$$\eqalign{ & f\left( x \right) = \frac{{2\,{{\cos }^2}\frac{x}{2} - 2\sin \frac{x}{2}\,\cos \frac{x}{2}}}{{2\,{{\cos }^2}\frac{x}{2} + 2\sin \frac{x}{2}\,\cos \frac{x}{2}}} \cr & = \frac{{\cos \frac{x}{2} - \sin \frac{x}{2}}}{{\cos \frac{x}{2} + \sin \frac{x}{2}}} \cr & = \tan \left( {\frac{\pi }{4} - \frac{x}{2}} \right){\text{ at }}x \cr & = \pi ,\,f\left( \pi \right) \cr & = - \tan \frac{\pi }{4} \cr & = - 1 \cr} $$

$$\eqalign{ & f\left( {1 + 0} \right) = \mathop {\lim }\limits_{h \to 0} \left\{ {\left[ {{{\left( {1 + h} \right)}^2}} \right] - {{\left[ {1 + h} \right]}^2}} \right\} = \mathop {\lim }\limits_{h \to 0} \left\{ {1 - 1} \right\} = 0 \cr & f\left( {1 - 0} \right) = \mathop {\lim }\limits_{h \to 0} \left\{ {\left[ {{{\left( {1 - h} \right)}^2}} \right] - {{\left[ {1 - h} \right]}^2}} \right\} = \mathop {\lim }\limits_{h \to 0} \left\{ {0 - 0} \right\} = 0 \cr & {\text{Also }}f\left( 1 \right) = 0.{\text{ So, }}\,f\left( x \right)\,\,{\text{is continuous at }}x = 1. \cr} $$

$$\eqalign{ & \left( {\bf{i}} \right)\,\,f\left( x \right) = \operatorname{sgn} \left( {{x^3} - x} \right) \cr & {\text{Here }}{x^3} - x = 0 \cr & \Rightarrow x = 0,\, - 1,\,1 \cr & {\text{Hence,}}\,f\left( x \right){\text{ is discontinuous at }}x = 0,\, - 1,\,1 \cr & \left( {{\bf{ii}}} \right)\,\,{\text{If }}f\left( x \right) = \operatorname{sgn} \left( {2\,\cos \,x - 1} \right) \cr & {\text{Here, }}2\,\cos \,x - 1 = 0 \cr & \Rightarrow \,\cos \,x = \frac{1}{2} \cr & \Rightarrow x = \,2n\pi + \left( {\frac{\pi }{3}} \right),\,n\, \in \,Z,\,{\text{where }}f\left( x \right){\text{ is discontinuous}}. \cr & \left( {{\bf{iii}}} \right)\,\,f\left( x \right) = \operatorname{sgn} \left( {{x^2} - 2x + 3} \right) \cr & {\text{Here,}}\,{x^2} - 2x + 3 > 0{\text{ for all }}x \cr & {\text{Thus, }}f\left( x \right) = 1{\text{ for all }}x \cr & {\text{Hence, continuous for all }}x. \cr} $$

\[\begin{array}{l} f\left( x \right) = \left\{ \begin{array}{l} \frac{{{x^2}}}{x},\,\,x \ne 0\\ \,\,\,0,\,\,\,x = 0 \end{array} \right.\\ = \left\{ \begin{array}{l} \frac{{{x^2}}}{x} = x,\,\,\,\,\,\,\,\,\,\,\,x > 0\\ \,\,\,\,\,0,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,x = 0\\ \frac{{{x^2}}}{{ - x}} = - x,\,\,\,\,\,x < 0 \end{array} \right. \end{array}\]
$$\eqalign{ & {\text{Now, }}\mathop {\lim }\limits_{x \to {0^ - }} f\left( x \right) = \mathop {\lim }\limits_{x \to 0} \left( { - x} \right) = 0 \cr & \mathop {\lim }\limits_{x \to {0^ + }} f\left( x \right) = \mathop {\lim }\limits_{x \to 0} \left( x \right) = 0{\text{ and }}f\left( 0 \right) = 0 \cr} $$
So, $$f\left( x \right)$$  is continuous at $$x = 0$$
Also, $$f\left( x \right)$$  is continuous for all other values of $$x.$$
Hence, $$f\left( x \right)$$  is continuous everywhere.

As $$f\left( x \right)$$  is continuous in $$\left[ {0,\,\frac{\pi }{2}} \right),$$  it is continuous at $$x = \frac{\pi }{4}$$
$$\eqalign{ & \therefore \mathop {\lim }\limits_{h \to 0} \frac{{1 - \tan \left( {\frac{\pi }{4} + h} \right)}}{{4\left( {\frac{\pi }{4} + h} \right) - \pi }} = \mathop {\lim }\limits_{h \to 0} \frac{{1 - \tan \left( {\frac{\pi }{4} - h} \right)}}{{4\left( {\frac{\pi }{4} - h} \right) - \pi }} = f\left( {\frac{\pi }{4}} \right) = \lambda \cr & \therefore \,\,\lambda = \mathop {\lim }\limits_{h \to 0} \frac{{1 - \frac{{1 + \tan \,h}}{{1 - \tan \,h}}}}{{4h}} = \mathop {\lim }\limits_{h \to 0} \frac{{ - 2\tan \,h}}{{\left( {1 - \tan \,h} \right)4h}} = - \frac{1}{2} \cr} $$

