If _ x → 1 ( (3 x ^ 2 - 4x + 1) - x ^ 2 + 1 2 x ^ 3 - 7 x ^ 2 + ax...

Given $\lim_{x \to 1}\left(\dfrac{\sin \left(3{x}^{2} - 4x + 1\right) - {x}^{2} + 1}{2{x}^{3} - 7{x}^{2} + ax + b}\right) = - 2 . . . \left(i\right)$

Putting $x = 1$ in limit we get

$⇒ \lim_{x \to 1}\left(\dfrac{\sin 0 - 1 + 1}{2 - 7 + a + b}\right) = - 2$ $⇒ \lim_{x \to 1}\left(\dfrac{0}{a + b - 5}\right) = - 2$

For limit to exist $a + b - 5$ should be zero

so $a + b - 5 = 0$ $⇒ a + b = 5 . . . \left(ii\right)$

Now Using L-Hospital in equation \left(i\right)$we get</p><p>$\lim_{x \to 1}\dfrac{\cos \left(3{x}^{2} - 4x + 1\right)\left(6x - 4\right) - 2x}{6{x}^{2} - 14x + a} = - 2$</p><p>Again putting$x = 1$we get</p><p>$⇒ \lim_{x \to 1}\dfrac{\left(\cos 0\right) \times \left(6 - 4\right) - 2}{6 - 14 + a} = - 2$</p><p>$⇒ \lim_{x \to 1}\dfrac{0}{a - 8} = - 2$</p><p>Again for limit to exist$a - 8 = 0 ⇒ a = 8$</p><p>So from equation \left(ii\right)$ we get

$8 + b = 5 ⇒ b = - 3$

So $a - b = 8 - \left(- 3\right) = 11$