For $$f\left( x \right)$$  to be continuous at $$x = 0,$$  we should have $$\mathop {\lim }\limits_{x \to 0} f\left( x \right) = f\left( 0 \right) = 12{\left( {\log \,4} \right)^3}$$
$$\eqalign{ & \mathop {\lim }\limits_{x \to 0} f\left( x \right) = \mathop {\lim }\limits_{x \to 0} {\left( {\frac{{{4^x} - 1}}{x}} \right)^3} \times \frac{{\left( {\frac{x}{p}} \right)}}{{\left( {\sin \frac{x}{p}} \right)}} \cdot \frac{{p{x^2}}}{{\log \left( {1 + \frac{1}{3}{x^2}} \right)}} \cr & = {\left( {\log \,4} \right)^3} \cdot 1 \cdot p.\mathop {\lim }\limits_{x \to 0} \left( {\frac{{{x^2}}}{{\frac{1}{3}{x^2} - \frac{1}{{18}}{x^4} + .....}}} \right) \cr & = 3p{\left( {\log \,4} \right)^3} \cr & {\text{Hence, }}p = 4 \cr} $$

$$f\left( x \right) = \frac{{\tan \left( {\pi \left[ {x - \pi } \right]} \right)}}{{1 + {{\left[ x \right]}^2}}}$$
By def. $$\left[ {x - \pi } \right]$$  is an integer whatever be the value of $$x.$$
And so $$\pi \left[ {x - \pi } \right]$$  is an integral multiple of $$\pi .$$
Consequently $$\tan \left( {\pi \left[ {x - \pi } \right]} \right) = 0,\,\,\forall \,x.$$
And since $$1 + {\left[ x \right]^2} \ne 0$$   for any $$x,$$ we conclude that $$f\left( x \right) = 0.$$
Thus $$f\left( x \right)$$  is constant function and so, it is continuous and differentiable any number of times, that is $$f'\left( x \right),\,f''\left( x \right),\,f'''\left( x \right),.....$$      all exist for every $$x,$$ their value being 0 at every pt. $$x.$$
Hence, out of all the alternatives only (D) is correct.

$$\eqalign{ & \mathop {\lim }\limits_{x \to 0} \frac{{2\left( {1 - \cos \,x} \right)}}{{{x^2}}} \cr & = \mathop {\lim }\limits_{x \to 0} \frac{{2.2{{\sin }^2}\frac{x}{2}}}{{{x^2}}} \cr & = 4\mathop {\lim }\limits_{x \to 0} \frac{{\sin \frac{x}{2}}}{{\frac{x}{2} \times 2}}.\mathop {\lim }\limits_{x \to 0} \frac{{\sin \frac{x}{2}}}{{\frac{x}{2} \times 2}} \cr & = \mathop {\lim }\limits_{x \to 0} \frac{{\sin \frac{x}{2}}}{{\frac{x}{2}}}.\mathop {\lim }\limits_{x \to 0} \frac{{\sin \frac{x}{2}}}{{\frac{x}{2}}} \cr & = 1 \times 1 \cr & = 1 \cr} $$

$$\eqalign{ & \left( {\bf{A}} \right)\,\,\mathop {\lim }\limits_{x \to 1} f\left( x \right) = {\text{ does not exist}}{\text{.}} \cr & \left( {\bf{B}} \right)\,\,\mathop {\lim }\limits_{x \to 1} f\left( x \right) = {\text{ does not exist}}{\text{.}} \cr & \left( {\bf{C}} \right)\,\,\mathop {\lim }\limits_{x \to 1} f\left( x \right) = {\text{ does not exist}}{\text{.}} \cr} $$
$$\left( {\bf{D}} \right)\,\,\mathop {\lim }\limits_{x \to 1} f\left( x \right) = \frac{{ - 1}}{{2\sqrt 2 }},$$     therefore $$f\left( x \right)$$  has removable discontinuity at $$x = 1.$$

