1. Given functions

and

We need the points where the composite function

is discontinuous.

2. First identify where $f$ is discontinuous

(i) For input $y<0$

Then $f(y)=[y]$ which is discontinuous at every integer $y<0$. So discontinuities from this branch occur at $y=-1,-2,-3,\dots$

(ii) For input $y\ge 0$

Then $f(y)=|1-y|$ which is continuous for all $y\ge 0$.

(iii) Check the joining point $y=0$

At $y=0$, $f(0)=|1-0|=1$ while $\lim_{y\to 0^-}f(y)=\lim_{y\to 0^-}[y]=-1$ So $f$ is also discontinuous at $y=0.$

Hence, $f$ is discontinuous exactly at $y=0,-1,-2,-3,\dots$

3. Now find where $g(x)$ takes these values

Since $f\circ g$ can be discontinuous when $g(x)$ hits a discontinuity value of $f$, we solve $g(x)\in\{0,-1,-2,-3,\dots\}.$

But first note the range of $g$.

For $x<0$:

$g(x)=e^x-x$ Now $e^x>0$ and $-x>0$, so $g(x)=e^x-x>0$ for all $x<0$. Thus this branch never gives $0,-1,-2,\dots$.

Also, for $x<0$, $g$ is continuous.

For $x\ge 0$:

$g(x)=x^2-2x=(x-1)^2-1$ This has minimum value $-1$ at $x=1$. So for $x\ge 0$, $g(x)\ge -1.$ Therefore $g(x)$ can only possibly equal $0$ or $-1$ among the discontinuity points of $f$.

So we only solve:

(a) $g(x)=0$

For $x<0$: impossible since $g(x)>0$.

For $x\ge 0$: $x^2-2x=0$ $x(x-2)=0$ $x=0,2.$

(b) $g(x)=-1$

For $x<0$: impossible.

For $x\ge 0$: $x^2-2x=-1$ $x^2-2x+1=0$ $(x-1)^2=0$ $x=1.$

So candidates are $x=0,1,2.$

4. Write $f\circ g$ explicitly on intervals

We examine $g(x)$ and then apply $f$.

Case 1: $x<0$

Then $g(x)=e^x-x>0$. Hence $f(g(x))=|1-g(x)|$ which is continuous for $x<0$ because $g$ is continuous and $|1-t|$ is continuous for $t>0$. So no discontinuity for $x<0$.

Case 2: $x\ge 0$

Then $g(x)=x^2-2x=(x-1)^2-1.$ Now split according to sign of $g(x)$:

• $g(x)\ge 0$ when $x\in\{0\}\cup[2,\infty)$
• $-1\le g(x)<0$ when $0<x<2$, except $x=1$ gives $-1$

If $0<x<2$, then $g(x)\in[-1,0)$. For values in $[-1,0)$, we have $[g(x)]=-1$ except at $g(x)=-1$, where still $[-1]=-1$. Thus for all $x\in(0,2]$ actually, whenever $g(x)<0$ or $g(x)=-1$, $f(g(x))=[g(x)]=-1.$ In particular, on $(0,2)$ it is constant $-1$, and at $x=1$ also $-1$.

If $x=0$ or $x\ge 2$, then $g(x)\ge 0$, so $f(g(x))=|1-g(x)|,$ continuous on such regions.

5. Check continuity at candidate points $x=0,1,2$

At $x=1$

Nearby for $x\in(0,2)$, $f(g(x))=-1.$ Also at $x=1$, since $g(1)=-1$, $f(g(1))=f(-1)=[-1]=-1.$ So the function is continuous at $x=1$.

At $x=0$

• For $x\to 0^-$: $g(x)=e^x-x\to 1$, so $f(g(x))\to |1-1|=0.$
• For $x\to 0^+$: $g(x)=x^2-2x\in(-1,0)$ for small $x>0$, hence $f(g(x))=-1.$
• Value at $x=0$: $g(0)=0$, so $f(g(0))=f(0)=1.$

Left and right limits are different, so discontinuous at $x=0$.

At $x=2$

• For $x\to 2^-$: $g(x)\to 0^-$, hence $f(g(x))=[g(x)]=-1.$
• For $x\to 2^+$: $g(x)\to 0^+$, so $f(g(x))=|1-g(x)|\to 1.$
• Value at $x=2$: $g(2)=0$, so $f(g(2))=f(0)=1.$

Left and right limits are different, so discontinuous at $x=2$.

6. Conclusion

The composite function $f\circ g$ is discontinuous exactly at $x=0,\;2.$ So it is discontinuous at exactly two points.

Therefore, the correct option is:

$\boxed{\text{B: two points}}$

7. Comparison with stored answer

Stored correct answer: B

Our derived answer: B

They agree.