If f is strictly increasing, then which of the following is necessarily WRONG?
C integral of f(x) from 0 to 1 = integral of f(x) from 1 to 2
E f'(1)=-f'(2)
I think "necessarily wrong" means "always wrong".
If f is strictly increasing, then for any x1<x2, f(x1)<f(x2).
So f'(x) should always be positive, right?
9th floor is right about option C~~ But the explanation of option c I guess should be like:
"f(x) is strictly increasing" does not mean the integral of f(x) as a function also has to be a strictly increasing function. And in fact, it can be even a decreasing on the domain. The point here is that although f(x) is increasing, it does not have to be positive on the whole space. Hence, if some qualitative change happen in between 0<x<2 like f(x) starts from a negative value and eventually reaches f(x)=0 and then becomes positive. The integral of f(x) can be the same value at some points in the left hand side of f(x)=0 and the right hand side of f(x)=0.
Example here: Set f(x)=x^2-b (x>=0) let b>0 be any number greater or equal to 2 in this example, then:
We all know that f(x) is strictly increasing here but the value of f(x) is less than 0 when x<square root of b, hence the integral of f(x) is decreasing. And this quality changes after x becomes greater than square root of b. Therefore...
C can be proved by Intermediate value of Integration Theorem, Integral of f(x) from 1 to 2 can be equal to 1*f(c) 1<c<2 and in the same way, Integral of f(x) from 0 to 1 is equal to 1*f(d),0<d<1,Since f(c)>f(d) can always be true, the statement is wrong