2009 EXAM 1 - HOW DID YOU FIND IT?

[sachinmachin]

LL easy enough except for the last part of last question, i just said because the h value was too large on the positive side thus causing the increase in result, any1 know if that would get the mark?

yeh i just said because it was an over estimate... im not sure if that was enough... but i guess for one mark it could be ok

yeh i said it was an under-approximate.....lol i dont even knw what i said....haha

[mystikal]

any1 realise no graph drawing? .. also last question = gay

[hard]

it was easy but i made some awful mistakes. i thought 2ln(x)=ln(x^2) always, i never got taught that it is not, argh!

WHATTTTT IT ISSSSSSSSSSSSSSSSSSSSSSSSSSSSSS ISNT IT  i thought 2ln(x)=ln(x)^2???

[tomygun_123]

it was easy but i made some awful mistakes. i thought 2ln(x)=ln(x^2) always, i never got taught that it is not, argh!

WHATTTTT IT ISSSSSSSSSSSSSSSSSSSSSSSSSSSSSS ISNT IT  i thought 2ln(x)=ln(x)^2???

it does equal that, stop stressing.... if you didnt take the 2 inside the log function you wouldnt have been able to solve it... chill bill

[kendraaaaa]

Easer than I expected, hopefully they don't get revenge on us in Exam 2. haha

[hard]

it was easy but i made some awful mistakes. i thought 2ln(x)=ln(x^2) always, i never got taught that it is not, argh!

WHATTTTT IT ISSSSSSSSSSSSSSSSSSSSSSSSSSSSSS ISNT IT  i thought 2ln(x)=ln(x)^2???

it does equal that, stop stressing.... if you didnt take the 2 inside the log function you wouldnt have been able to solve it... chill bill

OH SWEEET

[GoodGuys]

It was aiiight, however i just realised i did fucked up mistakes. ow well, its over now. Nothing i can do about it

[almostatrap]

it was easy but i made some awful mistakes. i thought 2ln(x)=ln(x^2) always, i never got taught that it is not, argh!

WHATTTTT IT ISSSSSSSSSSSSSSSSSSSSSSSSSSSSSS ISNT IT  i thought 2ln(x)=ln(x)^2???

it does equal that, stop stressing.... if you didnt take the 2 inside the log function you wouldnt have been able to solve it... chill bill

not ALWAYS. not when x=-1.

[simpak]

IT WAS TOO EASY AND NOW I'M GONNA FAIL.
h8 grade distribution.

[nels]

issues:
question 8: k=e^a(1-a) or k=e^a(a-1)?
question 9: was there two solutions? or is only 3/2 accepted?

[simpak]

Question 8 was just bad for me.

[krzysiek]

For Q9, i got two solutions... but they're probably wrong, because one was a negative - and in my state of mind, I somehow justified the negative to also be an answer. Hey, it made sense to me in the exam.. :p

Something to do with the 2loge(x) = loge(x^2) therefore a negative can be put in there because it will turn to a positive, once squared. Also, the other loge(x)'s were all positive still once putting in the negative, so for me it seemed to put so I also put a neg. I am prepared to lose a mark though on that one.

[Craxe]

QUESTION 8
YOU CAN'T MAKE THE GRADIENT EQUAL THE ORIGINAL FUNCTION !!
THE ANSWER IS e^a (a^2 - 1)

BECAUSE



Question 8
it should be
e^a (a^2-1)

because if you went
ae^2 = e^a + k
you are saying that the gradient equals f(x)

you have to find the equation of the gradient which is
y = (ae^a)x

and make that equal
y=  e^a + k

when x = a

HENCE
(ae^a) * a =  e^a + k

gives k = e^a (a^2-1)

All well and good... but what we did is make the tangent equation equal to the original function. Which is a correct method

[sachinmachin]

issues:
question 9: was there two solutions? or is only 3/2 accepted?

i personally think that x=3/2 and -1

if you sub -1 into 2loge(x), that will = loge1^2 which = 0
if you sub -1 into loge(x+2(or 3??)) that will = loge(-1+2 or 3) which is defined..

i cant see why not...thats what i put

[TonyZ]

if you type y=2loge(x) and y=loge(x^2) into your graphic calculator.. you will see that they are only the same in the first and fourth quadrants...

therefore you can change 2loge(x) to loge(x^2) only when x>0