Exam 1 - How did you all go?

[hard]

would u lose marks if you put c?

[sachinmachin]

i got 34/40... could i still get 40 raw with high b+ sac results....fml frickin stupid mistakes...

[avram_grant]

yeh man, if you pwn exam 2 you can still get 45+ raw

[littlecherry25]

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)

[krzysiek]

You don't need C, but C could be 0 so maybe you won't get punished :p

Also... for last question, i left it as 2406/1200 which is the same as 401/200, do you think I might get punished for it? I did all the other working out and showing the proper use of the formula. I hope I am alright with that.

Also, with the controversial log question... if you sub in x = -1 into them, wasn't it a positive number anyways? One was 2loge(x) which can be re-written as loge(x^2) so if you put -1 in there, turns to 1? If anyone gets what i'm on

[almostatrap]

would u lose marks if you put c?

it said "an antidirivitive" so no

[avram_grant]

i got k = ae^a - e^a

[littlecherry25]

i got k = ae^a - e^a

people you can't make the gradient equal the original function !!!

[BlueYoHo]

for that log one im pretty sure -1 was defined because the two logs that were there were:
2loge (x) = loge (x^2)                (which means ((-1)^2) is defined as that will be just positive 1)
and loge (x+3)                           for which if x =-1, it would be loge(2)

[KeyMan]

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)

f(x) = e^x + k
f'(x) = e^x
f'(a) = e^a
y = mx + c
0 = e^x(0) + c
c = 0
y = xe^a
y(a) = ae^a

simultaneous eqn ae^a = e^a + k
and then solve for k

[BlueYoHo]

You don't need C, but C could be 0 so maybe you won't get punished :p

Also... for last question, i left it as 2406/1200 which is the same as 401/200, do you think I might get punished for it? I did all the other working out and showing the proper use of the formula. I hope I am alright with that.

Also, with the controversial log question... if you sub in x = -1 into them, wasn't it a positive number anyways? One was 2loge(x) which can be re-written as loge(x^2) so if you put -1 in there, turns to 1? If anyone gets what i'm on

sorry i didn't see this post, but yes that's what i was thinking too

[sachinmachin]

for 1b, would i lose marks for leaving it as "2/(4pi^2+8pi+4)"

[BlueYoHo]

also for that question where u find the inverse function, u also have to give the domain im pretty sure
on all the prac exams ive done a Q like that is 2 marks
but because the question on this 09 exam was 3 marks i assumed it wanted something extra, for which i gave the domain
which was the range of the original function =  R\{-4}

[littlecherry25]

OH SHIT MY BAD

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


OH SHIT MY BAD

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)

f(x) = e^x + k
f'(x) = e^x
f'(a) = e^a
y = mx + c
0 = e^x(0) + c
c = 0
y = xe^a
y(a) = ae^a

simultaneous eqn ae^a = e^a + k
and then solve for k

[almostatrap]

for 1b, would i lose marks for leaving it as "2/(4pi^2+8pi+4)"

i left it factorized, it didn't ask for any particular form. i think it's fine


with the log question -1 is not valid, solve the equation with your calculator and see. its like saying (x/0).0 = x

i left -1 and I'll lose a mark, but it seems like a lot of people will have which is good