(i)   $f\left(x+1\right)=f\left(x+3\right)orf\left(x\right)=f\left(x+2\right)$

Thus, $f\left(x\right)$ is periodic with period 2.

Then, ${\int }_a^{a+b}f\left(x\right)dx$ is independent of a, for which b is multiple of 2. Thus, b=2,4,6….

(ii)let  $I=25{\int }_0^{\pi /4}\left({\tan }^6\left(x-\left[x\right]\right)+{\tan }^4\left(x-\left[x\right]\right.\right)dx$ $\left\{\because 0<\pi \le \frac{\pi }{4}\Rightarrow \left[x\right]=0\right\}$

$=25{\int }_0^{\pi /4}\left({\tan }^{6}x+{\tan }^{4}x\right)dx$

$=25{\int }_0^{\pi /4}{\tan }^{4}x\left(1+{\tan }^{2}x\right)dx$

$=25{\int }_0^{\pi /4}{\tan }^{4}x{\sec }^{2}dx$

$=25{\left(\frac{{\tan }^{5}x}{5}\right)}_0^{\pi /4}$

$=25\times \frac{1}{5}=5$

(iii)let   $I={\int }_1^4\frac{{\tan }^{-1}\left[x^2\right]}{{\tan }^{-1}\left[x^2\right]+{\tan }^{-1}\left[25+x^2-10x\right]}dx$ …(i)

On applying  ${\int }_a^{b}f\left(x\right)dx={\int }_a^{b}f\left(a+b-x\right)dx$ , we ger

$I={\int }_a^b\frac{{\tan }^{-1}\left[{\left(5-x\right)}^2\right]}{{\tan }^{-1}\left[{\left(5-x\right)}^2\right]+{\tan }^{-1}\left[x^2\right]}dx$    …(ii)

Adding Eqs, (i) and (ii), we get

$2I={\int }_1^{4}1dx$

$2I=3$

$I=\frac{3}{2}$

$\left[I\right]=\left[\frac{3}{2}\right]=1$

(iv)Let   $y=\sqrt{x+\sqrt{x+\sqrt{x+\mathrm{...}}}}$ 

$\Rightarrow y=\sqrt{x+y}\Rightarrow y^2-y-x=0$

$\Rightarrow y=\frac{1\pm \sqrt{1+4x}}{2}$

$y=\frac{1+\sqrt{1+4x}}{2}$   $\left[\because y>1\right]$

$\therefore l={\int }_0^2\frac{1+\sqrt{1+4x}}{2}dx$

$={\left[\frac{x}{2}+\frac{{\left(1+4x\right)}^{3/2}}{\frac{3}{2}.2\times 4}\right]}_0^2$

$=\left(1+\frac{27}{12}\right)-\left(0+\frac{1}{12}\right)$

$=1+\frac{26}{12}=\frac{19}{6}$

$\therefore \left[I\right]=3$