Let function $f(x) = {x^2} + x + \sin x - \cos x + \log (1 + |x|)$ be defined over the interval $[0, 1]$. The odd extensions of $f(x)$ to interval $[-1, 1]$ is

Domain of $f (x)$ = $\sqrt {{{\log }_2}\left( {\frac{{10x - 4}}{{4 - {x^2}}}} \right) - 1} $ , is

Statement $-1$ : The equation $x\, log\, x = 2 - x$ is satisfied by at least one value of $x$ lying between $1$ and $2$

Statement $-2$ : The function $f(x) = x\, log\, x$ is an increasing function in $[1, 2]$ and $g (x) = 2 -x$ is a decreasing function in $[ 1 , 2]$ and the graphs represented by these functions intersect at a point in $[ 1 , 2]$

The period of function

$f\left( x \right) = {\cos ^2}\left( {\sin x} \right) + {\sin ^2}\left( {\cos x} \right)$ is

Let, $f(x)=\left\{\begin{array}{l} x \sin \left(\frac{1}{x}\right) \text { when } x \neq 0 \\ 1 \text { when } x=0 \end{array}\right\}$ and $A=\{x \in R: f(x)=1\} .$ Then, $A$ has

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.