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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

1. $0$
2. $40$
3. $10$
4. $20$

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$: \text{D}$

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