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c3 25th January 2013 exam

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Can't believe I couldn't do the ln 9500 question I taught the whole class how to do it the day before :frown:
Reply 221
What did you guys get for finding the rate, last question part c?
Reply 222
Original post by StrawHat_Shifaz
well, i didnt show any workings, and just wrote 2. would i get full marks?


A correct answer is a correct answer
I think you'll get full marks ....


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Original post by Samira123
What did you guys get for finding the rate, last question part c?


check page 8
Reply 224
Oh gosh.. for that IDIOTIC range question.. people are getting root 5/5 but I got 4/9 which also rounds off to 0.44....
:frown: I hope that is rightt!!

Ugh.. stupid maximum value question, stupid range question.. STUPID STUPID STUPID :frown:

Sorry for the rant... highly irritated with the paper and myself.
for the 4/12+something i got 2/11 and the max value as cos(x=0.927 something) =1 ?
Reply 226
Original post by otrivine
for the 4/12+something i got 2/11 and the max value as cos(x=0.927 something) =1 ?


I got that too :frown: apparently, that is wrong.
Original post by KalSA
I got that too :frown: apparently, that is wrong.


what impossible? like everyone in my class put that answer:confused:
Original post by otrivine
for the 4/12+something i got 2/11 and the max value as cos(x=0.927 something) =1 ?


i did the same, but, WRONG!! Join the gang!:party2:
Original post by StrawHat_Shifaz
i did the same, but, WRONG!! Join the gang!:party2:


why? cause we had the R value from above calculation and had to just put the 10 in the fraction?

4/12+10 something like that
Reply 230
So stupid I put root 5 as the range and didn't put that x value into the original eq to get the y value.
Original post by otrivine
why? cause we had the R value from above calculation and had to just put the 10 in the fraction?

4/12+10 something like that



this is why:
Well, we wanted to find the maximum value of p(θ)


p(θ) = 4 / (12 + (6cosθ + 8sinθ))

In the previous question we found out that 6cosθ + 8sinθ = 10cos(θ - 0.927)

p(θ) = 4 / (12 + 10cos(θ - 0.927))

If we want to maximise p(θ), we want the denominator of the function to be as small as possible, therefore giving a higher value of p(θ)

So we can find the minimum of 10cos(θ - 0.927) which will be 10(-1) = -10

So now we get p(θ) = 4 / (12-10) = 4 / 2 = 2

The maximum value for p(θ) = 2
Original post by Nia_Beth
What grades are people aiming for/wanting to get after that exam?


well i was aiming for 100ums but after that its likely ill only get over 85ums if im lucky. 8b) ****ed with me. maximum question ****ed with but i knew exactly how to do it but i thought the wrong of maximums and minimums .
Original post by strawhat_shifaz
this is why:
Well, we wanted to find the maximum value of p(θ)


p(θ) = 4 / (12 + (6cosθ + 8sinθ))

in the previous question we found out that 6cosθ + 8sinθ = 10cos(θ - 0.927)

p(θ) = 4 / (12 + 10cos(θ - 0.927))

if we want to maximise p(θ), we want the denominator of the function to be as small as possible, therefore giving a higher value of p(θ)

so we can find the minimum of 10cos(θ - 0.927) which will be 10(-1) = -10

so now we get p(θ) = 4 / (12-10) = 4 / 2 = 2

the maximum value for p(θ) = 2


omg ! So they tricked us!
what do u think the grade boundaries will be? I am sure majority will have got 2/11 wrong well everyone in my school got 2/11.
Original post by otrivine
what do u think the grade boundaries will be? I am sure majority will have got 2/11 wrong well everyone in my school got 2/11.


59-62 for A? :colone:
Original post by otrivine
omg ! So they tricked us!


Yeah, that was a pretty tricky question, I almost fell for it myself. Good thing I noticed before I was done :beard:
Original post by StrawHat_Shifaz
59-62 for A? :colone:


That would be so good :biggrin:

The question on x=coty

does my method work i converted the coty in the form of cos/sin and applied the quotient rule and used the sin^2x+cos^2=1 formula/

Also, i used another method where i differentiated the coty and mentioned the 1+cot e.t.c formula
Original post by StrawHat_Shifaz
this is why:
Well, we wanted to find the maximum value of p(θ)


p(θ) = 4 / (12 + (6cosθ + 8sinθ))

In the previous question we found out that 6cosθ + 8sinθ = 10cos(θ - 0.927)

p(θ) = 4 / (12 + 10cos(θ - 0.927))

If we want to maximise p(θ), we want the denominator of the function to be as small as possible, therefore giving a higher value of p(θ)

So we can find the minimum of 10cos(θ - 0.927) which will be 10(-1) = -10

So now we get p(θ) = 4 / (12-10) = 4 / 2 = 2

The maximum value for p(θ) = 2


oh gosh... i put the value 1 for cos(theta - 0.927). do you remember what was the mark for this question, plus the part after this question altogether?
Original post by justinawe
Yeah, that was a pretty tricky question, I almost fell for it myself. Good thing I noticed before I was done :beard:


Yes, and the other one did u put cos(x-0.927something)=1 for max of deta

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