Question

Let $f\left(x\right)=x(2-x),0\le x\le 2$ If the definition of $f$ is extended over the set $R-[0,2]$ by $f(x+2)=f\left(x\right)$ then $f$ is a

• A. periodic function of period $1$
• B. non periodic function
• C. periodic function of period $2$
• D. periodic function of period $\frac{1}{2}$

Answer 1

The correct option is C periodic function of period $2$

$f(x+T)=f\left(x\right)$

$(x+T)(2-x-T)=x(2-x)$
$2x-x^2-xT+2T-xT-T^2=2x-x^2$

$2xT-2T=-T^2$
$2x-2=-T$

$2(x-1)=-T$ ...(i)

Now $f(x+2)=f\left(x\right)$

$2(x-1)=-2$ ...(ii)

Substituting in (i), we get
$-T=-2$
$T=2$

Hence $f\left(x\right)$ is function with a period of $2$.