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if f(x) satisfies f(7-x)=f(7+x) for all x belong to real numbers such that f(x) have exactly 5 real roots which are all distinct,such that the sum of real roots is s.Then find s/7

in my view the function must be symmetrical about x=7 and one of the roots must be at x=7 and two roots on right of x=7 and two roots on left of x=7 but i could not proceed further.

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  • $\begingroup$ Maybe sum of those two on right and left cancel out $\endgroup$
    – Archis Welankar
    Mar 12, 2017 at 18:34

1 Answer 1

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Hint: the given condition can be written as $f(x)=f(14-x)$ so if $x_1$ is a root such that $f(x_1)=0$ then $x_2=14-x_1$ is also a root since $f(x_2)=f(14-x_1)=f(x_1)=0$. Note that $x_1+x_2=14$.

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    $\begingroup$ thanks @dxiv , it solved the que.. $\endgroup$
    – Suryansh Pratap Singh
    Mar 12, 2017 at 19:07

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