We have $$f\left( x \right) = \frac{{{x^2} + x + 2}}{{{x^2} + x + 1}} = \frac{{\left( {{x^2} + x + 1} \right) + 1}}{{{x^2} + x + 1}}$$
$$ = 1 + \frac{1}{{{{\left( {x + \frac{1}{2}} \right)}^2} + \frac{3}{4}}}$$
We can see here that as $$x \to \infty ,f\left( x \right) \to 1$$    which is the min value of $$f\left( x \right)\,{\text{.i}}{\text{.e}}{\text{.}}\,{f_{\min }} = 1.$$     Also $$f\left( x \right)$$  is max when $${\left( {x + \frac{1}{2}} \right)^2} + \frac{3}{4}$$    is min which is so when $$x = - \frac{1}{2}$$
$$\eqalign{ & {\text{i}}{\text{.e when}}\,{\left( {x + \frac{1}{2}} \right)^2} + \frac{3}{4} = \frac{3}{4} \cr & \therefore {f_{\max }} = 1 + \frac{3}{4} = \frac{7}{3} \cr & \therefore {R_f} = \left( {1,\frac{7}{3}} \right] \cr} $$

$$\eqalign{ & {x^2} - 2x + 3 = {\left( {x - 1} \right)^2} + 1 > 0\,\forall \,x\, \in \,R \cr & {\text{or }}f\left( x \right) = \operatorname{sgn} \left( {{x^2} - 2x + 3} \right) = 1 \cr & {\text{Hence, the range is }}\left\{ 1 \right\} \cr} $$

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

$$f:Z \to Z$$   is defined as $$f\left( x \right) = a{x^2} + bx + c$$
which implies for integer inputs, the function gives integer outputs.
$$ \Rightarrow f\left( 0 \right) = c = {Z_1}......\left( 1 \right)$$     (where $${Z_1}$$ is some integer)
Similarly, $$f\left( 1 \right) = a + b + c = {Z_2}......\left( 2 \right)$$       (where $${Z_2}$$ is some integer)
Equation $$\left( 2 \right) - \left( 1 \right)$$   gives $$a + b = {Z_2} - {Z_1},$$    which is also an integer.

$$\eqalign{ & {\text{We have }}fog\left( x \right) = f\left( {g\left( x \right)} \right) = \sin \left( {\ln \left| x \right|} \right) \cr & \therefore {R_1} = \left\{ {u: - 1 \leqslant u \leqslant 1} \right\}\,\left( {\because - 1 \leqslant \sin \theta \leqslant 1,\forall \theta } \right) \cr & {\text{Also }}gof\,\left( x \right) = g\left( {f\left( x \right)} \right) = \ln \left| {\sin x} \right| \cr & \because 0 \leqslant \left| {\sin x} \right| \leqslant 1 \cr & \therefore - \infty < \ln \left| {\sin x} \right| \leqslant 0 \cr & \therefore {R_2} = \left\{ {v: - \infty < v \leqslant 0} \right\} \cr} $$

$$\eqalign{ & y = {\log _3}\left( {5 + 4x + {x^2}} \right) \cr & {\text{For }}y - {x^2} + 4x + 5{\text{ we have}} \cr & \frac{{ - b}}{{2a}} = \frac{{ - 4}}{{ - 2}} = 2 = x \cr & {\text{Put }}x = 2 \cr & - 4 + 8 + 5 = 9 \cr & {\text{Maximum value}} = 9 \cr & \therefore \,y = {\log _3}\left( 9 \right) = 2 \cr & {\text{Also minimum value}} \sim 1 \cr & {\text{Range}} = \left( {0,\,2} \right) \cr} $$

Given functions are : $$f\left( x \right) = x$$   and $$g\left( x \right) = \left| x \right|$$
$$\therefore \,\left( {f + g} \right)\left( x \right) = f\left( x \right) + g\left( x \right) = x + \left| x \right|$$
According to definition of modulus function, \[\left( {f + g} \right)\left( x \right) = \left\{ \begin{array}{l} x + x,\,\,\,x \ge 0\\ x - x,\,\,\,x < 0 \end{array} \right. = \left\{ \begin{array}{l} 2x,\,\,{\rm}x \ge 0\\ \,0,\,\,\,\,{\rm}x < 0 \end{array} \right.\]

$$\eqalign{ & {2^{403}} = {2^{400}} \cdot {2^3} \cr & = {2^{4\infty 100}} \cdot {2^3} \cr & = {\left( {{2^4}} \right)^{100}} \cdot 8 \cr & = 8{\left( {{2^4}} \right)^{100}} = 8{\left( {16} \right)^{100}} \cr & = 8{\left( {1 + 15} \right)^{100}} \cr & = 8 + 15\mu \cr} $$
When $${2^{403}}$$ is divided by 15, then remainder is 8. Hence, fractional part of the number is $$\frac{8}{{15}}$$ Therefore value of $$k$$ is 8

$$f\left( x \right) = \left| {x - 1} \right| = \left\{ {_{x - 1,}^{ - x + 1}} \right.\,_{x \geqslant 1}^{x < 1}$$
Consider $$f\left( {{x^2}} \right) = {\left( {f\left( x \right)} \right)^2}$$
If it is true it should be $$\forall x$$
$$\therefore $$ Put $$x = 2$$
$${\text{LHS}} = f\left( {{2^2}} \right) = \left| {4 - 1} \right| = 3\,{\text{and}}\,{\text{RHS}} = {\left( {f\left( 2 \right)} \right)^2} = 1$$
$$\therefore \left( A \right)$$  is not correct
Consider $$f\left( {x + y} \right) = f\left( x \right) + f\left( y \right)$$
Put $$x = 2,y = 5$$   we get $$f\left( 7 \right) = 6;f\left( 2 \right) + f\left( 5 \right) = 1 + 4 = 5$$
$$\therefore \left( B \right)$$  is not correct
Consider $$f\left( {\left| x \right|} \right) = \left| {f\left( x \right)} \right|$$
Put $$x = - 5$$  then $$f\left( {\left| { - 5} \right|} \right) = f\left( 5 \right) = 4$$
$$f\left( {\left| { - 5} \right|} \right) = \left| { - 5 - 1} \right| = 6$$
$$\therefore \left( C \right)$$  is not correct.
Hence (D) is the correct alternative.

$$\eqalign{ & {\text{Let}}\,y = {2^{x\left( {x - 1} \right)}} \Rightarrow {x^2} - x - {\log _2}y = 0; \cr & x = \frac{1}{2}\left( {1 \pm \sqrt {1 + 4{{\log }_2}y} } \right) \cr} $$
Since $$x$$ is + ive, we choose only + out of $$ \pm \,{\text{for}}\,\left( {y \geqslant 1,{{\log }_2},y \geqslant 0} \right)$$
$$\eqalign{ & \therefore x = \frac{1}{2}\left( {1 + \sqrt {1 + 4{{\log }_2}y} } \right) \cr & {\text{or}}\,{f^{ - 1}}\left( x \right) = \frac{1}{2}\left( {1 + \sqrt {1 + 4{{\log }_2}x} } \right) \cr} $$