If $f(x) = \begin{cases}
{x^{2}-1}, & 0

Q: If $f(x) = \begin{cases} {x^{2}-1}, & 0 < x < 2 \\ 2x+3, & 2 \le x < 3 \end{cases}$ then the quadratic equation whose roots are $lim_{x\rightarrow2^{-}}f(x)$ and $lim_{x\rightarrow2^{+}}f(x)$ is:

To determine the quadratic equation whose roots are $ \lim_{x \rightarrow 2^{-}} f(x) $ and $ \lim_{x \rightarrow 2^{+}} f(x) $, we need to evaluate the limits of the piecewise function $ f(x) $ as $ x $ approaches 2 from the left and from the right. 1. **Evaluate $ \lim_{x \rightarrow 2^{-}} f(x) $:** For $ 0 < x < 2 $, the function is defined as $ f(x) = x^2 - 1 $. Therefore, as $ x $ approaches 2 from the left, we have: \[ \lim_{x \rightarrow 2^{-}} f(x) = \lim_{x \rightarrow 2^{-}} (x^2 - 1) = 2^2 - 1 = 4 - 1 = 3 \] 2. **Evaluate $ \lim_{x \rightarrow 2^{+}} f(x) $:** For $ 2 \le x < 3 $, the function is defined as $ f(x) = 2x + 3 $. Therefore, as $ x $ approaches 2 from the right, we have: \[ \lim_{x \rightarrow 2^{+}} f(x) = \lim_{x \rightarrow 2^{+}} (2x + 3) = 2(2) + 3 = 4 + 3 = 7 \] The roots of the quadratic equation are $ \lim_{x \rightarrow 2^{-}} f(x) = 3 $ and $ \lim_{x \rightarrow 2^{+}} f(x) = 7 $. 3. **Form the quadratic equation:** A quadratic equation with roots $ a $ and $ b $ can be expressed as: \[ (x - a)(x - b) = 0 \] Substituting the roots 3 and 7, we have: \[ (x - 3)(x - 7) = 0 \] Expanding this, we get: \[ x^2 - 7x - 3x + 21 = x^2 - 10x + 21 = 0 \] Thus, the quadratic equation whose roots are the limits is $ x^2 - 10x + 21 = 0 $, which corresponds to Option 4. 4. **Explanation of why other options are incorrect:** - **Option 1: $ x^2 - 6x + 9 = 0 $** has roots 3 and 3, which does not match the roots 3 and 7. - **Option 2: $ x^2 - 7x + 8 = 0 $** has roots 4 and 2, which does not match the roots 3 and 7. - **Option 3: $ x^2 - 14x + 49 = 0 $** has roots 7 and 7, which does not match the roots 3 and 7. Therefore, Option 4 is the correct choice, as it correctly represents the quadratic equation with roots 3 and 7.

Question

If $f(x) = \begin{cases}
{x^{2}-1}, & 0 < x < 2 \
2x+3, & 2 \le x < 3
\end{cases}$ then the quadratic equation whose roots are $lim_{xightarrow2^{-}}f(x)$ and $lim_{xightarrow2^{+}}f(x)$ is:

Accepted Answer

To determine the quadratic equation whose roots are $ \lim_{x \rightarrow 2^{-}} f(x) $ and $ \lim_{x \rightarrow 2^{+}} f(x) $, we need to evaluate the limits of the piecewise function $ f(x) $ as $ x $ approaches 2 from the left and from the right.

1. **Evaluate $ \lim_{x \rightarrow 2^{-}} f(x) $:**

For $ 0 < x < 2 $, the function is defined as $ f(x) = x^2 - 1 $. Therefore, as $ x $ approaches 2 from the left, we have:
   $$
   \lim_{x \rightarrow 2^{-}} f(x) = \lim_{x \rightarrow 2^{-}} (x^2 - 1) = 2^2 - 1 = 4 - 1 = 3
   $$

2. **Evaluate $ \lim_{x \rightarrow 2^{+}} f(x) $:**

For $ 2 \le x < 3 $, the function is defined as $ f(x) = 2x + 3 $. Therefore, as $ x $ approaches 2 from the right, we have:
   $$
   \lim_{x \rightarrow 2^{+}} f(x) = \lim_{x \rightarrow 2^{+}} (2x + 3) = 2(2) + 3 = 4 + 3 = 7
   $$

The roots of the quadratic equation are $ \lim_{x \rightarrow 2^{-}} f(x) = 3 $ and $ \lim_{x \rightarrow 2^{+}} f(x) = 7 $.

3. **Form the quadratic equation:**

A quadratic equation with roots $ a $ and $ b $ can be expressed as:
   $$
   (x - a)(x - b) = 0
   $$
   Substituting the roots 3 and 7, we have:
   $$
   (x - 3)(x - 7) = 0
   $$
   Expanding this, we get:
   $$
   x^2 - 7x - 3x + 21 = x^2 - 10x + 21 = 0
   $$

Thus, the quadratic equation whose roots are the limits is $ x^2 - 10x + 21 = 0 $, which corresponds to Option 4.

4. **Explanation of why other options are incorrect:**

- **Option 1: $ x^2 - 6x + 9 = 0 $** has roots 3 and 3, which does not match the roots 3 and 7.
   - **Option 2: $ x^2 - 7x + 8 = 0 $** has roots 4 and 2, which does not match the roots 3 and 7.
   - **Option 3: $ x^2 - 14x + 49 = 0 $** has roots 7 and 7, which does not match the roots 3 and 7.

Therefore, Option 4 is the correct choice, as it correctly represents the quadratic equation with roots 3 and 7.