Arun Sharma Solutions for Exercise 2: XAT(2008-2017)

If x and y are real numbers, then the minimum value of ${\mathrm{x}}^2+4\mathrm{xy}+6{\mathrm{y}}^2-4\mathrm{y}+4$ is

The operation (x) is defined by $\left(\mathrm{i}\right)$ $\left(1\right)=2 \left(\mathrm{ii}\right) (\mathrm{x}+\mathrm{y})=\left(\mathrm{x}\right) .\left(\mathrm{y}\right)$ for all positive integers x and y. If $(\mathrm{x}+\mathrm{y})=\left(\mathrm{x}\right)·\left(\mathrm{y}\right)$ for all positive integers x and y. If $\sum _{1=1}^{\mathrm{n}}\left(\mathrm{x}\right)=1022,\text{ then }\mathrm{n}=?$

The domain of the function $\mathrm{f} \left(\mathrm{x}\right)={\log }_{10\mathrm{g}}\left(\log \right(20\mathrm{x}$$-\mathrm{x}^\left\{2\right\}-91) ?\text{ is}:$

If $\mathrm{f}\left({\mathrm{x}}^2-1\right)={\mathrm{x}}^4-7{\mathrm{x}}^2+{\mathrm{k}}_1$ and $\mathrm{f}\left({\mathrm{x}}^3-2\right)={\mathrm{x}}^6-9{\mathrm{x}}^3+{\mathrm{k}}_2$ then the value of $\left({\mathrm{k}}_2-{\mathrm{k}}_1\right)$ is 

For a positive intege x, define $\mathrm{f}\left(\mathrm{x}\right)$ such that $\mathrm{f}(\mathrm{x}+ \mathrm{a})$ $=\mathrm{f}\left(\mathrm{ax}\right)$, where a is an integer and $\mathrm{f}\left(1\right)=4$ . If the value of $\mathrm{f}\left(1003\right)=\mathrm{k},$ then the value of ' k ' will be:

f is a function for which $\mathrm{f}\left(1\right)=1$ and  $\mathrm{f}\left(\mathrm{x}\right)=2 \mathrm{x}+\mathrm{f}(\mathrm{x}-1)$ for each natural number $\mathrm{x}\ge 2$ Find $\mathrm{f}\left(31\right)$

If x and y are real numbers, the least possible value of the expression.$4(\mathrm{x}-2{)}^2+4(\mathrm{y}-3{)}^2-2(\mathrm{x}-3{)}^2$ is 

If $\mathrm{f}\left(\mathrm{x}\right)=\mathrm{a} \mathrm{x}+\mathrm{b}$,a and b  are positive real numbers $\mathrm{f}\left(\mathrm{f}\right(\mathrm{x}\left)\right)=9 \mathrm{x}+8$ then the value of $\mathrm{a}+\mathrm{b}$ is