SOLUTION: Let f(0) = 0,f(pi/2)=1,f(3pi/2)=-1 be a continous and twice differentiable function. statement 1:|f''(x)|<=1 for atleast one x belongsto (0,3pi/2). statement 2:according to rolle

Algebra ->  Functions -> SOLUTION: Let f(0) = 0,f(pi/2)=1,f(3pi/2)=-1 be a continous and twice differentiable function. statement 1:|f''(x)|<=1 for atleast one x belongsto (0,3pi/2). statement 2:according to rolle      Log On


   



Question 201254: Let f(0) = 0,f(pi/2)=1,f(3pi/2)=-1 be a continous and twice differentiable function.
statement 1:|f''(x)|<=1 for atleast one x belongsto (0,3pi/2).
statement 2:according to rolle's theorem if y=g(x) is continous and differenciable for all x belongs to [a,b] and g(a)=g(b) then there exists atleast one c such that g'(c)=a.
choose the correct answer-
A. statement 1 and 2 are true, 2 is correct explaination of 1.
B. statement 1 and 2 are true, 2 is not the correct explaination of 1.
C. statement 1 is true, 2 is false.
D. statement 1 is false, 2 is true.

Answer by vleith(2983) About Me  (Show Source):
You can put this solution on YOUR website!
C
1 is true since somewhere between moving from 0 to to -1 a curve must have at least two inflection points. The change in slope there is 0, so there must be some point where f'' is o or close to it.
2 is false by definition of rolle's theorem --> http://en.wikipedia.org/wiki/Rolle%27s_theorem