If 0 < x < 1, then it is true that

Concept: For a base b with 0 < b < 1, its positive integer powers form a strictly decreasing sequence as the exponent increases -- raising b to a higher power always gives a smaller value, because each extra multiplication by b (a proper fraction) shrinks the magnitude further. In general, for positive integers m < n with 0 < b < 1, bm > bn.

1. The base here is x, with 0 < x < 1 given in the question.

2. The exponents being compared are 100 and 101, and 100 < 101.

3. Since x101 = x100 × x, write the higher power as the lower power multiplied once more by x.

4. Because 0 < x < 1, multiplying x100 by x (a positive fraction less than 1) produces a strictly smaller positive number.

5. Therefore x101 < x100, i.e. x100 > x101.

Cross-check: take a concrete value, x = 0.5. Then x101 = x100 × 0.5, exactly half of x100. Half of a positive number is always smaller than the number itself, so 0.5100 > 0.5101, confirming the general result. (The other options fail too: since 0 < x < 1, every positive power of x -- including x100 and x101 -- stays below 1, so neither can exceed 1.)

Result: the true statement among those given is x100 > x101.