2.99 The number of real roots of the equation $\left|x^{2}-1\right|=\frac{1}{10}\left(x+\frac{9}{10}\right)$ is
(A) 4.
(B) 3.
(C) 2.
(D) 1.
(2nd "Hope Cup" National Mathematics Invitational, 1991)
［Solution］When $|x|>1$, the equation can be transformed into
$x^{2}-1=\frac{1}{10}\left(x+\frac{9}{10}\right)$, which is $x^{2}-\frac{1}{10} x-\frac{109}{100}=0$.
At this point, the discriminant $\Delta=\left(-\frac{1}{10}\right)^{2}+4\left(\frac{109}{100}\right)>0$, indicating that the equation has two real roots, and the positive root has a larger absolute value, and is greater than 1 (the product of the roots has an absolute value greater than 1), which meets the initial condition. The negative root has an absolute value less than 1, which does not meet the initial condition.
When $|x|<1$, the equation can be transformed into
$1-x^{2}=\frac{1}{10}\left(x+\frac{9}{10}\right)$, which is $x^{2}+\frac{1}{10} x-\frac{91}{100}=0$,
Similarly, the equation has two real roots of opposite signs, and the positive root meets the condition, while the negative root does not meet the condition.
When $|x|=1$, i.e., $x= \pm 1$, the equation is not satisfied.

In summary, the equation has two real roots.
Therefore, the answer is $(C)$.