$$f\left( x \right) = \frac{{x\left( {\sqrt {{x^2} + 1} + \sqrt x } \right)}}{{x + 1 - x}} = x\left( {\sqrt {{x^2} + 1} + \sqrt x } \right).$$           As $$f\left( x \right)$$  is real valued, $$x \geqslant 0.$$
$$\therefore \,f\left( x \right)$$   is differentiable at $$x=0$$  if $$\mathop {\lim }\limits_{h \to 0} \frac{{f\left( {0 + h} \right) - f\left( 0 \right)}}{h}$$     is finite and definite.
$$\mathop {\lim }\limits_{h \to 0} \frac{{f\left( {0 + h} \right) - f\left( 0 \right)}}{h} = \mathop {\lim }\limits_{h \to 0} \frac{{h\left( {\sqrt {{h^2} + 1} + \sqrt h } \right)}}{h} = 1.$$           So, (B) is correct.
A function which is finitely differentiable is also continuous.

Graph of $$y = \left| {\left| x \right| - 1} \right|$$   is as follows :

The graph has sharp turnings at $$x =- 1, \,0$$   and $$1;$$ and hence not differentiable at $$x =- 1,\,0, \,1$$

\[f:R \to R,\,f\left( x \right) = \left\{ \begin{array}{l} \,\,\,{x^2},\,\,\,\,\,x \ge 0\\ - x,\,\,\,\,\,x < 0 \end{array} \right.\]
For continuity at $$x = 0$$
$$\eqalign{ & f\left( {0 - 0} \right) = \mathop {\lim }\limits_{h \to 0} f\left( {0 - h} \right) \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \mathop {\lim }\limits_{h \to 0} \left[ {\left( {0 - h} \right)} \right] \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \mathop {\lim }\limits_{h \to 0} \,h = 0 \cr & f\left( {0 + 0} \right) = \mathop {\lim }\limits_{h \to 0} f\left( {0 + h} \right) \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \mathop {\lim }\limits_{h \to 0} {\left( {0 + h} \right)^2} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 0 \cr & {\text{and }}f\left( 0 \right) = 0 \cr & f\left( x \right){\text{ is continuous at }}x = 0 \cr & {\text{For differentiability at }}x = 0 \cr & \mathop {\lim }\limits_{h \to 0} \frac{{ - \left( { - h} \right) - 0}}{{ - h}} = \mathop {\lim }\limits_{h \to 0} \frac{h}{{ - h}} = - 1 \cr & {\text{and }}\mathop {\lim }\limits_{h \to 0} \frac{{f\left( {0 + h} \right) - f\left( 0 \right)}}{h} = \mathop {\lim }\limits_{h \to 0} \,h = 0 \cr & f\left( x \right){\text{ is not differentiable at }}x = 0 \cr} $$

$$\eqalign{ & y = \frac{{\left( {1 - x} \right)\left( {1 + x} \right)\left( {1 + {x^2}} \right).....\left( {1 + {x^{{2^n}}}} \right)}}{{1 - x}} = \frac{{1 + {x^{{2^{n + 1}}}}}}{{1 - x}} \cr & \therefore \frac{{dy}}{{dx}} = \frac{{ - {2^{n + 1}}.{x^{{2^{n + 1}} - 1}}.\left( {1 - x} \right) + 1 - {x^{{2^{n + 1}}}}}}{{{{\left( {1 - x} \right)}^2}}}\,; \cr & \therefore {\text{ at }}x = 0,\,\frac{{dy}}{{dx}} = \frac{{ - {2^{n + 1}}.0.1 + 1 - 0}}{{{1^2}}} = 1 \cr} $$

$$\eqalign{ & {\text{Given }}f''\left( x \right){\text{ is continuous at }}x = 0 \cr & = \mathop {\lim }\limits_{x \to 0} f''\left( x \right) = f''\left( 0 \right) = 4 \cr & {\text{Now, }}\mathop {\lim }\limits_{x \to 0} \frac{{2f\left( x \right) - 3f\left( {2x} \right) + f\left( {4x} \right)}}{{{x^2}}}\,\,\,\,\left[ {\frac{0}{0}{\text{ form}}} \right] \cr & = \mathop {\lim }\limits_{x \to 0} \frac{{2f'\left( x \right) - 6f'\left( {2x} \right) + 4f'\left( {4x} \right)}}{{2x}}\,\,\,\,\left[ {\frac{0}{0}{\text{ form}}} \right] \cr & = \mathop {\lim }\limits_{x \to 0} \frac{{2f''\left( x \right) - 12f''\left( x \right) + 16f''\left( {4x} \right)}}{2} \cr & \left[ {{\text{Using L'Hospital rule successively}}} \right] \cr & = \frac{{2f''\left( 0 \right) - 12f''\left( 0 \right) + 16f''\left( 0 \right)}}{2} \cr & = 12 \cr} $$

$$\eqalign{ & 3f\left( x \right) - 2f\left( {\frac{1}{x}} \right) = x.....(1) \cr & {\text{Put }}x = \frac{1}{x},{\text{ then }}3f\left( {\frac{1}{x}} \right) - 2f\left( x \right) = \left( {\frac{1}{x}} \right).....(2) \cr & {\text{Solving (1) and (2), we get}} \cr & 5f\left( x \right) = 3x + \frac{2}{x}\, \Rightarrow f'\left( x \right) = \frac{3}{5} - \frac{2}{{5{x^2}}} \cr & \therefore \,f'\left( 2 \right) = \frac{3}{5} - \frac{2}{{20}} = \frac{1}{2} \cr} $$

