Let f (x) = ax & sup2; + BX + C (a, B, C belong to R) satisfy the following conditions: (1) when x belongs to R, its minimum value is 0 and f (x-1) = f (- x-1) holds. (2) when x belongs to (0,5), X ≤ f (x) ≤ 2 | X-1 | + 1 holds. (1) find the value of F (1); (2) find the analytic expression of F (x); (3) find the largest real number m (M > 1) so that there exists t belongs to R, and f (x + 1) holds as long as X belongs to [1, M] | Math enthusiast | Mathematical art

Let f (x) = ax & sup2; + BX + C (a, B, C belong to R) satisfy the following conditions: (1) when x belongs to R, its minimum value is 0 and f (x-1) = f (- x-1) holds. (2) when x belongs to (0,5), X ≤ f (x) ≤ 2 | X-1 | + 1 holds. (1) find the value of F (1); (2) find the analytic expression of F (x); (3) find the largest real number m (M > 1) so that there exists t belongs to R, and f (x + 1) holds as long as X belongs to [1, M]

(1)
Because 1 belongs to (0,5), 1A = 1 / 4
=>f(x)=(x+1)^2/4
(3)
And (x + 1) ^ 2 / 4-x = (x-1) ^ 2 / 4 > = 0
So (x + 1) ^ 2 / 4 > = x
Obviously, when x belongs to [1, M], it is a monotone increasing interval. If x belongs to [1, M], it has f (x + T) (x + T + 1) ^ 2 / 4 = X
=>x^2+2(t-1)x+(t+1)^2=0 (a)
And the curve passes through point (1,1), so 1 is its solution
=》1+2(t-1)+(t+1)^2=0
=>t=-4
Substituting t = - 4 into (a)
=>x^2-10x+9=0
=>x1=1,x2=9
So m = x2 = 9
So the value of the largest real number m is 9

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