Exam 1 - How did you all go?

[hard]

Admin edit: This seemed to be the most populous thread, so I'm making it the proper disc. thread.

[KeyMan]

Q1)
a) log_e(x) + 1
b) 1/(2pi^2 + 4pi + 2)

Q2)
a)-1/2log_e|1-2x| + c
b) 23/3

Q3)
3/(x+4)

Q4)
pi/6, 2pi/3

Q5)
a) 1/12
b) 1/3
c) 1/4

Q6) 1/12pi mm/min

Q7)
a) 2/3
b) 1.2

Q8)
k = e^a(a-1)

Q9)
x = 3/2

Q10)
401/200

These are my answers. Sorry about the formatting, don't know how to use latex ><

[avram_grant]

yep, got the same answers as you

[sdhains]

Q1)
a) log_e(x) + 1
b) 1/(2pi^2 + 4pi + 2)

Q2)
a)-1/2log_e|1-2x| + c
b) 23/3

Q3)
3/(x+4)

Q4)
pi/6, 2pi/3

Q5)
a) 1/12
b) 1/3
c) 1/4

Q6) 1/12pi mm/min

Q7)
a) 2/3
b) 1.2

Q8)
k = e^a(a-1)

Q9)
x = 3/2

Q10)
401/200

These are my answers. Sorry about the formatting, don't know how to use latex ><

I think 9 was x=3/2, -1. -1 was defined. (I Only had 3/2 but someone was saying that -1 was defined)
and 10 i got something different i think. But other than that looks good.

[unknown12]

Yep can confrim them and 10 b) you had to state that;

As the value of the gradient was positive at the point x (, the tangent is always above the graph and therefore using the linear approximation method which takes the value of the tangent the approximate value will always have a larger y value.

[avram_grant]

you couldnt have -1, since loge(-1) isnt defined

[sdhains]

you couldnt have -1, since loge(-1) isnt defined

awesome .
If I got 37/40 is it still possible to get 40 raw with A/A+ SACS?

[THem]

What was question 3 again? Short term memory ;|
nvm it was find the inverse

[hard]

isn't q 5: 1/16 since 1/4 * 1/4 = 1/16 :S

[KeyMan]

No, the ball wasn't replaced so it was 1/4 * 1/3

[hard]

FMLLLLLLLLLLLLLLLLL

[Murph#3]

What was question 2b again?

[saymun]

Q2)
a)-1/2log_e|1-2x| + c
b) 23/3

for Q2a, it said find AN anti-derivative...doesnt that mean yyou dont need C?

[KeyMan]

Evaluate the integral from 0 to 4 i think... forgot ><

[avram_grant]

yeh you dont need c