Question 2.

Let ‘r’ be the root of the function. It follows that $f(r) = 0$. We can represent this as $f(r) = f\{5 - (5 - r)\}$

Based on the relation: $f(5 - x) = f(5 + x)$; $f(r) = f\{5 - (5 - r)\} = f\{5 + (5 - r)\}$

∴ $f(r) = f(10 - r)$

Thus, every root ‘r’ is associated with another root ‘(10-r)’ [these form a pair]. For even distinct roots, in this case four, let us assume the roots to be as follows: $r1$, $(10 - r1)$, $r2$, $(10 - r2)$

The sum of these roots = $r1 + (10 - r1) + r2 + (10 - r2) = 20$

Hence, Option D is the correct answer.