The doubtful points are $$x = \frac{1}{2},\,1,\,\frac{\pi }{2}$$
$$\left| {x - 1} \right|$$  is differentiable at $$\frac{1}{2},\,\frac{\pi }{2}$$  but not differentiable at $$1.$$
$$\left| {x - \frac{1}{2}} \right|$$  is differentiable at $$1,\,\frac{\pi }{2}$$  but not differentiable at $$\frac{1}{2}$$
Remember $$\left| {x - a} \right|$$  is continuous everywhere, and differentiable everywhere except at $$x=a$$
$$\tan \,x$$  is differentiable at $$\frac{1}{2},\,1$$  but not differentiable at $$\frac{\pi }{2}$$
$$\therefore \,f\left( x \right)$$   is not differentiable at $$\frac{1}{2},\,1,\,\frac{\pi }{2}$$

$$\eqalign{ & {\text{We have, }}y = {t^{10}} + 1 \cr & \therefore \,\frac{{dy}}{{dt}} = 10{t^9} \cr & {\text{and }}x = {t^8} + 1 \Rightarrow \frac{{dx}}{{dt}} = 8{t^7} \cr & \therefore \,\frac{{dy}}{{dx}} = \frac{{\frac{{dy}}{{dt}}}}{{\frac{{dx}}{{dt}}}} = \frac{{10{t^9}}}{{8{t^7}}} = \frac{5}{4}\frac{{{t^{10}}}}{{{t^8}}} = \frac{{5\left( {y - 1} \right)}}{{4\left( {x - 1} \right)}}.....({\text{i}}) \cr & {\text{Differentiate (i) with respect to }}x,{\text{ we get}} \cr & \therefore \,\frac{{{d^2}y}}{{dx}} = \frac{5}{4}.\frac{{\left( {x - 1} \right).\frac{{dy}}{{dx}} - \left( {y - 1} \right)}}{{{{\left( {x - 1} \right)}^2}}} \cr & = \frac{5}{4}.\frac{1}{{\left( {x - 1} \right)}}\left[ {\frac{{dy}}{{dx}} - \frac{{\left( {y - 1} \right)}}{{\left( {x - 1} \right)}}} \right] \cr & = \frac{5}{4}.\frac{1}{{\left( {x - 1} \right)}}\left[ {\frac{5}{4}.\frac{{\left( {y - 1} \right)}}{{\left( {x - 1} \right)}} - \frac{{\left( {y - 1} \right)}}{{\left( {x - 1} \right)}}} \right]\,\,\,\,\,\,\,\,\left[ {{\text{Putting (i)}}} \right] \cr & = \frac{5}{4}.\frac{{\left( {y - 1} \right)}}{{{{\left( {x - 1} \right)}^2}}}\left( {\frac{5}{4} - 1} \right) \cr & = \frac{5}{{16}}.\frac{{{t^{10}}}}{{{{\left( {{t^8}} \right)}^2}}} \cr & = \frac{5}{{16{t^6}}} \cr} $$

$$\eqalign{ & {\text{Using LMVT in}}\left[ {0,\,2} \right] \cr & \frac{{f\left( 2 \right) - f\left( 0 \right)}}{{2 - 0}} = f'\left( c \right),{\text{ where}}\,c\, \in \left( {0,\,2} \right) \cr & \frac{{f\left( 2 \right) + 3}}{2} \leqslant 5\,\,\, \Rightarrow f\left( 2 \right) \leqslant 7 \cr} $$

Since $$g\left( x \right)$$  is differentiable, it will be continuous at $$x=3$$
$$\eqalign{ & \therefore \mathop {\lim }\limits_{x \to {3^ - }} g\left( x \right) = \mathop {\lim }\limits_{x \to {3^ + }} g\left( x \right) \cr & 2k = 3m + 2.....(1) \cr} $$
Also $$g\left( x \right)$$  is differentiable at $$x = 0$$
$$\eqalign{ & \therefore \mathop {\lim }\limits_{x \to {3^ - }} g'\left( x \right) = \mathop {\lim }\limits_{x \to {3^ + }} g'\left( x \right) \cr & \frac{K}{{2\sqrt {3 + 1} }} = m \cr & k = 4m.....(2) \cr} $$
Solving (1) and (2), we get
$$\eqalign{ & m = \frac{2}{5},\,\,\,k = \frac{8}{5} \cr & \therefore k + m = 2 \cr} $$