$$\eqalign{ & f\left( {{e^ + }} \right) = \mathop {\lim }\limits_{h \to 0} \left( {e + h - e} \right){2^{ - {2^{\frac{1}{{e - \left( {e - h} \right)}}}}}} \cr & = \mathop {\lim }\limits_{h \to 0} \left( h \right){2^{ - {2^{ - \frac{1}{h}}}}} \cr & = 0 \times 1 \cr & = 0 \cr & \left( {{\text{As for }}h \to 0,\, - \frac{1}{h} \to - \infty \Rightarrow {2^{ - \frac{1}{h}}} \to 0} \right) \cr & f\left( {{e^ - }} \right) = \mathop {\lim }\limits_{h \to 0} \left( { - h} \right){2^{ - {2^{\frac{1}{h}}}}} \cr & = 0 \times 0 \cr & = 0 \cr & {\text{Hence, }}f\left( x \right){\text{ is continuous at }}x = e \cr & f'\left( {{e^ + }} \right) = \mathop {\lim }\limits_{h \to 0} \frac{{f\left( {e + h} \right) - f\left( e \right)}}{h} \cr & = \mathop {\lim }\limits_{h \to 0} \frac{{h \times {2^{ - {2^{ - \frac{1}{h}}}}} - 0}}{h} \cr & = \mathop {\lim }\limits_{h \to 0} {2^{ - {2^{\frac{1}{h}}}}} \cr & = 1 \cr & f'\left( {{e^ - }} \right) = \mathop {\lim }\limits_{h \to 0} \frac{{f\left( {e - h} \right) - f\left( 0 \right)}}{{ - h}} \cr & = \mathop {\lim }\limits_{h \to 0} \frac{{\left( { - h} \right) \times {2^{ - {2^{ - \frac{1}{h}}}}} - 0}}{{ - h}} \cr & = \mathop {\lim }\limits_{h \to 0} {2^{ - {2^{\frac{1}{h}}}}} \cr & = 0 \cr & {\text{Hence, }}f\left( x \right){\text{ is non - differentiable at }}x = e \cr} $$

$$\eqalign{ & f\left( x \right) = \frac{{2\sin \,x.\cos \,x.\cos \,2x.\cos \,4x.\cos \,8x.\cos \,16x}}{{2\sin \,x}} = \frac{{\sin \,32x}}{{{2^5}\sin \,x}} \cr & \therefore f'\left( x \right) = \frac{1}{{32}}.\frac{{32\cos \,32x.\sin \,x - \cos \,x.\sin \,32x}}{{{{\sin }^2}x}} \cr & \therefore f'\left( {\frac{\pi }{4}} \right) = \frac{{32.\frac{1}{{\sqrt 2 }} - \frac{1}{{\sqrt 2 }}.0}}{{32.{{\left( {\frac{1}{{\sqrt 2 }}} \right)}^2}}} \cr} $$

$$\eqalign{ & \because \,\ln \,x = {\log _e}x,\,\,{\text{so}} \cr & f\left( x \right) = {\log _x}\left( {{{\log }_e}x} \right) = \frac{{\log \left( {\log \,x} \right)}}{{\log \,x}} \cr & \Rightarrow f'\left( x \right) = \frac{{\log \,x\left( {\frac{1}{{x\,\log \,x}}} \right) - \log \left( {\log \,x} \right).\frac{1}{x}}}{{{{\left( {\log \,x} \right)}^2}}} \cr & \therefore \,f'\left( e \right) = \frac{{\frac{1}{e} - 0}}{{{{\left( 1 \right)}^2}}} = \frac{1}{e} \cr} $$