$$\eqalign{ & {\text{Let }}f\left( x \right) = {x^2} - 2x + 3 \cr & {\text{Since, }}f'\left( c \right) = \frac{{f\left( {\frac{3}{2}} \right) - f\left( 1 \right)}}{{\frac{3}{2} - 1}}\,\,\,\,\,\,\,\,\left( {{\text{given}}} \right) \cr & \Rightarrow 2c - 2 = \frac{{\frac{9}{4} - \frac{6}{2} + 3 - \left( {1 - 2 + 3} \right)}}{{\frac{3}{2} - 1}} \cr & \Rightarrow c = \frac{5}{4} \cr} $$

$$\eqalign{ & f\left[ {{{\left( {\frac{\pi }{2}} \right)}^ - }} \right] = \mathop {\lim }\limits_{h \to 0} \frac{{1 - {{\sin }^3}\left[ {\left( {\frac{\pi }{2}} \right) - h} \right]}}{{3\,{{\cos }^2}\left[ {\left( {\frac{\pi }{2}} \right) - h} \right]}} \cr & = \mathop {\lim }\limits_{h \to 0} \frac{{1 - {{\cos }^3}h}}{{3\,{{\sin }^2}h}} \cr & = \frac{1}{2} \cr & f\left[ {{{\left( {\frac{\pi }{2}} \right)}^ + }} \right] = \mathop {\lim }\limits_{h \to 0} \frac{{q\left[ {1 - \sin \left\{ {\left( {\frac{\pi }{2}} \right) + h} \right\}} \right]}}{{{{\left[ {\pi - 2\left\{ {\left( {\frac{\pi }{2}} \right) + h} \right\}} \right]}^2}}} \cr & = \mathop {\lim }\limits_{h \to 0} \frac{{q\left( {1 - \cos \,h} \right)}}{{4\,{h^2}}} \cr & = \frac{q}{8} \cr & \therefore \,p = \frac{1}{2} = \frac{q}{8} \cr & \Rightarrow p = \frac{1}{2},\,\,q = 4 \cr} $$

$$\eqalign{ & {\text{We have }}f\left( x \right) = {\left[ x \right]^2} - \left[ {{x^2}} \right] \cr & {\text{At }}x = 0, \cr & {\text{L}}{\text{.H}}{\text{.L}}{\text{. }} = \mathop {\lim }\limits_{h \to 0} f\left( { - h} \right) = \mathop {\lim }\limits_{h \to 0} {\left[ { - h} \right]^2} - \left[ {{{\left( { - h} \right)}^2}} \right] \cr & = \mathop {\lim }\limits_{h \to 0} f{\left( { - 1} \right)^2} - \left[ {{h^2}} \right] = \mathop {\lim }\limits_{h \to 0} 1 - 0 = 1 \cr & {\text{R}}{\text{.H}}{\text{.L}}{\text{. }} = \mathop {\lim }\limits_{h \to 0} f\left( h \right) = \mathop {\lim }\limits_{h \to 0} {\left[ h \right]^2} - \left[ {{h^2}} \right] \cr & = \mathop {\lim }\limits_{h \to 0} \,0 - 0 = 0 \cr & \therefore \,{\text{L}}{\text{.H}}{\text{.L}}{\text{.}} \ne {\text{R}}{\text{.H}}{\text{.L}}{\text{.}} \cr & \therefore f\left( x \right){\text{is not continuous at }}x = 0 \cr & {\text{At }}x = 1 \cr & {\text{L}}{\text{.H}}{\text{.L}}{\text{. }} = \mathop {\lim }\limits_{h \to 0} f\left( {1 - h} \right) = \mathop {\lim }\limits_{h \to 0} {\left[ {1 - h} \right]^2} - \left[ {{{\left( {1 - h} \right)}^2}} \right] \cr & = \mathop {\lim }\limits_{h \to 0} \,0 - 0 = 0 \cr & {\text{R}}{\text{.H}}{\text{.L}}{\text{. }} = \mathop {\lim }\limits_{h \to 0} f\left( {1 + h} \right) = \mathop {\lim }\limits_{h \to 0} {\left[ {1 + h} \right]^2} - \left[ {{{\left( {1 + h} \right)}^2}} \right] \cr & = \mathop {\lim }\limits_{h \to 0} \,1 - 1 = 0 \cr & f\left( 1 \right) = {\left[ 1 \right]^2} - \left[ {{1^2}} \right] = 1 - 1 = 0 \cr & \therefore \,{\text{L}}{\text{.H}}{\text{.L}}{\text{.}} = {\text{R}}{\text{.H}}{\text{.L}}{\text{.}} = f\left( 1 \right) \cr & \therefore f\left( x \right)\,\,{\text{is continuous at }}x = {\text{ }}1{\text{ }} \cr} $$
Clearly at other integral pts $$f\left( x \right)$$  is not continuous.

