Let $f(x) = \frac{x^2 - 4}{x^2 + 4}$ for $|x| > 2$. Then the function $f: (- \infty, -2] \cup [2, \infty) \to (-1, 1)$ is

Let $R$ denote the set of all real numbers and $R^{+}$ denote the set of all positive real numbers. For the subsets $A$ and $B$ of $R$, define $f: A \rightarrow B$ by $f(x) = x^2$ for $x \in A$. Match the following lists:
| Column $I$ | Column $II$ |
| :--- | :--- |
| $A$. $f$ is one-one and onto, if | $1$. $A = R^{+}, B = R$ |
| $B$. $f$ is one-one but not onto, if | $2$. $A = B = R$ |
| $C$. $f$ is onto but not one-one, if | $3$. $A = R, B = R^{+}$ |
| $D$. $f$ is neither one-one nor onto, if | $4$. $A = B = R^{+}$ |

If $f(x)=\frac{2^{x}+2^{-x}}{2}$,then $f(x+y) \cdot f(x-y)$ is

If $f(x) = \cos (\log x)$,then $f(x^2)f(y^2) - \frac{1}{2}\left[ f\left( \frac{x^2}{y^2} \right) + f(x^2y^2) \right]$ has the value

Which of the following pairs of functions are identical?

The function $f(x) = \sec \left[ \log \left( x + \sqrt{1 + x^2} \right) \right]$ is . . . . . . function.