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

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 $f(x) = 2x$ and $g$ is the identity function,then:

If $f(x)=3[x]+\{x+1\}$,where $[x]$ is the greatest integer function of $x$ and $\{x\}$ is the fractional part function of $x$,then $f(-1.32)=$

Let $f, g$ and $h$ be the real-valued functions defined on $\mathbb{R}$ as $f(x) = \begin{cases} \frac{x}{|x|}, & x \neq 0 \\ 1, & x=0 \end{cases}$,$g(x) = \begin{cases} \frac{\sin(x+1)}{x+1}, & x \neq -1 \\ 1, & x=-1 \end{cases}$ and $h(x) = 2[x] - f(x)$,where $[x]$ is the greatest integer $\leq x$. Then the value of $\lim_{x \rightarrow 1} g(h(x-1))$ is

Let $f:[0,1] \rightarrow \mathbb{R}$ be an injective continuous function that satisfies the condition $-1 < f(0) < f(1) < 1$. Then,the number of functions $g:[-1,1] \rightarrow [0,1]$ such that $(g \circ f)(x) = x$ for all $x \in [0,1]$ is