f ( x ) = ( a = − 1 0 0 ∑ 1 0 0 ∣ x + a ∣ ) − 1 0 1 0 0 ; g ( x ) = x 2 ; − 1 0 π ≤ x ≤ 1 0 0 − 1 0 0 ≤ x < 4 0
Find the number of distinct values of x that satisfy f ( x ) = g ( x ) .
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Awesome solution... Thank you Mark Hennings... Just one correction I think there should be is that F(x) > x^2 holds for any non-integer -101 <= x <= 101 (instead of -100<=x<=100).
Note: If we consider the domains of f(x) and g(x) to be all real numbers, then f(x) = g(x) holds for all integral values of x such that -101 <= x <=101.
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This is true, but since the question was focused on domains no larger than [ − 1 0 0 , 1 0 0 ] , I restricted my answer to that range.
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Consider the function F ( x ) = a = − 1 0 0 ∑ 1 0 0 ∣ x + a ∣ − 1 0 1 0 0 − 1 0 0 ≤ x ≤ 1 0 0 If n is an integer between − 1 0 0 and 1 0 0 inclusive, then F ( n ) = a = 0 ∑ 1 0 0 + n a + a = 0 ∑ 1 0 0 − n a − 1 0 1 0 0 = 2 1 ( 1 0 0 + n ) ( 1 0 1 + n ) + 2 1 ( 1 0 0 − n ) ( 1 0 1 − n ) − 1 0 1 0 0 = n 2 Since F is linear between any pair of successive integers and since x 2 is a convex function, it follows that F ( x ) > x 2 for any noninteger − 1 0 0 ≤ x ≤ 1 0 0 .
Thus the number of points where f ( x ) = g ( x ) is the number of integers between − 1 0 0 and 1 0 0 that are common to the domains of both functions, namely the integers n such that − 3 1 ≤ n ≤ 3 9 . There are 7 1 of these.