Functions

$f\left(x\right)=\{\begin{array}{l}x-\left[x\right]if\left[x\right]isodd\\ 1+\left[x\right]-x,if\left[x\right]iseven\end{array}$

Graph of f (x)

So

$=2{\pi }^2{\displaystyle \underset{0}{\overset{1}{\int }}\left(1-x\right)cos\pi xdx}$

=4

Case I

= = 0 then

f (x) = yx

$g\left(x\right)=\frac{x}{y}$

$\frac{1}{x}{\displaystyle \sum _{i=1}^{n}f\left(a_i\right)\Rightarrow \frac{y}{x}\left(a_1+a_2+....+a_x\right)=0}$

$\Rightarrow f\left(g\left(0\right)\right)\Rightarrow f\left(0\right)$

=0

Use [x + n] = n + [x], where $n\in z$  f (x) = [x] – 10 will be minimum for $x\in [0,10]$  break the limits as G.I.F. is discontinuous at integral points.

${\displaystyle \underset{0}{\overset{10}{\int }}f\left(x\right)dx=-10-9-....-1}$    

$=\frac{-10.11}{2}$

${\displaystyle \underset{0}{\overset{10}{\int }}\left|f\left(x\right)\right|dx=10+9+....+1}$

$f\left(x\right)=0\Rightarrow {\left(x-p\right)}^2-q=0$  

Roots are  $p+\sqrt[]{q},p-\sqrt[]{q}$

Now, $\left|f\left(a_i\right)\right|=500$  

$Let{a}_1,a_2,a_3...ara,a+d,a+2d,a+3d$               

=>  $\frac{9}{4}{d}^2-q=500$           …….(i)

and ${\left|f\left(a_1\right)\right|}^2={\left|f\left(a_2\right)\right|}^2$

From equation (i)

$\frac{9}{4}.\frac{4q}{5}-q=500$     

$\frac{4q}{5}=500$           

and  $2\sqrt[]{q}=2*\frac{50}{2}=50$

$f\left(x+y\right)=2f\left(x\right)f\left(y\right),x,y\in N$

$f\left(1\right)=2$

${\displaystyle \sum _{k=1}^{10}f\left(\alpha +k\right)={\displaystyle \sum _{k=1}^{10}2f\left(\alpha \right)f\left(k\right)}}$

$=2f\left(\alpha \right){\displaystyle \sum _{k=1}^{10}f\left(k\right)=2.2^{2\alpha -1}.\frac{2}{3}\left(4^{10}-1\right)}$

$=\frac{2^{2\alpha +1}}{3}.\left(4^{10}-1\right)$

$\Rightarrow 2\alpha +1=9$

=4

 $f\left(x\right)=\left[1+x\right]+\frac{{\alpha }^{2\left|x\right|}+\left\{x\right\}+\left[x\right]-1}{2\left[x\right]+\left\{x\right\}}$

$li{m}_{x\to 0}-f\left(x\right)=\alpha -\frac{4}{3}$

$\Rightarrow li{m}_{h\to 0}1-1+\frac{{\alpha }^{-h-1}-1-1}{-h-1}=\alpha -\frac{4}{3}$

$\therefore \frac{{\alpha }^{-1}-2}{-1}\alpha -\frac{4}{3}$

32 - 10 + 3 = 0

$\therefore \alpha =3or1/3$

$?\alpha$ in integer, hence = 3