Question

If $f(x)=\frac{x^2-bx+25}{x^2-7x+10}$ for $x\ne 5$ is continuous at $x=5$, then the value of $f(5)$ is

• A. 0
• B. 5
• C. 10
• D. 25

Answer 1

Answer: (A)

$f(x)=\frac{x^2-bx+25}{x^2-7x+10},x\ne 5$
$f(x)$ is continuous at $x=5$, only if $\underset{x\to 5}{\lim }\frac{x^2-bx+25}{x^2-7x+10}$ is finite.
Now $x^2-7x+10\to 0$ when $x\to 5$,
then we must have $x^2-bx$ $+25\to 0$ for which $b=10$
Hence, $\underset{x\to 5}{\lim }\frac{x^2-10x+25}{x^2-7x+10}$
$=\underset{x\to 5}{\lim }\frac{x-5}{x-2}=0$