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Author Topic: C1 and C2 DOUBTS HERE!!!!!  (Read 97533 times)

Offline Saladin

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Re: C2 DOUBTS HERE!!!!!
« Reply #120 on: May 28, 2010, 09:04:05 pm »
Quote from: halosh92 on May 28, 2010, 08:26:04 pm
if u look at the marking scheme the curve is staring from 0.5 not from 0

exactly. I didn't get it either....
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Offline Darkstar3000

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Re: C2 DOUBTS HERE!!!!!
« Reply #121 on: May 30, 2010, 10:19:30 am »
the function f, defined for x is an element of R, x>0, is such that

F'(x)=x^2 - 2 + 1/x^2

a) Find the value of f''(x) at x=4

b) Given that f(3) = 0, find f(x)

c) Prove that f is an increasing function

my problem is part c i can do the first two, this is from Mixed exercise 9D Q3 in the c2 text book


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Offline Saladin

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Re: C2 DOUBTS HERE!!!!!
« Reply #122 on: May 30, 2010, 10:29:53 am »
Quote from: Darkstar3000 on May 30, 2010, 10:19:30 am
the function f, defined for x is an element of R, x>0, is such that

F'(x)=x^2 - 2 + 1/x^2

a) Find the value of f''(x) at x=4

b) Given that f(3) = 0, find f(x)

c) Prove that f is an increasing function

my problem is part c i can do the first two, this is from Mixed exercise 9D Q3 in the c2 text book




On the case!

Enjoy.
« Last Edit: May 30, 2010, 10:34:09 am by Engraved »
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Offline Darkstar3000

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Re: C2 DOUBTS HERE!!!!!
« Reply #123 on: May 30, 2010, 12:52:58 pm »
thx but where is (x - 1/x)^2 coming from
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Offline Saladin

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Re: C2 DOUBTS HERE!!!!!
« Reply #124 on: May 30, 2010, 12:59:26 pm »
Quote from: Darkstar3000 on May 30, 2010, 12:52:58 pm
thx but where is (x - 1/x)^2 coming from

That is the factorized version of f(x).
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Offline Darkstar3000

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Re: C2 DOUBTS HERE!!!!!
« Reply #125 on: May 30, 2010, 01:03:43 pm »
Quote from: Engraved on May 30, 2010, 12:59:26 pm
That is the factorized version of f(x).

thank you
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Offline Saladin

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Re: C2 DOUBTS HERE!!!!!
« Reply #126 on: May 30, 2010, 01:16:34 pm »
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Offline halosh92

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Re: C2 DOUBTS HERE!!!!!
« Reply #127 on: May 30, 2010, 04:16:19 pm »
edexcel
hey
june 2005
Q10b
i got the gradient but i cant seem to do whats next , could someone do it ?
thx  :D
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Offline cooldude

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Re: C2 DOUBTS HERE!!!!!
« Reply #128 on: May 30, 2010, 04:51:51 pm »
Quote from: halosh92 on May 30, 2010, 04:16:19 pm
edexcel
hey
june 2005
Q10b
i got the gradient but i cant seem to do whats next , could someone do it ?
thx  :D

y'=2-(16/x^3)
the minimum point is-->
2-(16/x^3)=0
8=x^3
x=2
we can show it is minimum by finding the second derivate-->
y''=48/x^2
f''(2)=12
therefore since the second derivate is positive it is a minima
therefore x>2
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Offline halosh92

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Re: C2 DOUBTS HERE!!!!!
« Reply #129 on: May 30, 2010, 05:41:04 pm »
Quote from: cooldude on May 30, 2010, 04:51:51 pm
y'=2-(16/x^3)
the minimum point is-->
2-(16/x^3)=0
8=x^3
x=2
we can show it is minimum by finding the second derivate-->
y''=48/x^2
f''(2)=12
therefore since the second derivate is positive it is a minima
therefore x>2

thxxx :)
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everyday we wake up is a miracle, then how do we say miracles dont happen?????

Offline halosh92

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Re: C2 DOUBTS HERE!!!!!
« Reply #130 on: May 31, 2010, 10:22:35 am »
jan 2005 edexcel
Q10 c
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Offline cooldude

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Re: C2 DOUBTS HERE!!!!!
« Reply #131 on: May 31, 2010, 10:29:09 am »
Quote from: halosh92 on May 31, 2010, 10:22:35 am
jan 2005 edexcel
Q10 c

erm there isnt ne 10th q for the paper uve specified  ???
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Offline Saladin

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Re: C2 DOUBTS HERE!!!!!
« Reply #132 on: May 31, 2010, 10:29:26 am »
Quote from: halosh92 on May 31, 2010, 10:22:35 am
jan 2005 edexcel
Q10 c

R u sure?

There is no question 10.
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Offline halosh92

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Re: C2 DOUBTS HERE!!!!!
« Reply #133 on: May 31, 2010, 10:38:34 am »
Quote from: Engraved on May 31, 2010, 10:29:26 am
R u sure?

There is no question 10.


oh am so sorry ;D
i meant question 9c ;D
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everyday we wake up is a miracle, then how do we say miracles dont happen?????

Offline Saladin

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Re: C2 DOUBTS HERE!!!!!
« Reply #134 on: May 31, 2010, 10:41:59 am »
Quote from: halosh92 on May 31, 2010, 10:38:34 am
oh am so sorry ;D
i meant question 9c ;D

Find the stationary point at which \frac{d^2A}{d x^2}=-ve

This is the maximum point. Now simply equate the x value found into the equation.
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