If f(x) = \( \frac{x^2 - bx + 25}{x^2 - 7x + 10} \) for x \neq 5 and f is continuous at x = 5, then f(5) has the value equal to-
Step-by-Step Solution
Key Concept: For f to be continuous at x = 5, we need lim(x→5) f(x) = f(5). Since the denominator equals zero at x = 5, the numerator must also equal zero for the limit to exist (removable discontinuity). This allows us to find b, then compute the limit.
**Step 1:** Factor the denominator.
$$x^2 - 7x + 10 = (x - 5)(x - 2)$$
**Step 2:** For $f(x)$ to be continuous at $x = 5$, the limit $\lim_{x \to 5} f(x)$ must exist. Since the denominator is zero at $x = 5$, the numerator must also be zero at $x = 5$ for a removable discontinuity to exist.
Substitute $x = 5$ into the numerator:
$$5^2 - b(5) + 25 = 0$$
$$25 - 5b + 25 = 0$$
$$50 - 5b = 0$$
$$5b = 50$$
$$b = 10$$
**Step 3:** Substitute $b = 10$ into the numerator.
The numerator becomes:
$$x^2 - 10x + 25 = (x - 5)^2$$
**Step 4:** Simplify $f(x)$ for $x \neq 5$.
$$f(x) = \frac{(x - 5)^2}{(x - 5)(x - 2)} = \frac{x - 5}{x - 2}$$
**Step 5:** Since $f$ is continuous at $x = 5$, $f(5)$ is defined as the limit of $f(x)$ as $x \to 5$.
$$f(5) = \lim_{x \to 5} f(x) = \lim_{x \to 5} \frac{x - 5}{x - 2}$$
Substitute $x = 5$ into the simplified expression:
$$f(5) = \frac{5 - 5}{5 - 2} = \frac{0}{3} = 0$$
Correct Answer: B