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Hi, could somebody please help me with the following question, I have been stuck on it for ages.

**1. let f[0,1] -> R be continuous with f(0)=0, f(1)=1. Prove the following:**

a.(i) If for c in (0,1) f is differentiable at c with f'(c)<0 then there are exists points y such that f(x)=y has more than one solution.

(ii). If f(x)>0 for x>0 show there is [tex]\delta[/tex] in (0,1) with [tex]\delta[/tex] not equal to f(x) for all x in [0.5,1].

(iii). If f is also differentiable on (0,1) prove there is an a in (0,1) such that

f'(a)/cos(a[tex]\pi/2[/tex]) = [tex]2/\pi[/tex]

b. If f:[-1,1]->R is such that f(sin(x)) is continuous on R then is f continuous. (give a proof or counterexample).

a.(i) If for c in (0,1) f is differentiable at c with f'(c)<0 then there are exists points y such that f(x)=y has more than one solution.

(ii). If f(x)>0 for x>0 show there is [tex]\delta[/tex] in (0,1) with [tex]\delta[/tex] not equal to f(x) for all x in [0.5,1].

(iii). If f is also differentiable on (0,1) prove there is an a in (0,1) such that

f'(a)/cos(a[tex]\pi/2[/tex]) = [tex]2/\pi[/tex]

b. If f:[-1,1]->R is such that f(sin(x)) is continuous on R then is f continuous. (give a proof or counterexample).

**3. I am really stuck on this question and so have not got very far at all. I think part 1 must involve the application of the intermediate value theorem twice though I can't see how to do so. I don't even know how to start part 2. For part 3 I think I will need to use the mean value theorem (and probably a previous question) but can't see where they have got the cos function from.For part b I suspect that it is true but don't know how to write a proof.**
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