We need to check differentiability of $f(x)=a\cos(|x^3-x|)+b|x|\sin(|x^3+x|)$ at the points mentioned in the options.

1. Basic observations

We analyze the potentially problematic parts:

• $|x|$ can cause non-differentiability at $x=0$.
• $|x^3-x|$ can cause non-differentiability where $x^3-x=0$, i.e. $x(x-1)(x+1)=0 \implies x=0,\pm 1.$
• $|x^3+x|=|x(x^2+1)|=|x| (x^2+1)$ since $x^2+1>0$.

So $|x^3+x|=|x|(x^2+1).$ Hence $|x|\sin(|x^3+x|)=|x|\sin(|x|(x^2+1)).$

We now test each option.

2. Option A: differentiable at $x=0$ if $a=0, b=1$

If $a=0, b=1$, then $f(x)=|x|\sin(|x^3+x|)=|x|\sin(|x|(x^2+1)).$

To test differentiability at $0$, compute $f(0)=0.$ Then $\frac{f(h)-f(0)}{h}=\frac{|h|\sin(|h|(h^2+1))}{h}.$

Use $\sin t \sim t$ as $t\to 0$: $|h|\sin(|h|(h^2+1)) \sim |h|\cdot |h|(h^2+1)=h^2(h^2+1).$ So $\frac{|h|\sin(|h|(h^2+1))}{h} \sim h(h^2+1)\to 0.$

More rigorously, since $|\sin u|\le |u|$,

Thus derivative exists and equals $0$.

So Option A is correct.

3. Option B: differentiable at $x=1$ if $a=1, b=0$

If $a=1, b=0$, then $f(x)=\cos(|x^3-x|).$ Let $g(x)=|x^3-x|.$ At $x=1$, $g(1)=0$, so $f(1)=\cos 0=1.$

Now examine differentiability:

Use the standard expansion near $0$: $\cos u -1 = O(u^2).$ Also near $x=1$, $x^3-x=x(x-1)(x+1),$ so $|x^3-x|=|x(x+1)||x-1|=O(|x-1|).$ Hence $\cos(|x^3-x|)-1 = O(|x-1|^2).$ Therefore

So derivative exists at $x=1$.

Thus Option B is correct.

4. Option C: NOT differentiable at $x=0$ if $a=1, b=0$

If $a=1, b=0$, then again $f(x)=\cos(|x^3-x|).$ At $x=0$, $f(0)=\cos 0=1.$ Now

Near $h=0$,

And $\cos u -1 = O(u^2),$ so

Hence

Thus $f$ is differentiable at $0$.

So the statement “NOT differentiable” is false.

Therefore Option C is incorrect.

5. Option D: NOT differentiable at $x=1$ if $a=1, b=1$

If $a=1, b=1$, then $f(x)=\cos(|x^3-x|)+|x|\sin(|x^3+x|).$ We already know the first term is differentiable at $x=1$. Now check the second term near $x=1$.

Since $x$ is near $1>0$, we have $|x|=x$. Also $|x^3+x|=x^3+x$ for $x$ near $1$ because $x^3+x>0$ there. Thus near $x=1$ the second term becomes $x\sin(x^3+x),$ which is a composition/product of differentiable functions, hence differentiable at $x=1$.

Therefore the sum is differentiable at $x=1$. So the statement “NOT differentiable” is false.

Hence Option D is incorrect.

6. Final conclusion

The correct options are: $\boxed{A,\ B}$

This matches the stored correct answer.