We have $$y = \frac{{x + 6}}{{\left( {x - 2} \right)\left( {x - 3} \right)}}$$
At $$y$$ -axis, $$x = 0 \Rightarrow y = 1$$
On differentiating, we get
$$\eqalign{ & \frac{{dy}}{{dx}} = \frac{{\left( {{x^2} - 5x + 6} \right)\left( 1 \right) - \left( {x + 6} \right)\left( {2x - 5} \right)}}{{{{\left( {{x^2} - 5x + 6} \right)}^2}}} \cr & \frac{{dy}}{{dx}} = 1\,{\text{at}}\,{\text{point}}\left( {0,1} \right) \cr & \therefore \,{\text{Slope}}\,{\text{of}}\,{\text{normal}} = - 1 \cr & {\text{Now}}\,{\text{equation}}\,{\text{of}}\,{\text{normal}}\,{\text{is}}\,y - 1 = - 1\left( {x - 0} \right) \cr & \Rightarrow y - 1 = - x \cr & x + y = 1 \cr & \therefore \,\,\left( {\frac{1}{2},\frac{1}{2}} \right)\,{\text{satisfy}}\,{\text{it}}. \cr} $$

$$\eqalign{ & b\cos \,\theta = a \cr & \therefore \,b\cos \,\theta + b = 4 \cr & {\text{or, }}b = \frac{4}{{1 + \cos \,\theta }} \cr & \therefore \,a = \frac{{4\cos \,\theta }}{{1 + \cos \,\theta }} \cr & \therefore {\text{ area}} = \Delta = \frac{1}{2}\,ba\sin \,\theta = \frac{1}{2}.\frac{4}{{1 + \cos \,\theta }}.\frac{{4\cos \,\theta }}{{1 + \cos \,\theta }}.\sin \,\theta = \frac{{4\sin \,2\theta }}{{{{\left( {1 + \cos \,\theta } \right)}^2}}} \cr & \therefore \,\frac{{d\Delta }}{{d\theta }} = 4.\frac{{2\cos \,2\theta {{\left( {1 + \cos \,\theta } \right)}^2} + \sin \,2\theta .2\left( {1 + \cos \,\theta } \right)\sin \,\theta }}{{{{\left( {1 + \cos \,\theta } \right)}^4}}} \cr & \therefore \,\frac{{d\Delta }}{{d\theta }} = 0\,\,\, \Rightarrow \cos \,2\theta .\left( {1 + \cos \,\theta } \right) + \sin \,2\theta .\sin \,\theta = 0 \cr & {\text{or, }}\cos \,2\theta + \cos \,\theta = 0 \cr & \therefore \,\cos \,2\theta = - \cos \,\theta = \cos \,\left( {\pi - \theta } \right){\text{ or }}\theta = \frac{\pi }{3} \cr} $$
$$\therefore \,\Delta $$  is maximum when $$\theta = \frac{\pi }{3}.$$   (It cannot be minimum, since in the given situation minimum is 0.)

$$\eqalign{ & {\text{Given}}\,\frac{{dr}}{{dt}} = 3 \cr & {\text{Let }}A = {\text{Area of circle}} = \pi {r^2} \cr & \therefore \,\frac{{dA}}{{dt}} = 2\pi r.\frac{{dr}}{{dt}} = 6\pi r \cr & {\text{Now, }}{\left. {\frac{{dA}}{{dt}}} \right|_{r = 10}} = 6 \times 10 \times \pi = 60\pi \,c{m^2}/s \cr} $$

Clearly, domain of the function is $$\left[ {2,\,4} \right].$$
$$\eqalign{ & {\text{Now, }}f'\left( x \right) = \frac{1}{{\sqrt {x - 2} }} - \frac{1}{{2\sqrt {4 - x} }} \cr & {\text{or, }}f'\left( x \right) = 0 \cr & {\text{or, }}\sqrt {x - 2} = 2\sqrt {4 - x} \cr & {\text{or, }}x - 2 = 16 - 4x \cr & {\text{or, }}x = \frac{{18}}{5} \cr & {\text{Now, }}f\left( 2 \right) = \sqrt 2 , \cr & f\left( {\frac{{18}}{5}} \right) = 2\sqrt {\frac{{18}}{5} - 2} + \sqrt {4 - \frac{{18}}{5}} = \sqrt {10} , \cr & f\left( 4 \right) = 2\sqrt 2 \cr} $$
Hence, range of the function is $$\left[ {\sqrt 2 ,\,\sqrt {10} } \right].$$
Also, here $$x = \left( {\frac{{18}}{5}} \right)$$  is the point of global maxima.

Eccentricity of ellipse $$ = \frac{1}{2}$$
$$\eqalign{ & {\text{Now,}}\, - \frac{a}{e} = - 4 \Rightarrow a = 4 \times \frac{1}{2} = 2 \Rightarrow a = 2 \cr & {\text{We have }}{b^2} = {a^2}\left( {1 - {e^2}} \right) = {a^2}\left( {1 - \frac{1}{4}} \right) = 4 \times \frac{3}{4} = 3 \cr & \therefore \,{\text{Equation of ellipse is }}\frac{{{x^2}}}{4} + \frac{{{y^2}}}{3} = 1 \cr & {\text{Now differentiating, we get}} \cr & \Rightarrow \frac{x}{2} + \frac{{2y}}{3} \times y' = 0 \Rightarrow y' = - \frac{{3x}}{{4y}} \cr & y'\left| {_{\left( {1,\frac{3}{2}} \right)}} \right| = - \frac{3}{4} \times \frac{2}{3} = - \frac{1}{2} \cr & {\text{Slope}}\,{\text{of}}\,{\text{normal}} = 2 \cr & \therefore \,\,{\text{Equation}}\,{\text{of}}\,{\text{normal}}\,{\text{at}}\,\left( {1,\frac{3}{2}} \right)\,{\text{is}} \cr & y - \frac{3}{2} = 2\left( {x - 1} \right) \Rightarrow 2y - 3 = 4x - 4\therefore 4x - 2y = 1 \cr} $$

