C3 and C4 DOUBTS HERE!!!

[deadlyelder]

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?

[mdwael]

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..

[elemis]

Think about the e^x graph. The question contains a graph that is very similar to the e^x graph.

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

[deadlyelder]

Good Luck to everyone for this paper.

[hmh]

anybody have any revision notes or pointers?

Here are some..

[hmh]

Can anyone please do C4 January 2011 Question 7 C.

[abuelzouz]

Can anyone please do C4 January 2011 Question 7 C.

i will solve this paper now and post solution

[abuelzouz]

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

[hmh]

Thanks, I just couldn't get the limits correct but now I know how to do it.

[WARRIOR]

Hi guys , can anyone get me the answers of these questions ? all of them

thanks

[iluvme]

Hi guys , can anyone get me the answers of these questions ? all of them

thanks

Looks like its your homework

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

[WARRIOR]

Looks like its your homework

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

[Arthur Bon Zavi]

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)

[WARRIOR]

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 !

[WARRIOR]

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?