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Algebra Difficulty 5.1 AIME, harder Find the answer

Let pp be a real number and c0c \neq 0 an integer such that c0.1<xp(1(1+x)101+(1+x)10)<c+0.1c-0.1<x^{p}\left(\frac{1-(1+x)^{10}}{1+(1+x)^{10}}\right)<c+0.1 for all (positive) real numbers xx with 0<x<101000<x<10^{-100}. Find the ordered pair (p,c)(p, c).

A number or a short expression. Spacing and $ signs are ignored.

Solution

We are essentially studying the rational function f(x):=1(1+x)101+(1+x)10=10x+O(x2)2+O(x)f(x):=\frac{1-(1+x)^{10}}{1+(1+x)^{10}}=\frac{-10 x+O\left(x^{2}\right)}{2+O(x)}. Intuitively, f(x)10x2=5xf(x) \approx \frac{-10 x}{2}=-5 x for "small nonzero xx ". So g(x):=xpf(x)5xp+1g(x):= x^{p} f(x) \approx-5 x^{p+1} for "small nonzero xx ". If p+1=0,g5p+1=0, g \approx-5 becomes approximately constant as x0x \rightarrow 0. Since cc is an integer, we must have c=5c=-5 (as -5 is the only integer within 0.1 of -5 ).

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