$$\eqalign{ & {I_n} + {I_{n + 2}} = \int\limits_0^{\frac{\pi }{4}} {{{\tan }^n}x\left( {1 + {{\tan }^2}x} \right)\,dx} \cr & = \int\limits_0^{\frac{\pi }{4}} {{{\tan }^n}x\,{{\sec }^2}\,x\,dx = \left[ {\frac{{{{\tan }^{n + 1}}x}}{{n + 1}}} \right]_0^{\frac{\pi }{4}}} \cr & = \frac{{1 - 0}}{{n + 1}} \cr & = \frac{1}{{n + 1}} \cr & \therefore {I_n} + {I_{n + 2}} = \frac{1}{{n + 1}} \Rightarrow \mathop {\lim }\limits_{n \to \infty } n\left[ {{I_n} + {I_{n + 2}}} \right] \cr & = \mathop {\lim }\limits_{n \to \infty } \,n.\frac{1}{{n + 1}} \cr & = \mathop {\lim }\limits_{n \to \infty } \frac{n}{{n + 1}} \cr & = \mathop {\lim }\limits_{n \to \infty } \frac{n}{{n\left( {1 + \frac{1}{n}} \right)}} = 1 \cr} $$

$$\eqalign{ & {a_n} - {a_{n - 1}} = \int_0^{\frac{\pi }{2}} {\frac{{{{\sin }^2}nx - {{\sin }^2}\left( {n - 1} \right)x}}{{\sin \,x}}dx} \cr & = \frac{1}{2}\int_0^{\frac{\pi }{2}} {\frac{{\cos \,2\left( {n - 1} \right)x - \cos \,2nx}}{{\sin \,x}}dx} \cr & = \int_0^{\frac{\pi }{2}} {\sin \left( {2n - 1} \right)x\,dx} \cr & = \left[ { - \frac{{\cos \left( {2n - 1} \right)x}}{{2n - 1}}} \right]_0^{\frac{\pi }{2}} \cr & = \frac{1}{{2n - 1}} \cr} $$

According to question
$$\eqalign{ & f'\left( 1 \right) = \tan \frac{\pi }{6} = \frac{1}{{\sqrt 3 }} \cr & f'\left( 2 \right) = \tan \frac{\pi }{3} = \sqrt 3 \cr & {\text{and }}f'\left( 3 \right) = \tan \frac{\pi }{4} = 1 \cr & {\text{So, }}\int\limits_1^3 {f''\left( x \right)f'\left( x \right)dx} + \int\limits_1^3 {f''\left( x \right)dx} \cr & = \left[ {\frac{{{{\left\{ {f'\left( x \right)} \right\}}^2}}}{2}} \right]_1^3 + \left[ {f'\left( x \right)} \right]_2^3 \cr & = \frac{1}{2}\left[ {1 - \frac{1}{3}} \right] + \left[ {1 - \sqrt 3 } \right] \cr & = \frac{4}{3} - \sqrt 3 \cr} $$

$$\eqalign{ & I = \int\limits_a^b {x\,f\left( x \right)dx} = \int\limits_a^b {\left( {a + b - x} \right)f\left( {a + b - x} \right)dx} \cr & = \left( {a + b} \right)\int\limits_a^b {f\left( {a + b - x} \right)dx - \int\limits_a^b {x\,f\left( {a + b - x} \right)dx} } \cr & = \left( {a + b} \right)\int\limits_a^b {f\left( x \right)dx} - \int\limits_a^b {x\,f\left( x \right)} dx \cr & \left[ {\because {\text{ given that }}f\left( {a + b - x} \right) = f\left( x \right)} \right] \cr & 2I = \left( {a + b} \right)\int\limits_a^b {f\left( x \right)dx} \cr & \Rightarrow I = \frac{{\left( {a + b} \right)}}{2}\int\limits_a^b {f\left( x \right)dx} \cr} $$

$$\eqalign{ & {\text{Limit}} = \mathop {\lim }\limits_{n \to \infty } \sum\limits_{r = 1}^{4n} {\frac{1}{n}.\frac{1}{{1 + \frac{r}{n}}}} = \int_\alpha ^\beta {\frac{1}{{1 + x}}dx} \cr & {\text{where }}\alpha = \mathop {\lim }\limits_{n \to \infty } \frac{r}{n},\,{\text{when}}\,\,\,r = 1\,\,\,\,\,\,\,\, \Rightarrow \alpha = 0 \cr & \beta = \mathop {\lim }\limits_{n \to \infty } \frac{r}{n},\,{\text{when}}\,\,\,r = 4n\,\,\,\, \Rightarrow \beta = 4 \cr & \therefore {\text{ Limit}} = \int_0^4 {\frac{{dx}}{{1 + x}}} = \left[ {\log \left( {1 + x} \right)} \right]_0^4 = \log \,5 \cr} $$

