$$\eqalign{ & f\left( x \right) = \frac{{{x^2}}}{{1 - {x^2}}} \Rightarrow f\left( { - x} \right) = \frac{{{x^2}}}{{1 - {x^2}}} = f\left( x \right) \cr & f'\left( { - x} \right) = \frac{{2x}}{{{{\left( {1 - {x^2}} \right)}^2}}} \cr & \mathop {\lim }\limits_{x \to \pm \infty } f\left( x \right) = - 1 \cr & \therefore f\left( x \right)\,{\text{increases}}\,{\text{in}}\,x \in \left( {10,\infty } \right) \cr & {\text{Also}}\,f\left( 0 \right) = 0\,{\text{and}} \cr & {\text{and}}\,f\left( x \right)\,{\text{is}}\,{\text{even}}\,{\text{function}} \cr & \therefore {\text{Set}}\,A = R - \left[ { - 1,0} \right) \cr} $$
And the graph of function $$f\left( x \right)$$  is

Alternative For $$f$$ to be subjective $$A =$$  Range of $$f.$$
$$\eqalign{ & \frac{{{x^2}}}{{1 - {x^2}}} = y \Rightarrow {x^2} = y - {x^2}y \cr & \Rightarrow x = \pm \sqrt {\frac{y}{{1 + y}}} \Rightarrow y\left( {1 + y} \right) \geqslant 0{\text{ and }}y \ne - 1 \cr & \Rightarrow y \in \left( { - \infty , - 1} \right)U\left[ {0,\infty } \right) \Rightarrow y \in R - \left[ { - 1,0} \right) \cr & \Rightarrow A = R - \left[ { - 1,0} \right) \cr} $$

$$\eqalign{ & f\left( 1 \right) = 1 \cr & f\left( 2 \right) = 2\sum\limits_{x = 1}^1 {f\left( x \right)} \cr & = 2f\left( 1 \right) = 2 \times 1 = 2 \cr & f\left( 3 \right) = 2\sum\limits_{x = 1}^2 {f\left( x \right)} \cr & = 2f\left( 1 \right) + 2f\left( 2 \right) = 2 \times 1 + 2 \times 2 = 6 \cr & f\left( 4 \right) = 2\sum\limits_{x = 1}^3 {f\left( x \right)} \cr & = 2\left[ {f\left( 1 \right) + f\left( 2 \right) + f\left( 3 \right)} \right] \cr & = 2\left[ {1 + 2 + 6} \right] = 18 \cr & {\text{The sequence is}} = 1 + 2 + 6 + 18 + ..... \cr & \sum\limits_{n = 1}^m {f\left( n \right)} = 1 + 2\left( {1 + 3 + 9 + .....} \right) \cr & \sum\limits_{n = 1}^m {f\left( n \right)} = 1 + 2\frac{{\left( 1 \right)\left( {{3^{m - 1}} - 1} \right)}}{2} \cr & \sum\limits_{n = 1}^m {f\left( n \right)} = {3^{m - 1}} \cr} $$

$$nx + n - \left[ {nx + n} \right]$$     has the period 1 and $$\tan \frac{{\pi x}}{2}$$  has the period $$\frac{\pi }{{\frac{\pi }{2}}}$$  i.e., 2.
LCM of 1, 2 is 2.

The function must be even.
$$\eqalign{ & f\left( {0 + 0} \right) = f\left( 0 \right) + f\left( 0 \right)\,\,\,\,\,\therefore f\left( 0 \right) = 0 \cr & f\left( {x - x} \right) = f\left( x \right) + f\left( { - x} \right) \cr & {\text{or }}0 = f\left( x \right) + f\left( { - x} \right)\,\,\,\,\,\,\,\,\,\therefore f\left( x \right){\text{ is odd}}{\text{.}} \cr & {\text{Also,}}\,\,{\log _e}\left( {x + \sqrt {{x^2} + 1} } \right){\text{ }}\,{\text{is odd because}} \cr & {\log _e}\left( { - x + \sqrt {{{\left( { - x} \right)}^2} + 1} } \right) = - {\log _e}\left( {x + \sqrt {{x^2} + 1} } \right) \cr & f\left( x \right) = \cos \,x + \sin \,x = \sqrt 2 \,\cos \left( {x - \frac{\pi }{4}} \right) \cr} $$
As $$\cos \left( { - x - \frac{\pi }{4}} \right) \ne \cos \left( {x - \frac{\pi }{4}} \right)$$      for every $$x,\,\cos \,x + \sin \,x$$    is not even.

