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If $$f(x)$$ be a polynomial of degree $$8$$ such that $$f(x) = f(4 - x)$$ $$\forall{x}\epsilon{R}$$, $$f(x)$$ has $$6$$ distinct real and equal roots then sum of roots of $$f(x) = 0$$ is $$A$$. Then the number $$A$$ is:


Solution

$$f(x)=f(4-x)$$ ......replacing $$x $$ by $$ x+2$$
$$f(2+x)=f(2-x)$$ this show $$f(x)$$ is a symmetrical 
The graph is like shown in figure $$1 $$
$$f(x)$$ is a polynomial of degree $$8$$ so there are $$8$$ roots
It has two equal root which is $$ 2$$ , $$2 $$ and six distinct root in form of $$2+a$$ , $$2-a$$, $$2+b$$, $$2-b$$, $$2+c$$, $$2-c$$
Now sum of roots are $$ 2+2 +2+a+2-a+2+b+2-b+2+c+2-c$$ 
Sum of roots is $$16$$.
794804_769243_ans_6d5f94cd6f9a49918787672b33e98eeb.jpg

Mathematics

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