If the graph of a non-constant function f(x) | vedclass

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(A) function $f(x)$ is symmetric about the point $(a, b)$ if $f(a+x) + f(a-x) = 2b$.Here,the point of symmetry is $(3, 4)$,so $a=3$ and $b=4$.Thus,$f(

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$32$

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(A) function $f(x)$ is symmetric about the point $(a, b)$ if $f(a+x) + f(a-x) = 2b$.Here,the point of symmetry is $(3, 4)$,so $a=3$ and $b=4$.Thus,$f(3+x) + f(3-x) = 2(4) = 8$.We need to calculate $S = \sum\limits_{r = 0}^6 f(r) + f(3) = f(0) + f(1) + f(2) + f(3) + f(4) + f(5) + f(6) + f(3)$.Using the symmetry property:For $x=3$,$f(3+3) + f(3-3) = f(6) + f(0) = 8$.For $x=2$,$f(3+2) + f(3-2) = f(5) + f(1) = 8$.For $x=1$,$f(3+1) + f(3-1) = f(4) + f(2) = 8$.Also,at the point of symmetry $x=3$,$f(3) = 4$.Substituting these into the sum:$S = [f(0) + f(6)] + [f(1) + f(5)] + [f(2) + f(4)] + f(3) + f(3)$$S = 8 + 8 + 8 + 4 + 4 = 32$.

If the graph of a non-constant function $f(x)$ is symmetric about the point $(3,4)$,then the value of $\sum\limits_{r = 0}^6 {f(r) + f(3)}$ is equal to

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