If `lim_(x rarr 1) (sin (3x^(2) - 4x+1) -x^(2)+1)/(2x^(3)- 7x^(2)+ ax +b)= -2`, then the value of `(a-b)` is equal to ____

Answer

To solve the limit problem, we need to evaluate the limit as \( x \) approaches 1 for the expression given in the question. Let's break it down step by step.

### Step 1: Substitute \( x = 1 \)

First, we substitute \( x = 1 \) into the expression:

\[ \lim_{x \to 1} \frac{\sin(3x^2 - 4x + 1) - x^2 + 1}{2x^3 - 7x^2 + ax + b} \]

Calculating the numerator:

\[ 3(1)^2 - 4(1) + 1 = 3 - 4 + 1 = 0 \]
\[ 1^2 - 1 = 0 \]

So, the numerator becomes:

\[ \sin(0) - 0 = 0 \]

Now for the denominator:

\[ 2(1)^3 - 7(1)^2 + a(1) + b = 2 - 7 + a + b = a + b - 5 \]

Thus, we have:

\[ \frac{0}{a + b - 5} \]

### Step 2: Apply L'Hôpital's Rule

Since we have a \( \frac{0}{0} \) form, we can apply L'Hôpital's Rule. We differentiate the numerator and the denominator.

**Numerator:**

Using the chain rule:

\[ \frac{d}{dx}[\sin(3x^2 - 4x + 1)] = \cos(3x^2 - 4x + 1) \cdot (6x - 4) \]

The derivative of \( -x^2 + 1 \) is \( -2x \).

So, the derivative of the numerator is:

\[ \cos(3x^2 - 4x + 1)(6x - 4) - 2x \]

**Denominator:**

The derivative of \( 2x^3 - 7x^2 + ax + b \) is:

\[ 6x^2 - 14x + a \]

### Step 3: Evaluate the Limit Again

Now we evaluate the limit again:

\[ \lim_{x \to 1} \frac{\cos(3(1)^2 - 4(1) + 1)(6(1) - 4) - 2(1)}{6(1)^2 - 14(1) + a} \]

Calculating this gives:

Numerator:

\[ \cos(0)(6 - 4) - 2 = 1 \cdot 2 - 2 = 0 \]

Denominator:

\[ 6 - 14 + a = a - 8 \]

Thus, we have:

\[ \frac{0}{a - 8} \]

### Step 4: Set the Limit to -2

Since we need the limit to equal -2, we set up the equation:

\[ \frac{0}{a - 8} = -2 \]

For this to hold true, the denominator must also equal zero:

\[ a - 8 = 0 \implies a = 8 \]

### Step 5: Substitute \( a \) Back to Find \( b \)

Now we substitute \( a = 8 \) back into our earlier expression for the denominator:

\[ a + b - 5 = 0 \implies 8 + b - 5 = 0 \implies b = -3 \]

### Step 6: Find \( a - b \)

Now we can find \( a - b \):

\[ a - b = 8 - (-3) = 8 + 3 = 11 \]

Thus, the value of \( a - b \) is:

\[ \boxed{11} \]