Question

Let $f:R\to R$ is a function satisfying $f(2-x)=f(2+x)$ and $f(20-x)=f\left(x\right)\forall x\in R$
If $f\left(0\right)=5$ then the minimum possible no. of values of $x$ satisfying $f\left(x\right)=5$ for $x=[0.,70]$, is

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

Answer 1

The correct option is A $21$

Given, $f(2-x)=f(2+x)..\left(i\right)$

$\therefore f\left(x\right)$ is symmetric about $x=2$

$f(20-x)=f\left(x\right)..\left(ii\right)$

$x\to x+10$

$f(10-x)=f(10+x)$

$f\left(x\right)$ is symmetric about $x=10$

$x\to x+2$

$f(18-x)=f(x+2)..\left(ii\right)$

$f(2-x)=f(18-x)$

$x\to -x$

$f(2+x)=f(18+x)$

$\therefore x+2\to x$

$\therefore f\left(x\right)=f(x+16)$

$f\left(x\right)$ has period = $16$

$f\left(x\right)=5,x\in [0,170]$

put $x=2$ in (i) $f\left(0\right)=f\left(4\right)$

put $x=4$ in (ii) $f\left(4\right)=f\left(16\right)$

when $x\in [0,16]\to f\left(x\right)$ has two solution

when $x\in [0,160]$ has $20$ solution.

When $xin[0,170]$ has ${}^{\prime }{21}^{\prime }$ solution