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# Stuck on a C2 question watch

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1. Hey guys, I'm stuck on a question while doing C2 A3 Practice Paper.
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
2. So can you show some working?
3. (Original post by Pherix)
Hey guys, I'm stuck on a question while doing C2 A3 Practice Paper.
I assume that you can do the first part

What did you try for the second part
4. 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
5. (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
6. I expanded [2x^(3/2) - 3x^(-3/2)]2 and brought the 5 from the LHS to the RHS.
7. (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
8. 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?
9. (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?

No I was referring to the missing bracket in your OP which changed the question A lot
10. (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?
Multiply by x^3 and solve the quadratic
11. 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.
12. (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.
13. I still don't get it. My brain's not really processing things well at the moment.
14. (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?
15. AHHHHHHHHHHHHHHHHH. Thank you so much!
Finally manage to solve it. I feel like an idiot now.
Thanks again!

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Updated: December 31, 2012
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