C++ O(1) solution:

int maximizingXor(int l, int r) {
    int x = l ^ r, p = 1;
    while ((x & (p - 1)) != x) p <<= 1;
    return p - 1;
}

Why this works.. this looks for highest bit which is different between l and r: for example (11, 12) that is third bit with value 4.. all bits above it are identical in both l and r, and you can not produce any l<=a<=b<=r which have non zero result of a^b. But you can find value a which has third bit zero and all remaining bits set (11 in this example), and b = a + 1 will clear all lower bits and set third bit, so a^b is all ones from highest difference bit down to bottom bit = maximum xor difference. qed.