Question

The function $f\left(x\right)=0$ has eight distinct real solution and $f$ also satisfy $f(4+x)=f(4-x)$. The sum of all the eight solution of $f\left(x\right)=0$ is:

• A. $12$
• B. $32$
• C. $16$
• D. $15$

Answer 1

The correct option is B $32$

Given,

$f\left(x\right)=0$ has $8$ roots

and

$f(4+x)=f(4-x)$

It means that it’s an even function about $x=4$

So, it’s roots can be $(4+k)$ and $(4-k)$ form. (k,l,m,n is a constant number).

So roots $(4+k)$ and $(4-k),(4+l)$and $(4-l),(4+m)$and $(4-m),(4+n)$ and $(4-n)$

Therefore sum $=32$