Question

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

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

Answer 1

$f\left(x\right)=\frac{x^2-6x+5}{x^2-5x+6}=\frac{x^2-5x-x+5}{x^2-2x-3x+6}$

$\Rightarrow f\left(x\right)=\frac{x(x-5)-(x-5)}{x(x-2)-3(x-2)}=\frac{(x-1)(x-5)}{(x-3)(x-2)}$

(A). $-1<x<1$ then $0<f\left(x\right)<1$

$\therefore A-p$

(B). If $1<x<2$

$\Rightarrow f\left(x\right)<0$

(C). If $3<x<5$

$\Rightarrow f\left(x\right)<0$

$\therefore c-q$

(D). If $x>5$

$\Rightarrow 0<f\left(x\right)<1$

Hence, solved.