Question

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

• A. $0$
• B. $1$
• C. $2$
• D. $3$

Answer 1

Answer: (A)

$0$

$f\left(x\right)=\frac{x^2-bx+25}{x^2-7x+10}$

$=\frac{x^2-bx+25}{(x-2)(x-5)}$

$f\left(x\right)$ is continuous at $x=5$

so there should a factor of $(x-5)$ in $x^2-bx+25$

$(x^2-bx+25)=(x-5)(x-5)$

$f\left(x\right)=\frac{(x-5)(x-5)}{(x-2)(x-5)}=\frac{(x-5)}{(x-2)}$

$\Rightarrow f\left(5\right)=0$.