$$\eqalign{ & af\left( x \right) + bf\left( {\frac{1}{x}} \right) = \frac{1}{x} - 5 \cr & \Rightarrow af\left( {\frac{1}{x}} \right) + bf\left( x \right) = x - 5 \cr & {\text{From these, }}\left( {{a^2} - {b^2}} \right)f\left( x \right) = a\left( {\frac{1}{x} - 5} \right) - b\left( {x - 5} \right) \cr & \therefore \,\int_1^2 {f\left( x \right)dx} = \int_1^2 {\frac{1}{{{a^2} - {b^2}}}\left\{ {\frac{a}{x} - bx - 5a + 5b} \right\}dx} \cr & = \frac{1}{{{a^2} - {b^2}}}\left[ {a\log \,x - b\frac{{{x^2}}}{2} + 5\left( {b - a} \right)x} \right]_1^2 \cr & = \frac{{a\log \,2 - 2b + 10\left( {b - a} \right) + \frac{b}{2} - 5\left( {b - a} \right)}}{{{a^2} - {b^2}}} \cr} $$

$$\eqalign{ & \int\limits_0^\infty {\frac{{dx}}{{\left( {{x^2} + {a^2}} \right)\left( {{x^2} + {b^2}} \right)}}} \cr & {\text{ = }}\frac{1}{{{b^2} - {a^2}}}\int\limits_0^\infty {\frac{{\left( {{x^2} + {b^2}} \right) - \left( {{x^2} + {a^2}} \right)}}{{\left( {{x^2} + {a^2}} \right)\left( {{x^2} + {b^2}} \right)}}} \cr & {\text{ = }}\frac{1}{{{b^2} - {a^2}}}\int\limits_0^\infty {\left[ {\frac{1}{{{x^2} + {a^2}}} - \frac{1}{{{x^2} + {b^2}}}} \right]dx} \cr & {\text{ = }}\frac{1}{{{b^2} - {a^2}}}\left[ {\frac{1}{a}{{\tan }^{ - 1}}\frac{x}{a} - \frac{1}{b}{{\tan }^{ - 1}}\frac{x}{b}} \right]_0^\infty \cr & {\text{ = }}\frac{1}{{{b^2} - {a^2}}}\left[ {\frac{\pi }{{2a}} - \frac{\pi }{{2b}}} \right] \cr & {\text{ = }}\frac{\pi }{{2ab\left( {a + b} \right)}} \cr} $$

$$\mathop {\lim }\limits_{n \to \infty } \sum\limits_{r = 0}^{n - 1} {\frac{1}{n}.\frac{1}{{\sqrt {1 - {{\left( {\frac{r}{n}} \right)}^2}} }}} = \int_0^1 {\frac{{dx}}{{\sqrt {1 - {x^2}} }} = \left[ {{{\sin }^{ - 1}}x} \right]_0^1 = \frac{\pi }{2}} $$

$$\eqalign{ & I = \int\limits_3^6 {\frac{{\sqrt x }}{{\sqrt {9 - x} + \sqrt x }}dx} .....(1) \cr & I = \int\limits_3^6 {\frac{{\sqrt {9 - x} }}{{\sqrt {9 - x} + \sqrt x }}dx} .....(2) \cr & \left[ {{\text{using }}\int\limits_a^b {f\left( x \right)} dx = \int\limits_a^b {f\left( {a + b - x} \right)dx} } \right] \cr} $$
using equation (1) and (2)
$$2I = \int\limits_3^6 {dx} = 3\,\, \Rightarrow I = \frac{3}{2}$$

$$\eqalign{ & g\left( x \right) = \int\limits_0^x {f\left( t \right)dt} \cr & \Rightarrow g\left( 2 \right) = \int\limits_0^2 {f\left( t \right)dt} = \int\limits_0^1 {f\left( t \right)dt} + \int\limits_1^2 {f\left( t \right)dt} \cr & {\text{Now,}}\,\frac{1}{2} \leqslant f\left( t \right) \leqslant 1\,{\text{for }}t \in \left[ {0,\,1} \right] \cr & {\text{We get }}\int\limits_0^1 {\frac{1}{2}dt} \leqslant \int\limits_0^1 {f\left( t \right)dt} \leqslant \int\limits_0^1 {1dt} \cr} $$
(applying line integral on inequality)
$$\eqalign{ & \Rightarrow \frac{1}{2} \leqslant \int\limits_0^1 {f\left( t \right)dt} \leqslant 1.....(1) \cr & {\text{Again, 0}} \leqslant f\left( t \right) \leqslant \frac{1}{2}\,{\text{for }}t \in \left[ {1,\,2} \right] \cr & {\text{We get }}\int\limits_1^2 {0dt} \leqslant \int\limits_1^2 {f\left( t \right)dt} \leqslant \int\limits_1^2 {\frac{1}{2}dt} \cr} $$
(applying line integral on inequality)
$$ \Rightarrow 0 \leqslant \int\limits_1^2 {f\left( t \right)dt} \leqslant \frac{1}{2}.....(2)$$
From (1) and (2), we get
$$\frac{1}{2} \leqslant \int\limits_0^1 {f\left( t \right)dt} + \int\limits_1^2 {f\left( t \right)dt} \leqslant \frac{3}{2}\,\,{\text{or}}\,\,\frac{1}{2} \leqslant g\left( 2 \right) \leqslant \frac{3}{2}$$
$$ \Rightarrow 0 \leqslant g\left( 2 \right) \leqslant 2$$     is the most appropriate solution.