$$\eqalign{ & \mathop {\lim }\limits_{x \to \pi } \frac{{1 - \sin \,x + \cos \,x}}{{1 + \sin \,x + \cos \,x}} \cr & {\text{Using L'hospital's rules }} \cr & \Rightarrow \mathop {\lim }\limits_{x \to \pi } \frac{{ - \cos \,x - \sin \,x}}{{\cos \,x - \sin \,x}} = \frac{{ - \cos \,\pi - \sin \,\pi }}{{\cos \,\pi - \sin \,\pi }} = \frac{{ - \left( { - 1} \right) - 0}}{{ - 1 - 0}} = - 1 \cr} $$

Given function is : \[f\left( x \right) = \left\{ \begin{array}{l} 3x - 4,\,\,\,\,\,0 \le x \le 2\\ 2x + \ell ,\,\,\,\,\,\,2 < x \le 9 \end{array} \right.\]      and also given that $$f\left( x \right)$$  is continuous at $$x = 2$$
For a function to be continuous at a point $${\text{L}}{\text{.H}}{\text{.L}}{\text{.}} = {\text{R}}{\text{.H}}{\text{.L}}{\text{.}} = {\text{V}}{\text{.F}}{\text{.}}$$      at that point. $$f\left( 2 \right) = 2 = {\text{V}}{\text{.F}}{\text{.}}$$
$$\eqalign{ & \Rightarrow {\text{R}}{\text{.H}}{\text{.L}}{\text{.}}\,:\,\mathop {\lim }\limits_{x \to 2} \left( {2x + \ell } \right) = 3\left( 2 \right) - 4 \cr & \Rightarrow \mathop {\lim }\limits_{h \to 0} \left\{ {2\left( {2 + h} \right) + \ell } \right\} = 6 - 4 \cr & \Rightarrow 4 + \ell = 2 \cr & \Rightarrow \ell = - 2 \cr} $$

For Rolle's theorem in $$\left[ {a,\,b} \right],\,f\left( a \right) = f\left( b \right),\,{\text{ln}}\left[ {0,\,1} \right] \Rightarrow f\left( 0 \right) = f\left( 1 \right) = 0$$
$$\because $$  the function has to be continuous in $$\left[ {0,\,1} \right]$$
$$\eqalign{ & \Rightarrow f\left( 0 \right) = \mathop {\lim }\limits_{x \to {0^ + }} f\left( x \right) = 0 \cr & \Rightarrow \mathop {\lim }\limits_{x \to 0} \,{x^\alpha }\log \,x = 0 \cr & \Rightarrow \mathop {\lim }\limits_{x \to 0} \frac{{\log \,x}}{{{x^{ - \alpha }}}} = 0 \cr & {\text{Applying L}}{\text{.H}}{\text{. Rule, }}\mathop {\lim }\limits_{x \to 0} \frac{{\frac{1}{x}}}{{ - \alpha {x^{ - \alpha - 1}}}} = 0 \cr & \Rightarrow \mathop {\lim }\limits_{x \to 0} \frac{{ - {x^\alpha }}}{\alpha } = 0 \cr & \Rightarrow \alpha > 0 \cr} $$

For continuity at $$x = 0,\,\mathop {\lim }\limits_{h \to 0} \,\sin \frac{1}{{0 + h}} = \mathop {\lim }\limits_{h \to 0} \,\sin \frac{1}{{0 - h}} = f\left( 0 \right)$$
But $$\mathop {\lim }\limits_{h \to 0} \,\sin \frac{1}{{0 + h}} = \mathop {\lim }\limits_{h \to 0} $$     { a number between 1 and $$-1$$ } $$ \ne a$$  definite number.
$$\therefore \,f\left( 0 \right)$$   cannot be a definite number.

$$\eqalign{ & {\text{Here, }}\mathop {\lim }\limits_{h \to 0} {\left\{ {\sin \left( {\frac{\pi }{2} + h} \right)} \right\}^{\frac{1}{{\pi - 2\left( {\frac{\pi }{2} + h} \right)}}}} \cr & = \mathop {\lim }\limits_{h \to 0} {\left\{ {\sin \left( {\frac{\pi }{2} - h} \right)} \right\}^{\frac{1}{{\pi - 2\left( {\frac{\pi }{2} - h} \right)}}}} \cr & = f\left( {\frac{\pi }{2}} \right) \cr & {\text{Now,}}\,\,\mathop {\lim }\limits_{h \to 0} {\left( {\cos \,h} \right)^{\frac{1}{{\left( { - 2h} \right)}}}} \cr & = {e^{\mathop {\lim }\limits_{h \to 0} \frac{1}{{\left( { - 2h} \right)}}\log \,\cos \,h}} \cr & = {e^{\mathop {\lim }\limits_{h \to 0} \frac{{\left( {\frac{1}{{\cos \,h}}} \right)\left( { - \sin \,h} \right)}}{{ - 2}}}} \cr & = {e^0} = 1 \cr & \therefore \,f\left( {\frac{\pi }{2}} \right) = 1 \cr} $$