If `f(x)=(x^2-bx+25)/(x^2-7x+10)` for `x!=5` and `f` is continuous at `x=5` then `f(5)=`

Options

A. 0
B. 5
C. 10
D. 25

Answer

To solve the problem, we need to ensure that the function \( f(x) = \frac{x^2 - bx + 25}{x^2 - 7x + 10} \) is continuous at \( x = 5 \). This means that the limit of \( f(x) \) as \( x \) approaches 5 must equal \( f(5) \).

### Step-by-Step Solution:

1. **Identify the function and the point of continuity**:
We have:
\[ f(x) = \frac{x^2 - bx + 25}{x^2 - 7x + 10} \]
We need to check the continuity at \( x = 5 \).

2. **Find the limit as \( x \) approaches 5**:
We need to compute:
\[ \lim_{x \to 5} f(x) = \lim_{x \to 5} \frac{x^2 - bx + 25}{x^2 - 7x + 10} \]

3. **Factor the denominator**:
The denominator can be factored as:
\[ x^2 - 7x + 10 = (x - 2)(x - 5) \]
Thus, we rewrite \( f(x) \):
\[ f(x) = \frac{x^2 - bx + 25}{(x - 2)(x - 5)} \]

4. **Evaluate the denominator at \( x = 5 \)**:
At \( x = 5 \), the denominator becomes:
\[ (5 - 2)(5 - 5) = 3 \cdot 0 = 0 \]
Since the denominator is zero, the numerator must also be zero for the limit to exist.

5. **Set the numerator to zero at \( x = 5 \)**:
We need:
\[ 5^2 - 5b + 25 = 0 \]
Simplifying this gives:
\[ 25 - 5b + 25 = 0 \implies 50 - 5b = 0 \implies 5b = 50 \implies b = 10 \]

6. **Substituting \( b \) back into the function**:
Now we substitute \( b = 10 \) into the function:
\[ f(x) = \frac{x^2 - 10x + 25}{(x - 2)(x - 5)} \]
The numerator can be factored as:
\[ x^2 - 10x + 25 = (x - 5)^2 \]
Thus, we have:
\[ f(x) = \frac{(x - 5)^2}{(x - 2)(x - 5)} \]

7. **Cancel the common factor**:
Canceling \( (x - 5) \) from the numerator and denominator gives:
\[ f(x) = \frac{x - 5}{x - 2} \quad \text{for } x \neq 5 \]

8. **Evaluate the limit as \( x \) approaches 5**:
Now we can find the limit:
\[ \lim_{x \to 5} f(x) = \lim_{x \to 5} \frac{x - 5}{x - 2} = \frac{5 - 5}{5 - 2} = \frac{0}{3} = 0 \]

9. **Conclusion**:
Since \( f \) is continuous at \( x = 5 \), we have:
\[ f(5) = 0 \]

### Final Answer:
\[ f(5) = 0 \]