For (a) and (b) let if and if . For (d) suppose and (other possible cases are similar) and suppose there is a such that then by the intermediate value property there is a such that so is not one-one. Similarly there is no such that and so for all . Now pick any and show that (Consider the cases where and are disjoint and when they're not separately). Since a continous function satisfies the ivp, you're done.