Columm-1	Columm-2
A)Is the function $f\left(x\right)=\frac{\sqrt{x^2+kx+1}}{x^2-k}$Is continuous for every $x\in R$ Then $k\in [-n,0]$ then $n$is	p)4
B) Let $f\left(x\right)=\frac{\log (1+x+x^2)+\log (1-x+x^2)}{\sec x-\cos x},$ $x\ne 0$ then the value of f(o) so that f is continuous at x=0 is	q)3
C) If $x^{2005}+\frac{1}{1+{\sin }^{2}x}=2005$then $x\in [o,k]$then 2k is	r)2
D) The function $\text{f}\left(\text{x}\right)=\frac{x^3}{4-\sin \pi x+3}$ Takes the value a for$x\in [-2,2]$the 7/a is	s)1

a) $x^2+kx+1\ge 0$and $x^2-k$ must not have any real root

            $\therefore k^2-4\le 0$and $k<0$

            $\Rightarrow k\in [-2,2]$AND $K<0$$k\in [-2,0]$

            b) for f to be continous at x=0

            $f\left(0\right)=\underset{x\to 0}{\lim }f\left(x\right)$

            $\underset{x\to 0}{\lim }\frac{\log [{(1+x^2)}^2-x^2]}{1-{\cos }^{2}x}\cos x$

            $\underset{x\to 0}{\lim }\frac{\log (1+(x^3+x^4))}{x^2+x^4}\times \frac{x^2}{{\sin }^{2}x}\times (1+x^2)\cos x=1$

            Let $f\left(x\right)=x^{2005}+{(1+{\sin }^{2}x)}^{-2}$

            Thus f is continuous and $f\left(0\right)=1<2005$and and $f\left(2\right)=2^{2005}$hence from intermediate value thermo $f\left(c\right)=2005$ $f\left(x\right)$is continuous for $x\in [-2,2]$now $f(-2)=1$and $f\left(2\right)=5$

            By intermediate value theorem takes all valves between 1 and 5