As $$f\left( x \right)$$  is continuous in $$\left[ {1,\,3} \right],\,f\left( x \right)$$    will attain all values between $$f\left( 1 \right)$$  and $$f\left( 3 \right).$$  As $$f\left( x \right)$$  takes rational values for all $$x$$ and there are innumerable irrational values between $$f\left( 1 \right)$$  and $$f\left( 3 \right),\,f\left( x \right)$$   can take rational values for all $$x$$ if $$f\left( x \right)$$  has a constant rational value at all points between $$x=1$$  and $$x=3.$$  So, $$f\left( 2 \right) = f\left( {1.5} \right) = 10.$$

\[f\left( x \right) = \left\{ \begin{array}{l} \,\,\,\,\,\,\,\,ax - 2\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, - 2 < x < 1\\ \,\,\,\,\,\,\,\,\,\,\, - 1\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, - 1 \le x \le 1\\ a + 2{\left( {x - 1} \right)^2}\,\,\,\,\,\,\,\,\,\,\,\,\,\,1 < x < 2 \end{array} \right.\]
$$\eqalign{ & {\text{if }}f\left( x \right){\text{ is continuous at }}x = - 1 \cr & {\text{then, }}\mathop {\lim }\limits_{x \to - 1} \left( {ax - 2} \right) = \mathop {\lim }\limits_{x \to - 1} \left( { - 1} \right) \cr & \Rightarrow a\left( { - 1} \right) - 2 = - 1 \cr & \Rightarrow a = - 1 \cr & {\text{if }}f\left( x \right){\text{ is continuous at }}x = 1 \cr & {\text{then, }}\mathop {\lim }\limits_{x \to 1} a + 2{\left( {x - 1} \right)^2} = \mathop {\lim }\limits_{x \to 1} \left( { - 1} \right) \cr & \Rightarrow a + 2{\left( {1 - 1} \right)^2} = - 1 \cr & \Rightarrow a = - 1 \cr} $$

We know that $$\left| {1 - x} \right|$$  is continuous everywhere, $$\left[ x \right]$$ is continuous everywhere except at integers, $$\sqrt {1 + {x^2}} $$   and $$\sqrt 3 x - 1$$   are continuous in their respective intervals of definition. So, the only doubtful points of continuity are $$x = \sqrt 3 ,\,4,\,5$$
$$\eqalign{ & f\left( {\sqrt 3 - 0} \right) = \mathop {\lim }\limits_{h \to 0} \sqrt {1 + {{\left( {\sqrt 3 - h} \right)}^2}} = 2\,; \cr & f\left( {\sqrt 3 + 0} \right) = \mathop {\lim }\limits_{h \to 0} \left\{ {\sqrt 3 \left( {\sqrt 3 + h} \right) - 1} \right\} = 2\,;\,\,\,\,\,f\left( {\sqrt 3 } \right) = 2 \cr & f\left( {4 - 0} \right) = \mathop {\lim }\limits_{h \to 0} \left\{ {\sqrt 3 \left( {4 - h} \right) - 1} \right\} = 4\sqrt 3 - 1\,;\,\,\,f\left( {4 + 0} \right) = \mathop {\lim }\limits_{h \to 0} \left[ {4 + h} \right] = 4 \cr & f\left( {5 - 0} \right) = \mathop {\lim }\limits_{h \to 0} \left[ {5 - h} \right] = \mathop {\lim }\limits_{h \to 0} \,4 = 4\,; \cr & f\left( {5 + 0} \right) = \mathop {\lim }\limits_{h \to 0} \left| {1 - \left( {5 + h} \right)} \right| = 4\,;\,\,\,\,\,f\left( 5 \right) = 4 \cr & \therefore f\left( x \right){\text{ is discontinuous at }}x = 4{\text{ only}}{\text{.}} \cr} $$