Function f is symmetric to the origin and periodic with period 8. If f(2)=3, what is the value of f(4)+f(6)?

$- 3$

Explanation:

Let $f \left(x\right)$ be the function symmetric to the origin i.e. $f \left(x\right)$ is odd hence

$f \left(- x\right) = - f \left(x\right)$

Since function $f \left(x\right)$ is periodic with period $8$ hence we have

$f \left(x + 8\right) = f \left(x\right)$

setting $x = - 4$ in above equation we get

$f \left(- 4 + 8\right) = f \left(- 4\right)$

$f \left(4\right) = f \left(- 4\right)$

$f \left(4\right) = - f \left(4\right) \setminus \quad \left(\setminus \because f \left(- x\right) = - f \left(x\right)\right)$

$2 f \left(4\right) = 0$

$f \left(4\right) = 0$

Again setting $x = - 2$ in above function we get

$f \left(- 2 + 8\right) = f \left(- 2\right)$

$f \left(6\right) = - f \left(2\right) \setminus \quad \left(\setminus \because f \left(- x\right) = - f \left(x\right)\right)$

$f \left(6\right) = - 3 \setminus \quad \left(\setminus \because f \left(2\right) = 3\right)$

$\setminus \therefore f \left(4\right) + f \left(6\right)$

$= 0 - 3$

$= - 3$