$$\eqalign{ & {\text{ln}}{\left( {x + \sin \,x} \right)^1} = y\,\,\,\,\,\,\,\,\,\left( {{\text{say}}} \right) \cr & \frac{{dy}}{{dx}} = \frac{1}{{\left( {x + \sin \,x} \right)}}\left( {1 + \cos \,x} \right) = \frac{{\left( {1 + \cos \,x} \right)}}{{\left( {x + \sin \,x} \right)}} \cr & x + \cos \,x = z\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left( {{\text{say}}} \right) \cr & \frac{{dz}}{{dx}} = \left( {1 - \sin \,x} \right) \cr & {\text{derivative of ln}}\left( {x + \sin \,x} \right){\text{ w}}{\text{.r}}{\text{.t }}\left( {x + \cos \,x} \right){\text{ is}} \cr & \frac{{dy}}{{dz}} = \frac{{\left( {1 + \cos \,x} \right)}}{{\left( {x + \sin \,x} \right)\left( {1 - \sin \,x} \right)}} \cr} $$

$$\because \,f\left( x \right) = \left| x \right| + {x^2}$$
\[ \Rightarrow \,f\left( x \right) = \left\{ \begin{array}{l} {x^2} + x,\,\,\,\,\,x \ge 0\\ {x^2} - x,\,\,\,\,\,x < 0 \end{array} \right.\]
$$\eqalign{ & {\text{L}}{\text{.H}}{\text{.L}}{\text{. }} = \mathop {\lim }\limits_{x \to {0^ - }} f\left( x \right) \cr & = \mathop {\lim }\limits_{h \to 0} f\left( {0 - h} \right) \cr & = \mathop {\lim }\limits_{h \to 0} {\left( {0 - h} \right)^2} - \left( {0 - h} \right) \cr & = \mathop {\lim }\limits_{h \to 0} {h^2} + h \cr & = 0 \cr & {\text{and}} \cr & {\text{R}}{\text{.H}}{\text{.L}}{\text{. }} = \mathop {\lim }\limits_{x \to 0} f\left( x \right) \cr & = \mathop {\lim }\limits_{h \to 0} f\left( {0 + h} \right) \cr & = \mathop {\lim }\limits_{h \to 0} {\left( {0 + h} \right)^2} + \left( {0 + h} \right) \cr & = \mathop {\lim }\limits_{h \to 0} {h^2} + h \cr & = 0 \cr & \Rightarrow {\text{L}}{\text{.H}}{\text{.L}}{\text{.}} = {\text{R}}{\text{.H}}{\text{.L}}{\text{.}} = f\left( 0 \right) \cr & \Rightarrow \,f\left( x \right){\text{ is continuous at }}x = 0 \cr & {\text{Now, L}}{\text{.H}}{\text{.D}}{\text{.}} = \mathop {\lim }\limits_{h \to 0} \frac{{f\left( {0 - h} \right) - f\left( 0 \right)}}{{ - h}} \cr & = \mathop {\lim }\limits_{h \to 0} \frac{{{h^2} + h}}{{ - h}} \cr & = - \mathop {\lim }\limits_{h \to 0} \,h + 1 \cr & = - 1 \cr & {\text{and, R}}{\text{.H}}{\text{.D}}{\text{.}} = \mathop {\lim }\limits_{h \to 0} \frac{{f\left( {0 + h} \right) - f\left( 0 \right)}}{h} \cr & = \mathop {\lim }\limits_{h \to 0} \frac{{{h^2} + h}}{h} \cr & = \mathop {\lim }\limits_{h \to 0} \,h + 1 \cr & = 1 \cr & {\text{Thus, L}}{\text{.H}}{\text{.D}}{\text{.}} \ne {\text{R}}{\text{.H}}{\text{.D}}{\text{.}} \cr & \Rightarrow \,f\left( x \right){\text{ is not differentiable at }}x = 0 \cr} $$

The doubtful points are $$x=0,\,1$$   because these are the turning points of the definition. $$\left| x \right|$$ is not differentiable at $$x=0,$$   while $${\left( {x - 1} \right)^2}\cos \frac{1}{{x - 1}}$$    is differentiable. The algebraic sum of a differentiable function and a nondifferentiable function is nondifferentiable. So, $$f\left( x \right)$$  is not differentiable at $$x=0.$$  Now,
$$\eqalign{ & f'\left( {1 + 0} \right) = \mathop {\lim }\limits_{h \to 0} \frac{{{{\left( {1 + h - 1} \right)}^2}\cos \frac{1}{{1 + h - 1}} - \left| {1 + h} \right| - \left( { - 1} \right)}}{h} \cr & = \mathop {\lim }\limits_{h \to 0} \frac{{{h^2}\cos \frac{1}{h} - h}}{h} \cr & = \mathop {\lim }\limits_{h \to 0} \left( {h\cos \frac{1}{h} - 1} \right) \cr & = 0 - 1 = - 1 \cr} $$
Similarly, $$f'\left( {1 - 0} \right) = - 1.$$    So, $$f\left( x \right)$$  is differentiable at $$x=1.$$

