For $x\le1$, we have $F(x)=\frac{x+1}{x^2+1}$.
Now $F'(x)=\frac{(x^2+1)-2x(x+1)}{(x^2+1)^2}=\frac{-x^2-2x+1}{(x^2+1)^2}$.
So $F'(x)=0$ gives $-x^2-2x+1=0$, which gives $x=-1\pm\sqrt2$.
Both $-1-\sqrt2$ and $-1+\sqrt2$ lie in $(-\infty,1]$.
The sign of $F'(x)$ is negative on $(-\infty,-1-\sqrt2)$, positive on $(-1-\sqrt2,-1+\sqrt2)$, and negative on $(-1+\sqrt2,1]$.
Hence $x=-1-\sqrt2$ is a local minimum and $x=-1+\sqrt2$ is a local maximum for the known part.
Now $F(-1+\sqrt2)=\frac{\sqrt2}{(-1+\sqrt2)^2+1}=\frac{1+\sqrt2}{2}$.
Also $F(1)=1$.
Since $\frac{1+\sqrt2}{2}>1$, the value at $x=-1+\sqrt2$ is greater than $F(1)$.
For $x>1$, we are given $F'(x)<0$, so $F$ is decreasing after $x=1$. Therefore, for every $x>1$, $F(x)<F(1)=1$.
Thus no value for $x>1$ can exceed $F(-1+\sqrt2)=\frac{1+\sqrt2}{2}$. Hence $x=-1+\sqrt2$ is a point of global maximum. So (A) is true.
At $x=1$, the function is decreasing on the left near $1$, so values just before $1$ are greater than $F(1)$. Also, since $F'(x)<0$ for $x\ge1$, values just after $1$ are less than $F(1)$. Hence $x=1$ is neither a local maximum nor a local minimum. So (B) is false and (D) is true.
For (C), although $x=-1-\sqrt2$ is a local minimum on the known part, it may not be a global minimum because the unknown part for $x>1$ may keep decreasing and go below this value. Hence (C) is not necessarily true.
