Question

If $f(x+\frac{1}{2})+f(x-\frac{1}{2})=f\left(x\right)$ for all $xϵR$, then the period of $f\left(x\right)$ is:

• A. $1$
• B. $2$
• C. $3$
• D. $4$

Answer 1

The correct option is C $3$

If $T$ is the period of $f\left(x\right)$, then $f(x+T)=f\left(x\right)$.

Given : $f(x+\frac{1}{2})+f(x-\frac{1}{2})=f\left(x\right)$ --(1)

Replacing $x+\frac{1}{2}$ by $x$,

$⟹f(x+1)+f\left(x\right)=f(x+\frac{1}{2})$

From (1), $f(x+\frac{1}{2})=f\left(x\right)-f(x-\frac{1}{2})$

Substituting, $f(x+1)+f\left(x\right)=f\left(x\right)-f(x-\frac{1}{2})$

$⟹f(x+1)=-f(x-\frac{1}{2})$ --(2)

Hence $1$ is not the period of $f\left(x\right)$.

Replacing $x$ by $x+1$ in equation (2)

$f(x+2)=-f(x+\frac{1}{2})$ Hence $2$ is not a period of $f\left(x\right)$.

Replacing $x$ by $x+\frac{1}{2}$ in equation (2)

$f(x+\frac{3}{2})=-f\left(x\right)$ ---(3)

Replace $x$ by $x+\frac{3}{2}$ in equation (3)

$f(x+3)=-f(x+\frac{3}{2})=-(-f(x\left)\right)=f\left(x\right)$ (from (3))

Hence $3$ is the period of $f\left(x\right)$.