f is one-one but not onto mapping

f is onto but not one-one mapping

f is both one-one and onto

f is neither one-one nor onto

f is of period 2$$\pi$$

f is of period $$\sqrt {2\pi } $$

f is not periodic

f is of period $$\pi$$

I(n) < J(n)

I(n) > J(n)

I(n) = J(n)

I(n) + J(n) = 0

$$( - \infty , - 1) \cup (2,\infty )$$

$$[ - 1,1] \cup (2,\infty ) \cup ( - \infty , - 2)$$

$$( - \infty ,1) \cup (2,\infty )$$

$$[ - 1,1] \cup (2,\infty )$$

Here (a, b) $$ \equiv $$ {x : a < x < b} and [a, b] $$ \equiv $$ {x : a $$ \le $$ x $$ \le $$ b}