$$\eqalign{ & {2^x} + {2^y} = 2 \cr & {2^y} = 2 - {2^x} \cr & \because \,{2^y} > 0\,\,\,\left[ {{\text{exponential function}}} \right] \cr & \therefore \,2 - {2^x} > 0 \cr & {2^x} < 2 \cr & x < 1 \cr & \therefore \,x \in \left( { - \infty ,\,1} \right) \cr} $$

Clearly, $${x^2} + x + k \geqslant 0$$    for all $$x.$$
So, $$D = 1 - 4k \leqslant 0\,\,\, \Rightarrow k \geqslant \frac{1}{4}$$

Given $$f\left( x \right) = 2 - \left| {x - 5} \right|$$
Domain of $$f\left( x \right)$$  is defined for all real values of $$x$$
$$\eqalign{ & {\text{Since, }}\left| {x - 5} \right| \geqslant 0 \cr & \Rightarrow - \left| {x - 5} \right| \leqslant 0 \cr & \Rightarrow 2 - \left| {x - 5} \right| \leqslant 2 \cr & \Rightarrow f\left( x \right) \leqslant 2 \cr} $$
Hence, range of $$f\left( x \right)$$  is $$\left( { - \infty ,\,2} \right]$$

$$\eqalign{ & {\log _{10}}\left( {1 + {x^3}} \right) > 0 \cr & {\text{or, }}1 + {x^3} > {10^0} = 1 \cr & {\text{or, }}{x^3} > 0\,\, \Rightarrow x\, \in \left( {0,\, + \infty } \right) \cr} $$

Verify by trial.

In the definition of function
$$f\left( x \right) = \frac{{x\left( {x - p} \right)}}{{q - p}} + \frac{{x\left( {p - q} \right)}}{{p - q}} = p$$
Putting $$p$$ and $$q$$ in place of $$x,$$ we get
$$\eqalign{ & f\left( p \right) = \frac{{p\left( {p - p} \right)}}{{q - p}} + \frac{{p\left( {p - q} \right)}}{{p - q}} = p \cr & \Rightarrow f\left( p \right) = p{\text{ and}} \cr & f\left( q \right) = \frac{{q\left( {q - p} \right)}}{{q - p}} + \frac{{q\left( {p - q} \right)}}{{p - q}} = q \cr & \Rightarrow f\left( q \right) = q \cr & {\text{Putting }}x = \left( {p + q} \right) \cr & f\left( {p + q} \right) = \frac{{\left( {p + q} \right)\left( {p + q - p} \right)}}{{\left( {q - p} \right)}} + \frac{{\left( {p + q} \right)\left( {p + q - q} \right)}}{{\left( {p - q} \right)}} \cr & = \frac{{\left( {p + q} \right)q}}{{\left( {q - p} \right)}} + \frac{{\left( {p + q} \right)p}}{{\left( {p - q} \right)}} \cr & = \frac{{pq + {q^2} - {p^2} - pq}}{{\left( {q - p} \right)}} \cr & = \frac{{{q^2} - {p^2}}}{{\left( {q - p} \right)}} \cr & = \frac{{\left( {q - p} \right)\left( {q + p} \right)}}{{\left( {q - p} \right)}} \cr & = \left( {p + q} \right) \cr & = f\left( q \right) + f\left( p \right) \cr & {\text{So,}}\,f\left( p \right) + f\left( q \right) = f\left( {p + q} \right) \cr} $$