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(C) Given $\lim\limits _{x \rightarrow 1} \frac{\sin \left(3 x^{2}-4 x+1\right)-x^{2}+1}{2 x^{3}-7 x^{2}+a x+b}=-2$.Since the limit is finite and the

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$11$

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(C) Given $\lim\limits _{x \rightarrow 1} \frac{\sin \left(3 x^{2}-4 x+1\right)-x^{2}+1}{2 x^{3}-7 x^{2}+a x+b}=-2$.Since the limit is finite and the numerator approaches $0$ as $x \rightarrow 1$,the denominator must also approach $0$.$2(1)^3 - 7(1)^2 + a(1) + b = 0 \implies a + b - 5 = 0 \dots (1)$.Applying $L'H\hat{o}pital$ rule:$\lim\limits _{x \rightarrow 1} \frac{\cos \left(3 x^{2}-4 x+1\right)(6 x-4) - 2x}{6x^2 - 14x + a} = -2$.For the limit to be finite,the denominator must be $0$ at $x=1$:$6(1)^2 - 14(1) + a = 0 \implies a - 8 = 0 \implies a = 8$.Substituting $a=8$ into $(1)$:$8 + b - 5 = 0 \implies b = -3$.Therefore,$(a - b) = 8 - (-3) = 11$.

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

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