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?