orangetictac

Adventure wrote: »

For question 8 (b)(ii) any one get

16+ln5 ??

I think i got 12 +ln5

Cokehead Mother

jennyq wrote: »

Induction did come up in q.3 & you did have to know the rules for (n r) as regards binomial in fairness.

Did aaaanybody else at all do 5(c) :rolleyes: Here's my solution from the other thread, I'd like to know if anybody else did something different or the same though, doesn't seem like many did it.

http://www.boards.ie/vbulletin/showpost.php?p=56160914&postcount=81

Yep. Did it the exact same way and got n = -1 like you. Messiest question ever.

Decerto

for question 2(c) did anyone simplify a/b +b/a to a^2 +b^2/ab and say that is > then 2 and then multiply across and u end up with (a+b)^2> 0 which proves its true and do tht again with a minus sign

Peleus

ShogunWarrior wrote: »

Hey just wondering about Q6 (a), I keep doing it and by the chain rule it makes sense to me as 3x^2 / (sqrt(x^3))

Attached solution if someone would take a look.

ye thats right. everyone else got (3/2)rootX
thats equal to your answer: (3x^2)/(root(x^3)). so its right just in a different form.

Cokehead Mother

Peleus wrote: »

i dont think you're allowed to let u=x^3. because then you are not including the cube in the differentiation.

What do you mean? You let u = x^3, so du/dx = 3x^2 and then you multiply that by 1/2rootu.

The quick way is to you use your junior cert knowledge of indices and then differentiate. It's just x to the power of 3 halves.

dyle123

Cokehead Mother wrote: »

Yep. Did it the exact same way and got n = -1 like you. Messiest question ever.

I did something similar.

I got the common ratio twice and showed that they were not equal and so it could not be a GP if they were consecutive terms.

Timans

Cokehead Mother wrote: »

Yep. Did it the exact same way and got n = -1 like you. Messiest question ever.

I did the same thing but got n = -2.

I'd say I'll get a good 15/20 though.

Did everyone change (n) to (n)(n-1)
(r) (r)(r-1) etc etc?

Nanaki

Cokehead Mother wrote: »

What do you mean? You let u = x^3, so du/dx = 3x^2 and then you multiply that by 1/2rootu.

The quick way is to you use your junior cert knowledge of indices and then differentiate. It's just x to the power of 3 halves.

I wrote it ax x^(3/2)

orangetictac

What did y'all get for 4c)???

orangetictac

tabouli wrote: »

4)a) 3

c)ii) 2n^3 +4n^2 +7n +2^n -1

I think i got 2n^3 +4n^2 +7n +2(2^n -1)
???

King Ludvig

Q1(a) 2/(2-x)

Q2(a) (x+5)^2 +7
(b)(i) 47

Q3(b)(i) (1-i)

Q5(a) -2<x<5
(b)(i) x=-3 and x=9
(ii) x= -2 and x=11

Q6(a) (1/2)(x)^-3/2
(c)(i) -36

Q7(a) 2+2cos2x
(b)(i) (-5x-3y)/(3x+5y)

Q8(a) 2+(sin3x)/3
(b)(i) e-1
(c) 22.92 units squared

tabouli

orangetictac wrote: »

I think i got 2n^3 +4n^2 +7n +2(2^n -1)
???

Not sure, I tried to use Sn of a GP for that part, but got confused.

Calorimeterman

question 8 c, is around about 10 units, I drew it out, and I can draw graphs well, I got 6 in the calculation, but by counting the squares I managed to get it to about 10.5 units (give or take)

orangetictac

well i got 32/3 so ur not 2 far off i guess

Cunning

a/b+b/a = (a^2+b^2)/(ab)

now let b/a=C so that b=C.a , substitute C.a for b

now we get (a^2+C^2.a^2)/(a.C.a)

cancel the a^2 above and below the lines to get
(1+C^2) / C

let this = X, which is the anser to the sum a/b+b/a

now X.C = 1+C^2

rearrange for a quadratic
C^2 - C.X + 1 = 0


using the standard formula C= -X +/- sqrt(X^2 -4) all over 2

Remeber that b=C.A and that a and/or b must NOT be complex
now look at the parts under the square root
in order for C not to be a complex/imaginary number X^2 > 4
thus X must be < -2
or X must be > 2

nice
B2 in LC pass maths 2001

padkav

you could also get the answer in the form 3/2*square Root x, same as yours!

