If f(x) ={`x^2-1, 0 lt x lt 2` , `2x+3 , 2 le x lt 3`then the quadratic equation whose roots are `lim_(x->2^+)f(x)` and `lim_(x->2^-)f(x)` is

Options

A. `x^(2)-6x+9=0`
B. `x^(2)-7x+8=0`
C. `x^(2)-14x+49=0`
D. `x^(2)-10x+21=0`

Explanation

To solve the problem, we need to find the limits of the function \( f(x) \) as \( x \) approaches 2 from the right (denoted as \( \lim_{x \to 2^+} f(x) \)) and from the left (denoted as \( \lim_{x \to 2^-} f(x) \)). Then, we will use these limits to form a quadratic equation.

### Step 1: Find \( \lim_{x \to 2^+} f(x) \)

For \( x \) approaching 2 from the right (i.e., \( x \to 2^+ \)), we use the piece of the function defined for \( 2 \leq x < 3 \):
\[ f(x) = 2x + 3 \]
Now, substituting \( x = 2 \):
\[ \lim_{x \to 2^+} f(x) = 2(2) + 3 = 4 + 3 = 7 \]

### Step 2: Find \( \lim_{x \to 2^-} f(x) \)

For \( x \) approaching 2 from the left (i.e., \( x \to 2^- \)), we use the piece of the function defined for \( 0 < x < 2 \):
\[ f(x) = x^2 - 1 \]
Now, substituting \( x = 2 \):
\[ \lim_{x \to 2^-} f(x) = 2^2 - 1 = 4 - 1 = 3 \]

### Step 3: Identify the roots

We have found:
- \( \alpha = \lim_{x \to 2^+} f(x) = 7 \)
- \( \beta = \lim_{x \to 2^-} f(x) = 3 \)

### Step 4: Form the quadratic equation

The quadratic equation with roots \( \alpha \) and \( \beta \) can be expressed as:
\[ x^2 - (\alpha + \beta)x + \alpha \beta = 0 \]
Calculating \( \alpha + \beta \) and \( \alpha \beta \):
\[ \alpha + \beta = 7 + 3 = 10 \]
\[ \alpha \beta = 7 \cdot 3 = 21 \]

Thus, the quadratic equation is:
\[ x^2 - 10x + 21 = 0 \]

### Final Answer

The quadratic equation whose roots are \( \lim_{x \to 2^+} f(x) \) and \( \lim_{x \to 2^-} f(x) \) is:
\[ x^2 - 10x + 21 = 0 \]

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