We have,
$$\eqalign{ & {A_{n + 1}} - {A_n} = \int\limits_0^{\frac{\pi }{2}} {\frac{{\sin \left( {2n + 1} \right)x - \sin \left( {2n - 1} \right)x}}{{\sin \,x}}dx} \cr & = \int\limits_0^{\frac{\pi }{2}} {\frac{{2\,\cos \,2nx\,\sin \,x}}{{\sin \,x}}dx} \cr & = 2\int\limits_0^{\frac{\pi }{2}} {\cos \,2n\pi \,dx} \cr & = 0 \cr} $$
Again,
$$\eqalign{ & {B_{n + 1}} - {B_n} = \int\limits_0^{\frac{\pi }{2}} {\frac{{{{\sin }^2}\left( {n + 1} \right)x - {{\sin }^2}nx}}{{{{\sin }^2}x}}dx} \cr & = \int\limits_0^{\frac{\pi }{2}} {\frac{{\sin \left( {2n + 1} \right)x\,\sin \,x}}{{{\sin^2} \,x}}dx} \cr & = {A_{n + 1}} \cr} $$

$$\eqalign{ & {\text{Let }}I = \int\limits_0^\pi {x\,f\left( {\sin \,x} \right)} dx = \int\limits_0^\pi {\left( {\pi - x} \right)\,f\left( {\sin \,x} \right)} dx \cr & \therefore 2I = \pi \int\limits_0^\pi {f\left( {\sin \,x} \right)} dx = \pi .2\int\limits_0^{\frac{\pi }{2}} {f\left( {\sin \,x} \right)} dx \cr & \therefore I = \pi \int\limits_0^{\frac{\pi }{2}} {f\left( {\sin \,x} \right)} dx\,\,\,\,\,\, \Rightarrow A = \pi \cr} $$

$$\eqalign{ & {\text{We have, if }}{e^x} > 2,\,\frac{2}{{{e^x}}} < 1 \cr & {\text{Also, }}\frac{2}{{{e^x}}} > 0 \Rightarrow 0 < \frac{2}{{{e^x}}} < 1 \cr & \therefore \,{\text{If }}x > {\log _e}2,\,\left[ {\frac{2}{{{e^x}}}} \right] = 0 \cr & {\text{Again if }}0 < x < {\log _e}2{\text{ then }}1 < {e^x} < 2 \cr & \Rightarrow 1 > \frac{1}{{{e^x}}} > \frac{1}{2} \cr & \Rightarrow 2 > \frac{2}{{{e^x}}} > 1{\text{ or }}1 < \frac{2}{{{e^x}}} < 2 \cr & \therefore \,\left[ {\frac{2}{{{e^x}}}} \right] = 1 \cr & \therefore \,I = \int\limits_0^\infty {\left[ {\frac{2}{{{e^x}}}} \right]dx} \cr & = \int\limits_0^\infty {\left[ {2{e^{ - x}}} \right]} dx \cr & = \int\limits_0^{\log \,2} {\left[ {2{e^{ - x}}} \right]dx} + \int\limits_{\log \,2}^\infty {\left[ {2{e^{ - x}}} \right]} dx \cr & = \int\limits_0^{\log \,2} {\left( 1 \right)dx} + \int\limits_{\log \,2}^\infty {\left( 0 \right)} dx \cr & = {\log _e}2 \cr} $$

