monotone functions - Discontinuous derivative, positive on a dense set

Whether there exists a continuos monotone function $f: [0,1]\rightarrow [0,1]$ with the following properties:

(1) $f$ strictly increase, $f(0)=0$ and $f(1)=1$;

(2) there is no interval $A\subset [0,1]$, where the derivative $f'$ (a) exists, (b) is continuous and (c) is finite;

(3) there exists a set $B\subset [0,1]$ dense in $[0,1]$, where the derivative $f'$ (a) exists, (b) is positive and (c) is finite?

I can construct an example of function $f:[0,1]\rightarrow [0,1]$ with condition (1), whose derivative is either $0$, or $+\infty$ (at points, where the dewrivative exists). It follows from Lebesgues theorem that $f'$ is zero on some set $A\subset[0,1]$, which has Lebesgue measure 1 and, whence, is dence in $[0,1]$. Clearly, since the derivative is either $0$, or $+\infty$ (in the example, which I can construct), and the function is invertible, then $f'\neq 0$ at any interval.

In other words, I can answer "yes" to my question if ommit the word "positive" in the condition (3).

I hope that my question is "natural", but neither have found "in the internet" neither its proof, nor counter example, nor this question as it is (for example as open problem).