If $\underset{\mathrm{x}\to 1}{lim} \frac{\sin \left(3{\mathrm{x}}^2-4\mathrm{x}+1\right)-{\mathrm{x}}^2+1}{2{\mathrm{x}}^3-7{\mathrm{x}}^2+\mathrm{ax}+\mathrm{b}}=-2$ , then the value of (a - b) is equal to _________.

$\underset{\mathrm{x}\to 1}{lim} \frac{\sin \left(3{\mathrm{x}}^2-4\mathrm{x}+1\right)-{\mathrm{x}}^2+1}{2{\mathrm{x}}^3-7{\mathrm{x}}^2+\mathrm{ax}+\mathrm{b}}=-2$

For finite limit ,since numerator is 0  for x=1 , denominator must be 0 for x=1

$\Rightarrow \mathrm{a}+\mathrm{b}-5=0 ....\left(1\right)$

Apply L’H rule 

$\underset{\mathrm{x}\to 1}{lim} \frac{\cos \left(3{\mathrm{x}}^2-4\mathrm{x}+1\right)(6\mathrm{x}-4)-2\mathrm{x}}{\left(6{\mathrm{x}}^2-14\mathrm{x}+\mathrm{a}\right)}=-2$

For finite limit 

$6-14+\mathrm{a}=0$

$\mathrm{a}=8$

From (1) $\mathrm{b}=-3$

Now $\left(\mathrm{a}-\mathrm{b}\right)=11$