Let $P(x)$ be a polynomial with real coefficients such that $P\left(\sin ^2 x\right)=P\left(\cos ^2 x\right)$ for all $x \in[0, \pi / 2)$. Consider the following statements:

$I.$ $P(x)$ is an even function.

$II.$ $P(x)$ can be expressed as a polynomial in $(2 x-1)^2$

$III.$ $P(x)$ is a polynomial of even degree.

Then,

Consider the identity function $I _{ N }: N \rightarrow N$ defined as $I _{ N }$ $(x)=x$  $\forall $  $x \in N$ Show that although $I _{ N }$ is onto but $I _{ N }+ I _{ N }:$  $ N \rightarrow N$ defined as $\left(I_{N}+I_{N}\right)(x)=$ $I_{N}(x)+I_{N}(x)$ $=x+x=2 x$ is not onto.

The function $f(x) = \;|px - q|\; + r|x|,\;x \in ( - \infty ,\;\infty )$, where $p > 0,\;q > 0,\;r > 0$ assumes its minimum value only at one point, if

Let $R$ be the set of real numbers and $f: R \rightarrow R$ be defined by $f(x)=\frac{\{x\}}{1+[x]^2}$, where $[x]$ is the greatest integer less than or equal to $x$, and $\left\{x{\}}=x-[x]\right.$. Which of the following statements are true?

$I.$ The range of $f$ is a closed interval.

$II.$ $f$ is continuous on $R$.

$III.$ $f$ is one-one on $R$

Period of $f(x) = nx + n - [nx + n]$, $n \in N$

where [ ] denotes the greatest integer function is :

The function $f(x) = \frac{{{{\sec }^{ - 1}}x}}{{\sqrt {x - [x]} }},$ where $[.]$ denotes the greatest integer less than or equal to $x$  is defined for all  $x$  belonging to