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Author Topic: C3 and C4 DOUBTS HERE!!!  (Read 36146 times)

Offline T.Q

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C3 and C4 DOUBTS HERE!!!
« on: May 05, 2010, 10:28:37 am »
POST YOUR C3  DOUBTS  HERE
« Last Edit: November 11, 2011, 01:34:47 pm by Crooky :P »
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Offline T.Q

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Re: C3 and C4 Doubts Here!!!
« Reply #1 on: May 05, 2010, 10:29:31 am »
POST YOUR C4 DOUBTS HERE
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Offline Saladin

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Re: C3 and C4 Doubts Here!!!
« Reply #2 on: May 05, 2010, 10:32:32 am »
Good job man, this way, we can understand what they want!
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Offline pjb13

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Re: C3 and C4 Doubts Here!!!
« Reply #3 on: May 28, 2010, 02:11:38 pm »
1) Given h(x) = (x+1)(x-3). How do I know that h is differentiable on (-1,3) ?

2)How close to 2 must x be for the values of g(x) = x^2 to be within 0.2 from 4?

I know how to solve most of this...I reach an answer of -0.051<x-2<0.049
Then it says to insure both ends of the last compound inequality are satisfied take -0.049<x-2<0.049
My question is why cant we take -0.051<x-2<0.051....how do I know which one to take?

Thanx
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Offline Saladin

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Re: C3 and C4 Doubts Here!!!
« Reply #4 on: May 28, 2010, 03:32:58 pm »
Quote from: pjb13 on May 28, 2010, 02:11:38 pm
1) Given h(x) = (x+1)(x-3). How do I know that h is differentiable on (-1,3) ?

2)How close to 2 must x be for the values of g(x) = x^2 to be within 0.2 from 4?

I know how to solve most of this...I reach an answer of -0.051<x-2<0.049
Then it says to insure both ends of the last compound inequality are satisfied take -0.049<x-2<0.049
My question is why cant we take -0.051<x-2<0.051....how do I know which one to take?

Thanx

Where did you get this question from?
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Offline pjb13

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Re: C3 and C4 Doubts Here!!!
« Reply #5 on: May 29, 2010, 09:01:17 am »
Ah just realized tht its not required...extra stuff given by my school  >:(
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Offline Saladin

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Re: C3 and C4 Doubts Here!!!
« Reply #6 on: June 10, 2010, 06:38:17 am »
This exam is next!
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Offline T.Q

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Re: C3 and C4 Doubts Here!!!
« Reply #7 on: June 10, 2010, 11:46:15 am »
Quote from: Engraved on June 10, 2010, 06:38:17 am
This exam is next!

gd luck  :)

i have done c3 in jan
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Offline sweetest angel

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Re: C3 and C4 Doubts Here!!!
« Reply #8 on: June 11, 2010, 06:25:20 am »
can anyone xplain to me Q 4 in June 2005 pleeaassee?? i got the ms but i dnt get the second step.
Thanks :)
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Offline vanibharutham

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Re: C3 and C4 Doubts Here!!!
« Reply #9 on: June 11, 2010, 06:52:48 am »
For Q4 June 2005,

Integration by substitution...

x = sin A
x² = sin²A

1 - x² = 1 - sin²A

And since sin²A + cos²A = 1,

1 - sin²A = 1 - x² = cos²A

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Offline sweetest angel

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Re: C3 and C4 Doubts Here!!!
« Reply #10 on: June 11, 2010, 09:39:42 am »
umm ya i got that part its the part where the 1/(cos^2A)^3/2 X cosA is made into 1/cos^2A

isn't it supposed to be 1/cos^-1/2A?
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Offline sweetest angel

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Re: C3 and C4 Doubts Here!!!
« Reply #11 on: June 11, 2010, 09:43:08 am »
Jan '06 4-b :show did they do the last step??

i got till 1/8cos(arcsin(x/4)) but the next step is magic to me :/
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Offline astarmathsandphysics

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Re: C3 and C4 Doubts Here!!!
« Reply #12 on: June 11, 2010, 09:43:49 am »
1/(cos^2A)^3/2 X cosA =1/cos^3 A cosA= 1/cos^2 A
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Offline sweetest angel

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Re: C3 and C4 Doubts Here!!!
« Reply #13 on: June 11, 2010, 10:02:38 am »
Quote from: astarmathsandphysics on June 11, 2010, 09:43:49 am
1/(cos^2A)^3/2 X cosA =1/cos^3 A cosA= 1/cos^2 A

thank u thank u thank u!!! :)
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Offline astarmathsandphysics

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Re: C3 and C4 Doubts Here!!!
« Reply #14 on: June 11, 2010, 10:11:52 am »
x=4 sin(2y+6)
differentiate wrt y
\frac {dx}{dy} =8 cos(2y+6)
\frac {dy}{dx}=\frac {1}{8 cos(2y+6)} =\frac {1}{8 sqrt(1-sin^2 (2y+6))} =\frac {1}{8(1-(x/4)^2))
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