Peleus

Cunning wrote: »

rearrange for a quadratic
C^2 - C.X + 1 = 0


using the standard formula C= -X +/- sqrt(X^2 -4) all over 2

you got it right but [C^2 - C.X + 1 = 0] isn't a quadratic. you're missing an X squared.

Peleus

Cokehead Mother wrote: »

What do you mean? You let u = x^3, so du/dx = 3x^2 and then you multiply that by 1/2rootu.

The quick way is to you use your junior cert knowledge of indices and then differentiate. It's just x to the power of 3 halves.

ye, sorry you can use substitution. i just thought you couldn't, in a moment of silliness.

t-mobile1892

Woah im in JC nd im just lost readin dis......

Cokehead Mother

Peleus wrote: »

you got it right but [C^2 - C.X + 1 = 0] isn't a quadratic. you're missing an X squared.

Equations can have variables other than x!

fivetwenty

BTW - Did everyone here reach a stage of depression in this exam?

Mine came at the way too early 1(B) - I tried f(0),f(1),f(-1),f(2),(f-2), and said "**** this - It's never went this high up and made sense" . . . Lo and behold, 3 suddenly became my favourite number

Cokehead Mother wrote: »

Yep. Did it the exact same way and got n = -1 like you. Messiest question ever.

Yes! I did the same, except forgot to finish it off, I just realised no values they wanted would work, so stupidly assumed the examiner would!

ZorbaTehZ

Cunning, how can you use the quadratic formula when you have two variables in there, if you write it as an explicit function you can see that it isn't a quadratic. Plus I don't think you're taking into account the situation where a and b have opposite signs.

Peslo

Peleus wrote: »

[C^2 - C.X + 1 = 0] isn't a quadratic. you're missing an X squared.

Is this guy doing HL LC Maths??

For pete's sake loike.

Peslo

fivetwenty wrote: »

BTW - Did everyone here reach a stage of depression in this exam?

Mine came at the way too early 1(B) - I tried f(0),f(1),f(-1),f(2),(f-2), and said "**** this - It's never went this high up and made sense" . . . Lo and behold, 3 suddenly became my favourite number



Yes! I did the same, except forgot to finish it off, I just realised no values they wanted would work, so stupidly assumed the examiner would!

You are supposed to use one of the factors of the term indepdant of x!!!!
it was 9, so 3 should have been the first one you try! if not after 1 and -1 like!

Cokehead Mother

ZorbaTehZ wrote: »

Cunning, how can you use the quadratic formula when you have two variables in there, if you write it as an explicit function you can see that it isn't a quadratic. Plus I don't think you're taking into account the situation where a and b have opposite signs.

I don't understand what you mean. He's just solving a quadratic in terms of a/b. Why can't you do that?

His solution works on the basis of a and b being any real numbers so I think he's got the possibility of them having opposite signs covered.

fivetwenty

Peslo wrote: »

You are supposed to use one of the factors of the term indepdant of x!!!!
it was 9, so 3 should have been the first one you try! if not after 1 and -1 like!

Ooooh, Ah well, s/he can't take anything off me now!

ZorbaTehZ

Well if you have a function like this x^2 + xy - 3 = 0, could you use the quadratic equation on that? You need to write it explicitly, ie y = (something) - ie it would need to look like y = ax^2 + bx + c. If you think about an equation like cunning described, it wouldn't even look like a typical quadratic, explicitly it would be something like x = 1/C + C, which I think would be asymptotic.

Regarding the as and bs, if they are of opposite signs, then you have negative on the left -a/b-b/a, there's a solution given either on this thread or the other one which takes into account.