Is this a parabola?

Algebra Level 5

f ( x ) = ( a = 100 100 x + a ) 10100 ; 10 π x 100 g ( x ) = x 2 ; 100 x < 40 \begin{array} {lc} \displaystyle f\left( x \right) = \left( \sum _{ a=-100 }^{ 100 }{ \left| x+a \right| } \right) -10100; & -10\pi \le x \le 100 \\ g(x) = { x }^{ 2 }; & -100 \le x < 40 \end{array}

Find the number of distinct values of x x that satisfy f ( x ) = g ( x ) f(x) =g (x) .


The answer is 71.

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1 solution

Mark Hennings
Jul 22, 2017

Consider the function F ( x ) = a = 100 100 x + a 10100 100 x 100 F(x) \; = \; \sum_{a=-100}^{100}|x+a| - 10100 \hspace{2cm} -100 \le x \le 100 If n n is an integer between 100 -100 and 100 100 inclusive, then F ( n ) = a = 0 100 + n a + a = 0 100 n a 10100 = 1 2 ( 100 + n ) ( 101 + n ) + 1 2 ( 100 n ) ( 101 n ) 10100 = n 2 \begin{aligned} F(n) & = \; \sum_{a=0}^{100+n}a + \sum_{a=0}^{100-n} a - 10100 \; = \; \tfrac12(100+n)(101+n) + \tfrac12(100-n)(101-n) - 10100 \; = \; n^2 \end{aligned} Since F F is linear between any pair of successive integers and since x 2 x^2 is a convex function, it follows that F ( x ) > x 2 F(x) > x^2 for any noninteger 100 x 100 -100 \le x \le 100 .

Thus the number of points where f ( x ) = g ( x ) f(x) = g(x) is the number of integers between 100 -100 and 100 100 that are common to the domains of both functions, namely the integers n n such that 31 n 39 -31 \le n \le 39 . There are 71 \boxed{71} of these.

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.

Saransh Dave - 3 years, 10 months ago

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This is true, but since the question was focused on domains no larger than [ 100 , 100 ] [-100,100] , I restricted my answer to that range.

Mark Hennings - 3 years, 10 months ago

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Yes... even that would be fine...

Saransh Dave - 3 years, 10 months ago

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