$$f'\left( x \right) = {e^x}\left( {\sin \,x + \cos \,x} \right).$$     Therefore, the slope $$m$$ of the tangent is given by $$m = {e^x}\left( {\sin \,x + \cos \,x} \right).$$
Now, $$\frac{{dm}}{{dx}} = {e^x}\left\{ {\sin \,x + \cos \,x + \cos \,x - \sin \,x} \right\} = 2{e^x}.\cos \,x$$
and $$\frac{{{d^2}m}}{{d{x^2}}} = 2{e^x}\left( {\cos \,x - \sin \,x} \right)$$
At $$x=a,\,m$$   is the maximum
$$\eqalign{ & \therefore \,\,{\left. {\frac{{dm}}{{dx}}} \right)_{x = a}} = 0,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{\left. {\frac{{{d^2}m}}{{d{x^2}}}} \right)_{x = a}} < 0 \cr & {\left. {\frac{{dm}}{{dx}}} \right)_{x = a}} = 0\,\,\,\,\,\, \Rightarrow 2{e^a}\cos \,a = 0 \cr & \Rightarrow a = \frac{\pi }{2},\,\frac{{3\pi }}{2}{\text{ because }}0 \leqslant a \leqslant 2\pi \cr & {\left. {\frac{{{d^2}m}}{{d{x^2}}}} \right)_{x = a}} < 0\,\,\,\,\,\,\, \Rightarrow 2{e^a}\left( {\cos \,a - \sin \,a} \right) < 0 \cr & {\text{This is satisfied by }}a = \frac{\pi }{2} \cr} $$

$$y = {x^3} + ax - b$$
Since, the point (1, -5) lies on the curve.
$$\eqalign{ & \Rightarrow 1 + a - b = - 5 \cr & \Rightarrow a - b = - 6\,.......\left( 1 \right) \cr & {\text{Now, }}\frac{{dy}}{{dx}} = 3{x^2} + a \Rightarrow {\left( {\frac{{dy}}{{dx}}} \right)_{{\text{at }}x = 1}} = 3 + a \cr} $$
Since, required line is perpendicular to $$y = x - 4,$$   then slope of tangent at the point $$P\left( {1, - 5} \right) = - 1$$
$$\therefore 3 + a = - 1 \Rightarrow a = - 4 \Rightarrow b = 2$$       the equation of the curve is $$y = {x^3} - 4x - 2$$
$$ \Rightarrow \left( {2, - 2} \right)$$   lies on the curve.

$$\eqalign{ & {\text{Let }}y = {x^3} - px + q \Rightarrow \frac{{dy}}{{dx}} = 3{x^2} - p \cr & {\text{For }}\frac{{dy}}{{dx}} = 0 \Rightarrow 3{x^2} - p = 0 \Rightarrow x = \pm \sqrt {\frac{p}{3}} \cr & \frac{{{d^2}y}}{{d{x^2}}} = 6x \cr & {\left. {\frac{{{d^2}y}}{{d{x^2}}}} \right|_{x = \sqrt {\frac{p}{3}} }} = + ve{\text{ and }}{\left. {\frac{{{d^2}y}}{{d{x^2}}}} \right|_{x = - \sqrt {\frac{p}{3}} }} = - ve \cr & \therefore y{\text{ has minima at }}x = \sqrt {\frac{p}{3}} {\text{ and maxima at }}x = - \sqrt {\frac{p}{3}} \cr} $$

$$\eqalign{ & {\text{Here, }}h\left( x \right){\text{ }} = \frac{{{x^2} + \frac{1}{{{x^2}}}}}{{x - \frac{1}{x}}} = \left( {x - \frac{1}{x}} \right) + \frac{2}{{x - \frac{1}{x}}} \cr & {\text{When }}x - \frac{1}{x} < 0 \cr & \therefore \quad x - \frac{1}{x} + \frac{2}{{x - \frac{1}{x}}} \leqslant - 2\sqrt 2 \cr & {\text{Hence, }} - 2\sqrt 2 \,{\text{will be local maximum value of}}\,h\left( x \right). \cr & {\text{When}}\,x - \frac{1}{x} > 0 \cr & \therefore x - \frac{1}{x} + \frac{2}{{x - \frac{1}{x}}} \geqslant 2\sqrt 2 \cr} $$

$$\eqalign{ & {\text{In }} - \in < x \leqslant 0,\,f'\left( x \right) = \cos \left( {{x^2} - 3x} \right).\left( {2x - 3} \right) < 0 \cr & \therefore \,f\left( x \right)\,{\text{is decreasing}}{\text{.}} \cr & {\text{In }}x > 0,\,f'\left( x \right) = 6 + 10x > 0 \cr} $$
$$\therefore \,f\left( x \right)$$   is increasing. So, at $$x = 0,\,f\left( x \right)$$   has a local minimum.
$$\therefore $$  at $$x = 0,\,f\left( x \right)$$   is continuous because $$f\left( {0 - \in } \right) = f\left( 0 \right) = f\left( {0 + \in } \right)$$