If $f\left ( 5+x \right )= f\left ( 5-x \right )$ for every real $x,$ and $f\left ( x \right )=0$ has four distinct real roots, then the sum of these roots is

- $0$
- $40$
- $10$
- $20$

## 1 Answer

Given that, $ f(5+x) = f(5-x) ; \forall x \in R \quad \longrightarrow (1) $

And, $ f(x) = 0 $

Let $ 5 + x = k $

$ \Rightarrow \boxed{x = k – 5} $

Now, from equation $(1),$

$ f(k) = f(5-k+5) $

$ \Rightarrow \boxed{ f(k) = f(10-k)} $

If $f(k)=0,$ then $f(10-k)=0$

Let $\alpha,\beta$ be two roots of $f(k)=0,$ then $10- \alpha, 10 – \beta$ are also be roots of $f(k)=0.$

$\therefore$ The sum of roots $ = \alpha + \beta + 10 – \alpha + 10 – \beta = 20 $

$\textbf{Second Method:}$ Given that, $f(5+x) = f(5-x) ; \forall x \in R \quad \longrightarrow (1) $

And, $ f(x) = 0 $

Put, $ 5 – x = x $ in equation $(1)$

$ f( 5 + 5 – x) = f(x) $

$ \Rightarrow \boxed{ f(x) = f( 10-x )} $

Let $\alpha,\beta$ be two roots of $f(x)=0,$ then $10- \alpha, 10 – \beta$ are also be roots of $f(x)=0.$

$\therefore$ The sum of roots $ = \alpha + \beta + 10 – \alpha + 10 – \beta = 20 $

Correct Answer$: \text{D}$