Question

A polynomial of 6th degree f(x) satisfies f(x) = f(2-x), for all $ {\text{x}} \in {\text{R}} $ , if f(x)=0 has 4 distinct and two equal roots, then sum of the roots of f(x)=0 is:
$
  {\text{A}}{\text{. 4}} \\
  {\text{B}}{\text{. 5}} \\
  {\text{C}}{\text{. 6}} \\
  {\text{D}}{\text{. 7}} \\
$

Answer 1

Hint: If x=a is one root then 2-a is another root. This way if b is one root then 2-b is another root. Equate the values of x in the two functions to get the equal root. The roots of the function f(x) are obtained so, summate them and get the answer.

Complete step-by-step answer:
As it is given that f(x) = f(2-x), one can conclude that when f(x)=0 then f(2-x) is also equal to 0.
Also, for any value of x the function f(x) and f(2-x) give the same result. So, for the function f(x) we can say that if x is one root then 2-x will be the other root.
This way, considering x=a as one root, (2-a) is the other corresponding root of f(x).
And if b is taken as the third root, then (2-b) becomes the fourth root.
Since there are two equal roots present, to get the exact value of the double root, equate the inputs of the two functions. So,
As f(x)=f(2-x), for a double root we have, x=2-x
This gives x=1 as the double root.
As x=1 is the double root this implies that the graph of the function will be symmetric about x=1.
So, the roots are a, 2-a, b, 2-b, 1, 1.
The sum of roots therefore is,
a+2-a+b+2-b+1+1=6
The correct option is C.

Note: In these types of questions, asking for the sum of roots for any degree polynomial, the roots are assumed to be any variable and then added as the assumed variables get cancelled, giving a constant as the answer.