For $0 \leq x \leq 1$,what type of function is $f(x) = |x| + |x - 1|$?

The real-valued function $f: R \rightarrow [ \frac{5}{2}, \infty )$ defined by $f(x) = | 2x + 1 | + | x - 2 |$ is

$A$ function $f: R \rightarrow R$ defined by $f(x) = \begin{cases} 2x+3, & x \leq \frac{4}{3} \\ -3x^2+8x, & x > \frac{4}{3} \end{cases}$ 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 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	$4$. $A = B = R^{+}$

Given that $f: S \rightarrow R$ is said to have a fixed point at $c \in S$ if $f(c)=c$. Let $f:[1, \infty) \rightarrow R$ be defined by $f(x)=1+\sqrt{x}$. Then:

If $f : R \rightarrow R$ such that $f(x) = 5x - 3\cos x - 4\sin x$,then the function $f(x)$ is