Stuck on a C2 question
Hey guys, I'm stuck on a question while doing C2 A3 Practice Paper.
Please give me some tips on solving em. Thanks in advance
Please give me some tips on solving em. Thanks in advance
Given that f(x) = [2x^(3/2) - 3x^(-3/2)2 +5, x>0,
(a) find, to 3 s.f., the value of x for which f(x) =5.
(b) Show that f(x) may be written in the form Ax3 + B/x3 + C, where A, B and C are constants to be found.
(c) Hence evaluate ∫f(x),1,2 dx
(a) find, to 3 s.f., the value of x for which f(x) =5.
(b) Show that f(x) may be written in the form Ax3 + B/x3 + C, where A, B and C are constants to be found.
(c) Hence evaluate ∫f(x),1,2 dx
So can you show some working?
Original post by Pherix
Hey guys, I'm stuck on a question while doing C2 A3 Practice Paper.
Please give me some tips on solving em. Thanks in advance
Please give me some tips on solving em. Thanks in advance
I assume that you can do the first part
What did you try for the second part
I'm actually stuck on part A.
f(x) = = [2x^(3/2) - 3x^(-3/2)]2 +5
[2x^(3/2) - 3x^(-3/2)]2 + 5 = 5
4x^3 + 9x^(-3) -12 = 0 <-------(I'm not sure if this is related to part A or not but I'm stuck here for A)
4x^3 + 9/x^(3) -12 = 0
A=4, B=9, C=-12
As for part C, I manage to get 11.375 after integrating it.
Still stuck on part A
f(x) = = [2x^(3/2) - 3x^(-3/2)]2 +5
[2x^(3/2) - 3x^(-3/2)]2 + 5 = 5
4x^3 + 9x^(-3) -12 = 0 <-------(I'm not sure if this is related to part A or not but I'm stuck here for A)
4x^3 + 9/x^(3) -12 = 0
A=4, B=9, C=-12
As for part C, I manage to get 11.375 after integrating it.
Still stuck on part A
Original post by Pherix
I'm actually stuck on part A.
f(x) = = [2x^(3/2) - 3x^(-3/2)]2 +5
[2x^(3/2) - 3x^(-3/2)]2 + 5 = 5
How did you go from there to
4x^3 + 9x^(-3) -12 = 0 <-------(I'm not sure if this is related to part A or not but I'm stuck here for A)
4x^3 + 9/x^(3) -12 = 0
A=4, B=9, C=-12
As for part C, I manage to get 11.375 after integrating it.
Still stuck on part A
f(x) = = [2x^(3/2) - 3x^(-3/2)]2 +5
[2x^(3/2) - 3x^(-3/2)]2 + 5 = 5
How did you go from there to
4x^3 + 9x^(-3) -12 = 0 <-------(I'm not sure if this is related to part A or not but I'm stuck here for A)
4x^3 + 9/x^(3) -12 = 0
A=4, B=9, C=-12
As for part C, I manage to get 11.375 after integrating it.
Still stuck on part A
Original post by Pherix
I expanded [2x^(3/2) - 3x^(-3/2)]2 and brought the 5 from the LHS to the RHS.
Aaaaaaah
That was what you missed in the OP
Original post by Pherix
I missed? I thought it's okay to expand them mentally. Been taught to do so
Anyway, how do I get to solve for part A?
Anyway, how do I get to solve for part A?
No I was referring to the missing bracket in your OP which changed the question A lot
Original post by Pherix
I missed? I thought it's okay to expand them mentally. Been taught to do so
Anyway, how do I get to solve for part A?
Anyway, how do I get to solve for part A?
Multiply by x^3 and solve the quadratic
Original post by Pherix
Ahhhhhh. I just notice that I forgot to add that particular bracket. My bad 
Anyway, multiplying x^3 would give me 4x^6 - 12x^3 = -9
I just can't solve it now. I'm confuse now.
Anyway, multiplying x^3 would give me 4x^6 - 12x^3 = -9
I just can't solve it now. I'm confuse now.
It is a quadratic
Original post by Pherix
I still don't get it. My brain's not really processing things well at the moment.
Make a substitution . Can you see the quadratic now?
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