If f(x) satisfies f(7 - x) = f(7 + x) for all | vedclass

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(C) The condition $f(7 - x) = f(7 + x)$ implies that the function $f(x)$ is symmetric about the line $x = 7$.Let the $5$ distinct real roots of $f(x)

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$5$

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(C) The condition $f(7 - x) = f(7 + x)$ implies that the function $f(x)$ is symmetric about the line $x = 7$.Let the $5$ distinct real roots of $f(x) = 0$ be $x_1, x_2, x_3, x_4, x_5$.Since the function is symmetric about $x = 7$,if $x_i$ is a root,then $14 - x_i$ must also be a root.For $5$ distinct roots,one root must be the axis of symmetry itself,so $x_3 = 7$.The remaining roots must form pairs symmetric about $7$,such that $\frac{x_1 + x_5}{2} = 7$ and $\frac{x_2 + x_4}{2} = 7$.This gives $x_1 + x_5 = 14$ and $x_2 + x_4 = 14$.The sum of the roots $S = x_1 + x_2 + x_3 + x_4 + x_5 = (x_1 + x_5) + (x_2 + x_4) + x_3 = 14 + 14 + 7 = 35$.Therefore,$S/7 = 35/7 = 5$.

If $f(x)$ satisfies $f(7 - x) = f(7 + x)$ for all $x \in R$ such that $f(x)$ has exactly $5$ real roots which are all distinct,and the sum of the real roots is $S$,then $S/7$ is equal to:

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