Let  $f\left(x\right)=\frac{x^2-6x+5}{x^2-5x+6}$

Match of expressions/statements in Column I with expressions/statements in Column II and indicate your answer by darkening the appropriate bubbles in the  matrix given in the ORS

	Column-I		Column-II
A)	If $-1<x<1$, then $f\left(x\right)$ satisfies	p)	$0<f\left(x\right)<1$
B)	If $1<x<2$, then $f\left(x\right)$ satisfies	q)	$f\left(x\right)<0$
C)	If $3<x<5$, then $f\left(x\right)$ satisfies	r)	$f\left(x\right)>0$
D)	If $x>5$,then $f\left(x\right)$ satisfies	s)	$f\left(x\right)<1$

 $f\left(x\right)=\frac{x^2-6x+5}{x^2-5x+6}=\frac{(x-5)(x-1)}{(x-2)(x-3)}$
 $\left(A\right)\text{\hspace{0.17em}If }-1<x<1\text{ then }f\left(x\right)=\frac{(-ve)(-ve)}{(-ve)(-ve)}=+ve$
 $\therefore f\left(x\right)>0\left(r\right)$
 $\text{Also }f\left(x\right)-1=\frac{-x-1}{x^2-5x+6}=-\frac{(x+1)}{(x-2)(x-3)}$
 $\text{For }f\left(x\right)-1<0\Rightarrow f\left(x\right)<1\left(s\right)$
 $\therefore 0<f\left(x\right)<1\left(p\right)$
 $\left(B\right)\text{\hspace{0.17em}\hspace{0.17em}If }1<x<2\text{ then }f\left(x\right)=\frac{(-ve)(+ve)}{(-ve)(-ve)}=-ve$
 $\therefore f\left(x\right)<0\left(q\right)\text{\hspace{0.17em} and so }f\left(x\right)<1\left(s\right)$ 
 $\left(C\right)\text{\hspace{0.17em} If }3<x<5\text{ then }f\left(x\right)=\frac{(-ve)(+ve)}{(+ve)(+ve)}=-ve$ 
 $\therefore f\left(x\right)<0\left(q\right)\text{ and so }f\left(x\right)<1\left(s\right)$
 $\left(D\right)\text{\hspace{0.17em}For }f\left(x\right)-1=\frac{-(x+1)}{(x-2)(x-3)}<0$
$\text{Also }x>5,f\left(x\right)<1\left(s\right)$

$\therefore 0<f\left(x\right)<1\left(p\right)$