If f(x) `{:(=(x^(3)+x^2-16x+20)/((x-2)^(2)) ", if " x!= 2 ), (=k ", if " x = 2 ) :}` is continuous at x = 2 , then

Options

A. `k=2`
B. `k=0`
C. `k=20`
D. `k=7`

Answer

To determine the value of \( k \) that makes the function \( f(x) \) continuous at \( x = 2 \), we need to ensure that the limit of \( f(x) \) as \( x \) approaches 2 is equal to \( f(2) \).

### Step-by-Step Solution:

1. **Define the function**:
\[ f(x) = \begin{cases} \frac{x^3 + x^2 - 16x + 20}{(x-2)^2} & \text{if } x \neq 2 \\ k & \text{if } x = 2 \end{cases} \]

2. **Set up the continuity condition**:
For \( f(x) \) to be continuous at \( x = 2 \), we need:
\[ \lim_{x \to 2} f(x) = f(2) = k \]

3. **Calculate the limit**:
We first need to evaluate \( \lim_{x \to 2} \frac{x^3 + x^2 - 16x + 20}{(x-2)^2} \). Plugging \( x = 2 \) directly into the function gives:
\[ \frac{2^3 + 2^2 - 16 \cdot 2 + 20}{(2-2)^2} = \frac{8 + 4 - 32 + 20}{0} = \frac{0}{0} \]
This is an indeterminate form, so we can apply L'Hôpital's Rule.

4. **Apply L'Hôpital's Rule**:
Differentiate the numerator and the denominator:
- Derivative of the numerator \( x^3 + x^2 - 16x + 20 \):
\[ 3x^2 + 2x - 16 \]
- Derivative of the denominator \( (x-2)^2 \):
\[ 2(x-2) \]
Now we have:
\[ \lim_{x \to 2} \frac{3x^2 + 2x - 16}{2(x-2)} \]

5. **Evaluate the limit again**:
Plugging \( x = 2 \) into the new limit:
\[ \frac{3(2^2) + 2(2) - 16}{2(2-2)} = \frac{12 + 4 - 16}{0} = \frac{0}{0} \]
This is still an indeterminate form, so we apply L'Hôpital's Rule again.

6. **Differentiate again**:
- Derivative of the new numerator \( 3x^2 + 2x - 16 \):
\[ 6x + 2 \]
- Derivative of the new denominator \( 2(x-2) \):
\[ 2 \]
Now we have:
\[ \lim_{x \to 2} \frac{6x + 2}{2} \]

7. **Evaluate the limit**:
Plugging \( x = 2 \):
\[ \frac{6(2) + 2}{2} = \frac{12 + 2}{2} = \frac{14}{2} = 7 \]

8. **Set the limit equal to \( k \)**:
Since we found that:
\[ \lim_{x \to 2} f(x) = 7 \]
We set this equal to \( k \):
\[ k = 7 \]

### Final Answer:
Thus, the value of \( k \) that makes \( f(x) \) continuous at \( x = 2 \) is:
\[ \boxed{7} \]