1. We must determine which statement is true for every function $f:[0,1]\to\mathbb R$ that is continuous on $[0,1]$ and differentiable on $(0,1)$.

So we test each option carefully.

2. Option C

It states that there exists $c\in(0,1)$ such that $|f(c)-f(1)|<(1-c)|f'(c)|.$

Now apply Lagrange’s Mean Value Theorem on the interval $[c,1]$. Since $f$ is continuous on $[c,1]$ and differentiable on $(c,1)$, there exists some $\xi\in(c,1)$ such that $f(1)-f(c)=(1-c)f'(\xi).$ Hence $|f(c)-f(1)|=(1-c)|f'(\xi)|.$

But option C requires the same point $c$ on both sides, i.e. $|f(c)-f(1)|<(1-c)|f'(c)|,$ which is much stronger and not guaranteed by MVT.

To disprove it, take a simple function: $f(x)=x.$ Then $f'(x)=1$ and

while

So for every $c\in(0,1)$,

not strict inequality. Therefore Option C is false.

3. Option A

It states that there exists $c\in(0,1)$ such that $|f(c)-f(1)|<|f'(c)|.$

Again use the same counterexample: $f(x)=x.$ Then for any $c\in(0,1)$,

Here indeed $1-c<1$, so this example does not disprove A. We need a better test.

Take instead a constant function: $f(x)=k.$ Then

and

So the inequality becomes $0<0,$ which is false. Thus there is no such $c$ for a constant function. Hence Option A is false.

4. Option B

It states that there exists $c\in(0,1)$ such that $|f(c)+f(1)|<(1+c)|f'(c)|.$

Again test with a constant function: $f(x)=k.$ Then $f'(x)=0.$ So the inequality becomes

that is,

This is impossible for every constant function, and in particular for $k=1$. Therefore Option B is false.

5. Since A, B, and C are all false, the correct choice is $\boxed{\text{D: None}}.$

6. Comparison with stored answer

Stored correct answer: D.

Our derived answer is also D, so they agree.