$$\eqalign{ & {\text{We know that }}\frac{{\sin \,x}}{x} < 1,\,{\text{for }}x\, \in \left( {0,\,1} \right) \cr & \Rightarrow \frac{{\sin \,x}}{{\sqrt x }} < \sqrt x {\text{ on }}x\, \in \left( {0,\,1} \right) \cr & \Rightarrow \int\limits_0^1 {\frac{{\sin \,x}}{{\sqrt x }}dx < } \int\limits_0^1 {\sqrt x \,dx = \left[ {\frac{{2{x^{\frac{3}{2}}}}}{3}} \right]_0^1} \cr & \Rightarrow \int\limits_0^1 {\frac{{\sin \,x}}{{\sqrt x }}dx < } \frac{2}{3} \cr & \Rightarrow I < \frac{2}{3}{\text{ Also }}\frac{{\cos \,x}}{{\sqrt x }} < \frac{1}{{\sqrt x }}\,\,{\text{for }}x\, \in \left( {0,\,1} \right) \cr & \Rightarrow \int\limits_0^1 {\frac{{\cos \,x}}{{\sqrt x }}dx < \int\limits_0^1 {{x^{ - \frac{1}{2}}}} dx = \left[ {2\sqrt x } \right]_0^1 = 2} \cr & \Rightarrow \int\limits_0^1 {\frac{{\cos \,x}}{{\sqrt x }}dx < 2} \,\,\, \Rightarrow J < 2 \cr} $$

$$\eqalign{ & {\text{Given, }}F\left( x \right) = f\left( x \right) + f\left( {\frac{1}{x}} \right), \cr & {\text{where }}f\left( x \right) = \int_1^x {\frac{{\log \,t}}{{1 + t}}dt} \cr & \therefore \,F\left( e \right) = f\left( e \right) + f\left( {\frac{1}{e}} \right) \cr & \Rightarrow F\left( e \right) = \int_1^e {\frac{{\log \,t}}{{1 + t}}dt} + \int_1^{\frac{1}{e}} {\frac{{\log \,t}}{{1 + t}}dt} ......\left( {\text{A}} \right) \cr & {\text{Now for solving, }}I = \int_1^{\frac{1}{e}} {\frac{{\log \,t}}{{1 + t}}dt} \cr & \therefore \,{\text{Put }}\frac{1}{t} = z \Rightarrow - \frac{1}{{{t^2}}}dt = dz \Rightarrow dt = - \frac{{dz}}{{{z^2}}} \cr & {\text{and limit for }}t = 1 \Rightarrow z = 1{\text{ and for }}t = \frac{1}{e} \Rightarrow z = e \cr & \therefore \,I = \int_1^e {\frac{{\log \left( {\frac{1}{z}} \right)}}{{1 + \frac{1}{z}}}\left( { - \frac{{dz}}{{{z^2}}}} \right)} \cr & \Rightarrow I = \int_1^e {\frac{{\left( {\log \,1 - \log \,z} \right).z}}{{z + 1}}\left( { - \frac{{dz}}{{{z^2}}}} \right)} \cr & \Rightarrow I = \int_1^e { - \frac{{\log \,z}}{{\left( {z + 1} \right)}}\left( { - \frac{{dz}}{{{z^2}}}} \right)} \cr & \Rightarrow I = \int_1^e {\frac{{\log \,z}}{{z\left( {z + 1} \right)}}dz} \cr & \therefore \,I = \int_1^e {\frac{{\log \,t}}{{t\left( {t + 1} \right)}}dt} \cr & {\text{Equation}}\left( {\text{A}} \right){\text{ becomes :}} \cr & F\left( e \right) = \int_1^e {\frac{{\log \,t}}{{1 + t}}dt + \int_1^e {\frac{{\log \,t}}{{t\left( {1 + t} \right)}}dt} } \cr & \Rightarrow F\left( e \right) = \int_1^e {\frac{{t.\log \,t + \log \,t}}{{t\left( {1 + t} \right)}}dt} \cr & \Rightarrow F\left( e \right) = \int_1^e {\frac{{\left( {\log \,t} \right)\left( {t + 1} \right)}}{{t\left( {1 + t} \right)}}dt} \cr & \Rightarrow F\left( e \right) = \int_1^e {\frac{{\log \,t}}{t}dt} \cr & {\text{Let }}\log \,t = x \cr & \therefore \,\frac{1}{t}dt = dx \cr & \left[ {{\text{for limit }}t = 1,{\text{ }}x = 0{\text{ and }}t = e,\,x = \log \,e = 1} \right] \cr & \therefore \,F\left( e \right) = \int_0^1 {x\,dx} \cr & \Rightarrow F\left( e \right) = \left[ {\frac{{{x^2}}}{2}} \right]_0^1 \cr & \Rightarrow F\left( e \right) = \frac{1}{2} \cr} $$