$$\eqalign{ & I = \int\limits_{ - \frac{{3\pi }}{2}}^{ - \frac{\pi }{2}} {\left[ {{{\left( {x + \pi } \right)}^3} + {{\cos }^2}\left( {x + 3\pi } \right)} \right]} dx \cr & {\text{Put }}x + \pi = t \cr & I = \int\limits_{ - \frac{\pi }{2}}^{\frac{\pi }{2}} {\left[ {{t^3} + {{\cos }^2}t} \right)dt} = 2\int\limits_{ - \frac{\pi }{2}}^{\frac{\pi }{2}} {{{\cos }^2}t\,dt} \cr} $$
[ using the property of even and odd function]
$$ = \int\limits_0^{\frac{\pi }{2}} {\left( {1 + \cos \,2t} \right)dt} = \frac{\pi }{2} + 0$$

$${\text{Let, }}I = \int\limits_0^{\frac{\pi }{2}} {\frac{{{{\sin }^3}\,x}}{{\sin \,x + \cos \,x}}} .....(1)$$
Use the property $$\int_0^a {f\left( x \right)dx} = \int_0^a {f\left( {a - x} \right)dx} $$
$$ \Rightarrow I = \int\limits_0^{\frac{\pi }{2}} {\frac{{{{\cos }^3}x\,dx}}{{\sin \,x + \cos \,x\,}}} .....(2)$$
Adding equations (1) and (2), we get
$$\eqalign{ & 2I = \int\limits_0^{\frac{\pi }{2}} {\left( {1 - \frac{1}{2}\sin \left( {2x} \right)} \right)dx} \cr & \Rightarrow I = \frac{1}{2}\left[ {x + \frac{1}{4}\cos \,2x} \right]_0^{\frac{\pi }{2}} \cr & \Rightarrow I = \frac{{\pi - 1}}{4} \cr} $$

$$\eqalign{ & I = \int\limits_2^4 {\frac{{\log \,{x^2}}}{{\log \,{x^2} + \log \,\left( {36 - 12x + {x^2}} \right)}}} \cr & I = \int\limits_2^4 {\frac{{\log \,{x^2}}}{{\log \,{x^2} + \log \,{{\left( {6 - x} \right)}^2}}}} .....(1) \cr & I = \int\limits_2^4 {\frac{{\log \,{{\left( {6 - x} \right)}^2}}}{{\log {{\left( {6 - x} \right)}^2} + \log \,{x^2}}}} .....(2) \cr & {\text{Adding (1) and (2), we get}} \cr & 2I = \int\limits_2^4 {dx} = \left[ x \right]_2^4 = 2\,\,\, \Rightarrow I = 1 \cr} $$

$$\mathop {\lim }\limits_{n \to \infty } \sum\limits_{r = 1}^n {\frac{1}{n}{e^{\frac{r}{n}}}} = \int_0^1 {{e^x}dx} = \left[ {{e^x}} \right]_0^1 = e - 1$$

$$\eqalign{ & {\text{Let }}\int\limits_1^2 {\left\{ {{K^2} + \left( {4 - 4K} \right)x + 4{x^3}} \right\}} dx \leqslant 12 \cr & \Rightarrow \left. {{K^2}x + \frac{{\left( {4 - 4K} \right){x^2}}}{2} + \frac{{4{x^4}}}{4}} \right|_1^2 \leqslant 12 \cr & \Rightarrow \left[ {2{K^2} + \left( {2 - 2K} \right)\left( 4 \right) + 16} \right] - \left[ {{K^2} + \left( {2 - 2K} \right) + 1} \right] \leqslant 12 \cr & \Rightarrow \left( {2{K^2} + 8 - 8K + 16} \right) - \left( {{K^2} - 2K + 3} \right) \leqslant 12 \cr & \Rightarrow {K^2} - 6K + 21 \leqslant 12 \cr & \Rightarrow {K^2} - 6K + 9 \leqslant 0 \cr & \Rightarrow {\left( {K - 3} \right)^2} \leqslant 0 \cr & \Rightarrow K = 3 \cr} $$