If `lim_(x to 1) (ax^(2)+bx+c)/((x-1)^(2))=2`, then (a, b, c) is

Options

A. (2, -4, 2)
B. (2, 4, 2)
C. (2, 4, -2)
D. (2, -4, -2)

Explanation

To solve the problem, we need to find the values of \(a\), \(b\), and \(c\) such that

\[ \lim_{x \to 1} \frac{ax^2 + bx + c}{(x-1)^2} = 2. \]

### Step 1: Understanding the Limit Condition
For the limit to exist as \(x\) approaches 1, the numerator \(ax^2 + bx + c\) must equal \(0\) at \(x = 1\) (since the denominator \((x-1)^2\) approaches \(0\)). Therefore, we have:

\[ a(1)^2 + b(1) + c = 0 \implies a + b + c = 0. \]

### Step 2: Setting Up the Equality for the Limit
Since the limit equals \(2\), we can write:

\[ \lim_{x \to 1} \frac{ax^2 + bx + c}{(x-1)^2} = 2. \]

This implies that \(ax^2 + bx + c\) must be equal to \(2(x-1)^2\) for values of \(x\) near \(1\). Expanding \(2(x-1)^2\):

\[ 2(x-1)^2 = 2(x^2 - 2x + 1) = 2x^2 - 4x + 2. \]

### Step 3: Equating Coefficients
Now we can equate the polynomial \(ax^2 + bx + c\) with \(2x^2 - 4x + 2\):

\[ ax^2 + bx + c = 2x^2 - 4x + 2. \]

From this, we can compare the coefficients of \(x^2\), \(x\), and the constant term:

1. Coefficient of \(x^2\): \(a = 2\)
2. Coefficient of \(x\): \(b = -4\)
3. Constant term: \(c = 2\)

### Step 4: Final Values
Thus, we have:

\[ (a, b, c) = (2, -4, 2). \]

### Conclusion
The triplet \((a, b, c)\) is \((2, -4, 2)\).

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