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# Desperate C3 help!!!!!!!!!!!!! HELP! watch

1. I need help guys. I've been set about 100! C3 questions, literally! However, there are a few I just can't do. Could anybody point me in the right direction or show me how to do it.

1) Prove that: sec²x + cosec²x = sec²x . cosec²x

2) Solve: cot²2x + cosec2x -1 = 0

3) Find the x-cordinates of the turning points on the curve y = (x² + 4) / (2x - x²)

4) Differentiate and simplify y = squareroot(1-x) / squareroot(1+x)
y = squareroot(3x +2) / 2x
y = squareroot(x-5) / x

5) Find f '(4) when f(x)= x(e^squarerootx)

6) The volume, V, of a solid is given by V = x²(squarerrot(8-x)). fing the maximum value of V and the value of x at which this occurs. This one has given me nightmares, literally. How on EARTH!

It all has to be in tomorrow. AARRRGGGHHH! Maths isn't my strongest point. Thanks for any help/guidence!!!!!
2. Check that;

sec²x + cosec²x - sec²xcosec²x = 0
3. 1) sec²x + cosec²x = sec²x . cosec²x
1/cos²x + 1/sin²x = sec²x . cosec²x
(cos²x + sin²x)/sin²xcos²x = sec²x . cosec²x

as cos²x + sin²x = 1

1/sin²xcos²x = sec²x . cosec²x
4. I did things along that line, but forgot about that last part. Wow - thanks!
5. (Original post by hi there)
I need help guys. I've been set about 100! C3 questions, literally! However, there are a few I just can't do. Could anybody point me in the right direction or show me how to do it.

1) Prove that: sec²x + cosec²x = sec²x . cosec²x

2) Solve: cot²2x + cosec2x -1 = 0

3) Find the x-cordinates of the turning points on the curve y = (x² + 4) / (2x - x²)

4) Differentiate and simplify y = squareroot(1-x) / squareroot(1+x)
y = squareroot(3x +2) / 2x
y = squareroot(x-5) / x

5) Find f '(4) when f(x)= x(e^squarerootx)

6) The volume, V, of a solid is given by V = x²(squarerrot(8-x)). fing the maximum value of V and the value of x at which this occurs.

It all has to be in tomorrow. AARRRGGGHHH! Maths isn't my strongest point. Thanks for any help/guidence!!!!!
For 1, my initial thought is to try and get a common denominator on the bottom and see if you can do anything from there... it does work out.

For 2) - use the identity that to try an equation involving cot^2 and cosec^2 . Substitute that into the original, and solve. Should end up being a quadratic.

3 - Factorise top and bottom, simplify, differentiate using the quotient rule and take it from there.

4 - Differentiate using combinations of the chain rule and quotient rule, i would have thought.

5 - differentiate using a combo of the product rule and chain rule, then sub x = 4 into the equation.

6 - once again, differentiate using a combination of product rule and chain rule, then you should be able to take it from there, keeping in mind that max/min values is a C2 skill.
6. (Original post by chidona)
3 - Factorise top and bottom, simplify, differentiate using the quotient rule and take it from there.

4 - Differentiate using combinations of the chain rule and quotient rule, i would have thought.

5 - differentiate using a combo of the product rule and chain rule, then sub x = 4 into the equation.

6 - once again, differentiate using a combination of product rule and chain rule, then you should be able to take it from there, keeping in mind that max/min values is a C2 skill.
I've tried doing all of these but the maths and woking is all too complicated for me
7. 2) Solve: cot²2x + cosec2x -1 = 0

cot²x= 1 + cosec²x
therefore
cot²2x= 1 + cosec²2x

1 + cosec²2x + cosec2x -1 = 0
cosec²2x + cosec2x = 0
cosec2x(cosec2x + 1)= 0

etc
8. (Original post by hi there)
I've tried doing all of these but the maths and woking is all too complicated for me
Post your working and we can help you further, we discourage giving out full solutions on here.
9. (Original post by hi there)
3) Find the x-cordinates of the turning points on the curve
y = (x² + 4) / (2x - x²)
y(2x - x²) = (x² + 4)
2xy - x²y = x² + 4,
2y+2xy' - x²y' - 2xy = 2x (by implicit differentiation and y' = dy/dx)
y' = 2(x+xy-y)/(2x-x²), set y' = 0
0 = 2x+2xy-2y
0 = x+y(x-1)
0 = x+((x² + 4)(x-1))/(2x - x²)
0 = x(2x - x²) + (x² + 4)(x-1)
0 = 2x² - x^3 + x^3 - 4
0 = x² - 2 = (x-rt2)(x+rt2)

x = rt2 and x = -rt2
10. For 3 use quotient rule
y = (x² + 4) / (2x - x²)

where y=u/v

let u=x² + 4 and v=2x - x²

thus du/dx=2x and dv/dx=2-2x

then use the formula to find dy/dx

then let dy/dx=0 and rearrange for x

once you have found x sub into original equation to find corresponding y value thus getting stationary coordinates!
11. I really ought to start neg repping people for this.
12. This may be dumb question, but what does "rt" mean?
13. I'm sorry 'generalebreity'. Please don't give anybody neg rep!! I didn't know it was "wrong", but saying that i haven't read your sticky thread on the maths forum. I'll post my solutions in future.
14. (Original post by hi there)
I'm sorry 'generalebreity'. Please don't give anybody neg rep!! I didn't know it was "wrong", but saying that i haven't read your sticky thread on the maths forum. I'll post my solutions in future.
"rt" = square root, I assume.

I'm not talking about you, I'm talking about people like Decota who blatantly know better and are posting full solutions just to piss me off.
15. (Original post by Thaiprawns)
For 3 use quotient rule
y = (x² + 4) / (2x - x²)

where y=u/v

let u=x² + 4 and v=2x - x²

thus du/dx=2x and dv/dx=2-2x

then use the formula to find dy/dx

then let dy/dx=0 and rearrange for x

once you have found x sub into original equation to find corresponding y value thus getting stationary coordinates!
I did all of that, but the problem is when you put it altogether, whith one big messy equation which is hard to rearrange!
16. question 4 uses the same concepts as question 3

As for question 5
Find f '(4) when f(x)= x(e^squarerootx)

y=xe^squarerootx

use product rule y=uv
let u=x and v=e^squarerootx

find du/dx and dv/dx and use product rule to find dy/dx

once you have found dy/dx sub in x=4 and thus get your answer!
17. (Original post by generalebriety)
"rt" = square root, I assume.

I'm not talking about you, I'm talking about people like Decota who blatantly know better and are posting full solutions just to piss me off.
lol not at all, I was just following suit of the other people in the thread, see I even gave a hint in my first post but then people started posting solutions ..
18. (Original post by hi there)
I did all of that, but the problem is when you put it altogether, whith one big messy equation which is hard to rearrange!
As I said, post your working and we'll help you further. Or Decota will keep blatantly ruining the spirit of the forum. Whichever comes first, really.
19. (Original post by Decota)
lol not at all, I was just following suit of the other people in the thread, see I even gave a hint in my first post but then people started posting solutions ..
No one posted solutions apart from you.
20. (Original post by hi there)
I did all of that, but the problem is when you put it altogether, whith one big messy equation which is hard to rearrange!
What equation did you get? I will see if it corresponds with mine

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