Author Topic: C3 and C4 DOUBTS HERE!!!  (Read 35713 times)

Offline deadlyelder

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Re: C3 and C4 Doubts Here!!!
« Reply #90 on: June 14, 2011, 02:08:03 pm »
The Trinometry and Further Trinometry are truly nightmare. Please share tips for those. Do not have any problem with last chapter. Am solving all the practice papers that are available. How are you guys remembering the trinometry formulas? ???

Offline mdwael

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Re: C3 and C4 Doubts Here!!!
« Reply #91 on: June 14, 2011, 05:24:39 pm »
january 2007 question 4 ii

it says find the value of dy/dx:

In the end you get this..   9 x 4^0.5

this should give +18 and -18 but why in the markscheme only +18 is given? They didnt mention anything that the value is greater than 0..
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Re: C3 and C4 Doubts Here!!!
« Reply #92 on: June 15, 2011, 10:50:14 am »
Think about the ex graph. The question contains a graph that is very similar to the ex graph.

Imagine it in your head... does the gradient ever have a negative value ? Nope, its an increasing function.

Offline deadlyelder

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Re: C3 and C4 Doubts Here!!!
« Reply #93 on: June 15, 2011, 08:38:19 pm »
Good Luck to everyone for this paper. :)

Offline hmh

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Re: C3 and C4 Doubts Here!!!
« Reply #94 on: June 19, 2011, 05:16:13 pm »
anybody have any revision notes or pointers?

Here are some..
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Offline hmh

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Re: C3 and C4 Doubts Here!!!
« Reply #95 on: June 19, 2011, 05:22:36 pm »
Can anyone please do C4 January 2011 Question 7 C.
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Offline abuelzouz

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Re: C3 and C4 Doubts Here!!!
« Reply #96 on: June 19, 2011, 07:07:58 pm »
Can anyone please do C4 January 2011 Question 7 C.

i will solve this paper now and post solution

Offline abuelzouz

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Re: C3 and C4 Doubts Here!!!
« Reply #97 on: June 19, 2011, 08:04:45 pm »
Can anyone please do C4 January 2011 Question 7 C.

k so here is it
dx/du = 2(u-4)
dx= 2(u-4) du
x-1 = (u-4)^2
so root x-1 = u-4
and u= root x-1 +4

going back to equation
substitute
integral = integral 1/u & 2(u-4) du
= integral (2-8/u) du
= 2u- 8ln u
x=2 so u=5
x=5 so  u=6
substitute these values to get area
thats it hope i helped

Offline hmh

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Re: C3 and C4 Doubts Here!!!
« Reply #98 on: June 19, 2011, 08:26:36 pm »
Thanks, I just couldn't get the limits correct but now I know how to do it.
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Offline WARRIOR

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Re: C3 and C4 Doubts Here!!!
« Reply #99 on: September 19, 2011, 01:58:45 pm »
Hi guys , can anyone get me the answers of these questions ? all of them :P

thanks
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Offline iluvme

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Re: C3 and C4 Doubts Here!!!
« Reply #100 on: September 19, 2011, 02:50:39 pm »
Hi guys , can anyone get me the answers of these questions ? all of them :P

thanks

Looks like its your homework :P

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Offline WARRIOR

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Re: C3 and C4 Doubts Here!!!
« Reply #101 on: September 19, 2011, 05:05:24 pm »
Looks like its your homework :P

You've got any doubts or are all these questions your doubts?

well ill be specific in a moment of which questions i need anyway. this is important how do u solve this

find the inverse function of  f (x) =  2x + 3/x-1
NO secrets to SUCCESS , it is the result of 1.HARD WORK 2.GOOD PREPARATION 3.LEARNING FROM FAILURE
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Offline Arthur Bon Zavi

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Re: C3 and C4 Doubts Here!!!
« Reply #102 on: September 19, 2011, 05:14:24 pm »
well ill be specific in a moment of which questions i need anyway. this is important how do u solve this

find the inverse function of  f (x) =  2x + 3/x-1

Let f-1(x) be y and y, x.

y = (2x + 3) / (x - 1)
y (x - 1) = (2x + 3)
yx - y = 2x + 3
yx - 2x = 3 + y
x (y - 2) = 3 + y
x = (3 + y) / (y - 2)

Now replace the values again to get :

f-1(x) = (3 + x) / (x - 2)

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Offline WARRIOR

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Re: C3 and C4 Doubts Here!!!
« Reply #103 on: September 19, 2011, 05:50:24 pm »
Let f-1(x) be y and y, x.

y = (2x + 3) / (x - 1)
y (x - 1) = (2x + 3)
yx - y = 2x + 3
yx - 2x = 3 + y
x (y - 2) = 3 + y
x = (3 + y) / (y - 2)

Now replace the values again to get :

f-1(x) = (3 + x) / (x - 2)

thanks it turned out to be really easy !
NO secrets to SUCCESS , it is the result of 1.HARD WORK 2.GOOD PREPARATION 3.LEARNING FROM FAILURE
But it ain't how hard you hit; it's about how hard you can get hit, and keep moving forward-Balboa

Offline WARRIOR

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Re: C3 and C4 Doubts Here!!!
« Reply #104 on: September 19, 2011, 06:02:24 pm »
Let f-1(x) be y and y, x.

y = (2x + 3) / (x - 1)
y (x - 1) = (2x + 3)
yx - y = 2x + 3
yx - 2x = 3 + y
x (y - 2) = 3 + y
x = (3 + y) / (y - 2)

Now replace the values again to get :

f-1(x) = (3 + x) / (x - 2)

anothr question


what is the range and domain of the inverse that you just got out?

for  f(x) the  domain is x>1 , x E R
                 range    y  >=7
 right?

so for f^-1 = should be the opposite?
                domain y>=7
              range x>1?
NO secrets to SUCCESS , it is the result of 1.HARD WORK 2.GOOD PREPARATION 3.LEARNING FROM FAILURE
But it ain't how hard you hit; it's about how hard you can get hit, and keep moving forward-Balboa