Question

A polynomial of 6th degree $f\left(x\right)$ satisfies $f\left(x\right)=f(2-x),\forall xϵR$, if $f\left(x\right)=0$ has 4 distinct and two equal roots, then sum of the roots of $f\left(x\right)=0$ is:

• A. $4$
• B. $5$
• C. $6$
• D. $7$

Answer 1

Answer: (C)

$6$

In the functional relation replace $x$ with $x+1$.

We have,

$f(1+x)=f(1-x)$

This shows that the function is symmetric about $x=1$.

There is one and only one double root. If the double root exists at any value $x_0$ other than at $x=1$, then a double root will also exist at a value of $2-x_0$.

Hence, the double root exists at $x=1$

Say two other roots are $\alpha$ and $\beta$

$f\left(\alpha \right)=f(2-\alpha )=0$

$\therefore 2-\alpha$ is also a root.

And similarly, $2-\beta$ is also a root.

$\therefore$ the roots are $1,1,\alpha ,\beta ,2-\alpha ,2-\beta$

Hence, sum of the roots is $6$