Question

Let $f:R\to R$ is a function satisfying $f(2-x)=f(2+x)$ and $f(20-x)=f\left(x\right),\forall x\in R$. For this function $f$ answer the following question.If $f\left(0\right)=5$, then minimum possible number of values of $x$ satisfying $f\left(x\right)=5$, for $x\in [0,170]$, is

• A. $21$
• B. $12$
• C. $11$
• D. $22$

Answer 1

The correct option is C $11$

$f(2-x)=f(2+x)$ ........ $\left(i\right)$
and $f(20-x)=f\left(x\right)$ ...... $\left(ii\right),\forall xϵR$

Replacing $x$ by $2-x$ in equation $\left(i\right)$
$f\left(x\right)=f(4-x)$ ........ $\left(iii\right)$

By comparing $\left(ii\right)$ & $\left(iii\right)$
$f(20-x)=f(4-x)$

Replacing x by $(4-x)$
$\Rightarrow f(4-(4-x))=f(20-(4-x\left)\right)$
$\Rightarrow f\left(x\right)=f(x+16)$

Hence $f\left(x\right)$ is a periodic function of period $16$

Therefore minimum possible values of $x$ satisfying $f\left(x\right)=5$, if $f\left(0\right)=5$ is $\frac{170}{16}\approx 11$

Thus, the correct answer will be $11$.