If $f(x) = (\frac{3}{5})^x + (\frac{4}{5})^x - 1$,$x \in R$,then the equation $f(x) = 0$ has

The function $f: R \to R$ defined by $f(x) = x^2$ for all $x \in R$ is

Let $g(x) = 1 + x - [x]$ and $f(x) = \begin{cases} -1, & x < 0 \\ 0, & x = 0 \\ 1, & x > 0 \end{cases}$. Then for all $x$,$f(g(x))$ is equal to

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 items in Column-$I$ with the items in Column-$II$.

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	$4$. $A = B = R^{+}$

The function $f(x) = \frac{x^2+2x-15}{x^2-4x+9}$,$x \in R$ is

The function $f:[0,3] \rightarrow [1,29]$,defined by $f(x)=2x^3-15x^2+36x+1$,is