The number of ordered pair(s) (a, b) for which the function $f\left(x\right)=\mathrm{sgn}\left((x^2-ax+1)(b{x}^2-2bx+1)\right)$  is discontinuous at exactly one point 1 (Where
a,b are integers and sgn(x) denotes signum function of x) is/are _____

 $f\left(x\right)=\mathrm{sgn}\left((x^2-ax+1)(b{x}^2-2bx+1)\right)$
 $P\left(x\right)=x^2-ax+1,Q\left(x\right)=b{x}^2-2bx+1$
$P\left(x\right)Q\left(x\right)=0$ at exactly one value of x
Case (i) ${\Delta }_1=0$ and  ${\Delta }_2=0$
$a^2=4$ and  $4b^2-4b<0 \Rightarrow b\in (0,1)$
$a=2,-2$ no integral value of  b
Case (ii) ${\Delta }_1<0$ and  ${\Delta }_2=0$
$a^2-4<0 4b^2-4b=0 \Rightarrow b=0$, 1
$a\in (-2,2) (-1,1)(0,1)(1,1)$  are possible
Case (ii) ${\Delta }_1=0$ and  ${\Delta }_2=0$
 $a^2-4=0 4b^2-4b=0$
$a=2,-2 b=0,1$ 
$b=0 (2,0)(-2,0)$
$b=1 (2,1)(-2,1)$ (Not possible)
Total number of ordered pairs (a, b) = 3 + 3 = 6