1. The number of integers $x$ that satisfy the inequality $|x-1|+|x-9|<\sqrt{98}$ is ( ).
(A) 8
(B) 9
(C) 10
(D) 11
-1. (B).
When $x\frac{10-\sqrt{98}}{2}$. Therefore, when $\frac{10-\sqrt{98}}{2}<x<9$, the original inequality becomes $10-2x<\sqrt{98}$, i.e., $x>\frac{10-\sqrt{98}}{2}$. So, $\frac{10-\sqrt{98}}{2}<x<9$, the integer solutions are $x=1, 2, 3, 4, 5, 6, 7, 8$.
When $x>9$, the original inequality becomes $2x-10<\sqrt{98}$, i.e., $x<\frac{10+\sqrt{98}}{2}$. So, $9<x<\frac{10+\sqrt{98}}{2}$, no integer solutions.
In summary, the original inequality has 9 integer solutions.
Note: This problem can be solved more simply using the geometric meaning of absolute value.