C3 and C4 DOUBTS HERE!!!

[M-H]

what for the cubic equations..?

And, i didn't want the solution. thanks tho. but i have it. what i really wanted to know how i'd solve for any kind of function i'm given

[S.M.A.T]

what for the cubic equations..?

And, i didn't want the solution. thanks tho. but i have it. what i really wanted to know how i'd solve for any kind of function i'm given

Give me a question i will show you how to do it with explanation
For different type of function there are different methods.

By the way don't worry with these type(write in terms of l(x),m(x),....etc.) of questions, at least i can assure you that
you see like you i use to worry with these type of question
ay type r question C3 exam a asa na, faw faw book a dia raksa, do the mixed exercise that is important.
ay type r question question paper a paba na not even in Solomon

[astarmathsandphysics]

Work down from the highest power of x.
Because you have  4x^2 in the question, part of the answer will be 4m(x)
Now do 4x^2 +4x -4m =4x^2+4x-4(x^2-1)=4x+4
Because you have x ihere part of the answer will be l
4x+4 =2(2x+1)+2=2l+2
4x^2+4=4m+2l+2

[M-H]

Sir, the ans is, as asif's mentioned, ml(x). They want the answers expressed only as m(x), l(x) etc.

Asif-- Doubt ase jokon, its gotta be solved. Irrespective of whether it comes in the exam or not .Anyway thanks for that info, i haven't seen the Q papers as yet. So that's removes part of my worries

[M-H]

4x(^2) +4x
=4((x^2)+x)
=4(((x+0.5)^2)-(0.5)^2)
=4((x+0.5)^2)-1
=4(((2x+1)^2)/4)-1
=((2x+1)^2)-1
=l^2(x)-1
=ml(x)

Asif, how did reach the 4th step --as in, after x+.5 it becomes 2x+1..kibhabe???

[S.M.A.T]

Asif, how did reach the 4th step --as in, after x+.5 it becomes 2x+1..kibhabe???

4((x+0.5)^2)
=4((x+(1/2))^2)
=4(((2x+1)/2)^2)
=4(((2x+1)^2)/(2^2))
=4(((2x+1)^2)/4)
=(2x+1)

If you don't understand then tell me

[astarmathsandphysics]

I will look at that.

[M-H]

Find dy/dx for-
x=1/(2t-1), y=(t^2)/(2t-1)

I don't know how to differentiate functions with reciprocals...

Thanks a bunch in advance.

[iluvme]

Find dy/dx for-
x=1/(2t-1), y=(t^2)/(2t-1)

I don't know how to differentiate functions with reciprocals...

Thanks a bunch in advance.

i) x=1/(2t-1),
   dy/dx= 1(2t-1)^-1
             =-1.(2t-1)^-1-1.(2)
             =-1.(2t-1)^-2.(2)
             =-2/(2t-1)²
 
ii) y=t²/(2t-1)
       I'm sure you know the UV method, therefore,
     
    dy/dx=2t(2t-1)-t².(2)/(2t-1)²

[S.M.A.T]

Find dy/dx for-
x=1/(2t-1), y=(t^2)/(2t-1)

I don't know how to differentiate functions with reciprocals...

Thanks a bunch in advance.

Here

[S.M.A.T]

i) x=1/(2t-1),
   dy/dx= 1(2t-1)^-1
             =-1.(2t-1)^-1-1.(2)
             =-1.(2t-1)^-2.(2)
             =-2/(2t-1)²
 
ii) y=t²/(2t-1)
       I'm sure you know the UV method, therefore,
     
    dy/dx=2t(2t-1)-t².(2)/(2t-1)²

I think that would be dx/dt 

your answer is correct  +rep

[M-H]

Asif, I was expecting your photos
Thanks, +rep!!

Iluvme- ILY
+rep

[S.M.A.T]

You welcome

[iluvme]

I think that would be dx/dt  

your answer is correct   +rep

Yeah, habit I guess.

Thank you.

[iluvme]

Iluvme- ILY
+rep

<3 ya too.
Glad to be of help.