Apply L'Hospital Rule
We have,
$$\eqalign{ & \mathop {\lim }\limits_{x \to 2} \frac{{xf\left( 2 \right) - 2f\left( x \right)}}{{x - 2}}\,\,\left( {\frac{0}{0}} \right) \cr & = \mathop {\lim }\limits_{x \to 2} f\left( 2 \right) - 2f'\left( x \right) \cr & = f\left( 2 \right) - 2f'\left( 2 \right) \cr & = 4 - 2 \times 4 \cr & = - 4 \cr} $$

$${\text{Let }}s = {\sin ^{ - 1}}\left( {\frac{{2x}}{{1 + {x^2}}}} \right)\,{\text{and }}t = {\cos ^{ - 1}}\left( {\frac{{1 - {x^2}}}{{1 + {x^2}}}} \right)$$
we have to find out $$\frac{{ds}}{{dt}}$$  ; Putting $$x = \tan \,\theta ,$$   we get
$$\eqalign{ & s = {\sin ^{ - 1}}\left[ {\frac{{2\,\tan \,\theta }}{{1 + {{\tan }^2}\theta }}} \right] \cr & = {\sin ^{ - 1}}\left( {\sin \,2\theta } \right) \cr & = 2\theta \cr & = 2\,{\tan ^{ - 1}}x \cr & \therefore \,\frac{{ds}}{{dx}} = \frac{{2x}}{{1 + {x^2}}} \cr & {\text{and }}t = {\cos ^{ - 1}}\left( {\frac{{1 - {x^2}}}{{1 + {x^2}}}} \right) \cr & = {\cos ^{ - 1}}\left( {\frac{{1 - {{\tan }^2}\theta }}{{1 + {{\tan }^2}\theta }}} \right) \cr & = {\cos ^{ - 1}}\left( {\cos \,2\theta } \right) \cr & = 2\,\theta \cr & = 2\,{\tan ^{ - 1}}x \cr & \therefore \,\frac{{dt}}{{dx}} = \frac{2}{{1 + {x^2}}} \cr & \therefore \,\frac{{ds}}{{dt}} = \frac{{\frac{{ds}}{{dx}}}}{{\frac{{dt}}{{dx}}}} = \frac{2}{{1 + {x^2}}} \times \frac{{1 + {x^2}}}{2} = 1 \cr} $$

$$\eqalign{ & f\left( x \right) = \min \left\{ {x + 1,\,\left| x \right| + 1} \right\} \cr & \Rightarrow f\left( x \right) = x + 1\,\forall \,x \in R \cr} $$

Hence, $$f\left( x \right)$$  is differentiable everywhere for all $$x \in R.$$

$$\eqalign{ & {\text{Let }}f\left( x \right) = a{x^2} + bx + c \cr & {\text{Then }}g\left( x \right) = a{x^2} + bx + c + 2ax + b + 2a \cr & = a{x^2} + \left( {2a + b} \right)x + 2a + b + c \cr & {\text{As }}f\left( x \right) > 0\,{\text{for all }}x, \cr & a > 0\,{\text{and }}D = {b^2} - 4ac < 0 \cr & g\left( x \right) > 0\,{\text{for all }}x, \cr & {\text{if }}a > 0\,{\text{and }}D = {\left( {2a + b} \right)^2} - 4a\left( {2a + b + c} \right) < 0 \cr & {\text{Now, }}{\left( {2a + b} \right)^2} - 4a\left( {2a + b + c} \right) = - 4{a^2} + {b^2} - 4ac < 0{\text{ }} \cr & {\text{Hence, }}g\left( x \right) > 0 \cr} $$