If $$n = 0$$  then $$h\left( n \right)$$ is not defined, so, $$'h'$$ is not a function. All other are functions.

$$\eqalign{ & f\left( x \right) = \log \left( {x + \sqrt {{x^2} + 1} } \right) \cr & f\left( { - x} \right) = \log \left\{ { - x + \sqrt {{x^2} + 1} } \right\} = \log \left\{ {\frac{{ - {x^2} + {x^2} + 1}}{{x + \sqrt {{x^2} + 1} }}} \right\} \cr & = - \log \left( {x + \sqrt {{x^2} + 1} } \right) = - f\left( x \right) \cr & \Rightarrow f\left( x \right){\text{ is an odd function}}{\text{.}} \cr} $$

$${x^2} + \lambda x + \mu = $$    integer for all integer $$x.$$
When $$x = 0,\,\,{x^2} + \lambda x + \mu = 0 = $$      integer.
So, $$x\left( {x + \lambda } \right)$$   is an integer for all integral values of $$x.$$
Therefore, $$\lambda $$ is an integer.

If $$x < 1,\,f\left( x \right) = - \left( {x - 1} \right) - \left( {x - 2} \right) = - 2x + 3.$$         In this interval, $$f\left( x \right)$$  is decreasing.

If $$1 \leqslant x < 2,\,f\left( x \right) = x - 1 - \left( {x - 2} \right) = 1$$
In this interval, $$f\left( x \right)$$  is constant.
If $$2 \leqslant x \leqslant 3,\,f\left( x \right) = x - 1 + x - 2 = 2x - 3$$
In this interval, $$f\left( x \right)$$  is increasing.
$$\therefore \max \,f\left( x \right) = $$    the greatest among $$f\left( { - 1} \right)$$  and $$f\left( 3 \right) = 5,\,\,\min \,f\left( x \right) = f\left( 1 \right) = 1$$
So, range $$ = \left[ {1,\,5} \right].$$

$$\eqalign{ & f:{\bf{R}} \to {\bf{R}}\,\& \,g:{\bf{R}} \to {\bf{R}} \cr & {\text{We know that min}}\left\{ {{f_1}\left( x \right),\,{f_2}\left( x \right)} \right\} \cr & = \frac{{\left( {{f_1}\left( x \right) + {f_2}\left( x \right)} \right) - \left| {{f_1}\left( x \right) - {f_2}\left( x \right)} \right|}}{2} \cr & \therefore \,\min \left\{ {f\left( x \right) - g\left( x \right),\,0} \right\} \cr & = \frac{{\left( {f\left( x \right) - g\left( x \right) + 0} \right) - \left| {f\left( x \right) - g\left( x \right) - 0} \right|}}{2} \cr & = \frac{{\left( {f\left( x \right) - g\left( x \right)} \right) - \left| {f\left( x \right) - g\left( x \right)} \right|}}{2} \cr} $$

$$f\left( x \right) = \left( {x - 1} \right)\left( {x - 2} \right)\left( {x - 3} \right)$$
 $$\therefore f\left( 1 \right) = f\left( 2 \right) = f\left( 3 \right) = 0.$$      So, the function is many-one.

Clearly, for $$x < 0,\,f\left( x \right) < 0$$    and goes on decreasing as $$x$$ decreases.
For $$x > 3,\,f\left( x \right) > 0$$    and goes on increasing as $$x$$ increases.
$$\therefore \,f\left( x \right)$$  can have all real values. So, $$f$$ is onto.

Checking the options
$$\eqalign{ & \left( {\bf{A}} \right){\text{ L}}{\text{.H}}{\text{.S}}{\text{.}} = x\,\operatorname{sgn} \,x = x \times \left( { + 1} \right) = x{\text{ if }}x > 0 \cr & x \times 0 = 0{\text{ if }}x = 0 \cr & x \times \left( { - 1} \right) = - x{\text{ if }}x < 0 \cr & = \left| x \right| = {\text{R}}{\text{.H}}{\text{.S}}{\text{.}} \cr} $$
Therefore, statement is correct.
$$\eqalign{ & \left( {\bf{B}} \right){\text{ L}}{\text{.H}}{\text{.S}}{\text{.}} = \left| x \right|\operatorname{sgn} \,x = \left( { + x} \right) \times \left( { + 1} \right) = x{\text{ if }}x > 0 \cr & 0 \times 0 = 0{\text{ if }}x = 0 \cr & \left( { - x} \right) \times \left( { - 1} \right) = x{\text{ if }}x < 0 \cr} $$
Therefore, statement is correct.
$$\eqalign{ & \left( {\bf{C}} \right){\text{ L}}{\text{.H}}{\text{.S}}{\text{.}} = \left( {x\,\left( {\operatorname{sgn} \,x} \right)} \right) \times \operatorname{sgn} \,x \cr & = \left| x \right|\operatorname{sgn} \,x\,\,\,\,\,\left( {{\text{from }}\left( {\text{A}} \right)} \right) \cr & = x\,\,\,\,\,\left( {{\text{from }}\left( {\text{B}} \right)} \right) = {\text{R}}{\text{.H}}{\text{.S}}{\text{.}} \cr} $$
Therefore, statement is correct.
$$\eqalign{ & \left( {\bf{D}} \right){\text{ L}}{\text{.H}}{\text{.S}}{\text{.}} = \left( {\left| x \right|\operatorname{sgn} \,x} \right){\left( {\operatorname{sgn} \,x} \right)^2} \cr & = x{\left( {\operatorname{sgn} \,x} \right)^2}\,\,\,\,\,\left( {{\text{from}}\left( {\text{B}} \right)} \right) \cr & = \left| x \right| \times \left( {\operatorname{sgn} \,x} \right)\,\,\,\,\,\left( {{\text{from}}\left( {\text{A}} \right)} \right) \cr & = x\,\,\,\,\,\left( {{\text{from}}\left( {\text{B}} \right)} \right) \ne {\text{R}}{\text{.H}}{\text{.S}}{\text{.}} \cr} $$
Therefore, statement is incorrect.

From $$E$$ to $$F$$ we can define, in all, $$2 \times 2 \times 2 \times 2 = 16$$     functions (2 options for each element of $$E$$ ) out of which 2 are into, when all the elements of $$E$$ either map to 1 or to 2.
∴ No. of onto functions $$= 16 - 2 = 14$$

Given function is : $$f\left( {x + 1} \right) = {x^2} - 3x + 2$$
This function is valid for all real values of $$x.$$ So, putting $$x - 1$$  in place of $$x,$$ we get
$$\eqalign{ & f\left( x \right) = f\left( {x - 1 + 1} \right) \cr & \Rightarrow f\left( x \right) = {\left( {x - 1} \right)^2} - 3\left( {x - 1} \right) + 2 \cr & \Rightarrow f\left( x \right) = {x^2} - 2x + 1 - 3x + 3 + 2 \cr & \Rightarrow f\left( x \right) = {x^2} - 5x + 6 \cr} $$

$$\eqalign{ & \left| {{{\log }_4}{x^2}} \right| \leqslant 1 \cr & \Rightarrow - 1 \leqslant {\log _4}{x^2} \leqslant 1 \cr & \Rightarrow \frac{1}{4} \leqslant {x^2} \leqslant 4 \cr & \Rightarrow \frac{1}{2} \leqslant x \leqslant 2{\text{ or }} - 2 \leqslant x \leqslant - \frac{1}{2} \cr} $$