Question

If $f\left(x\right)=\frac{1}{x^2-17x+66}$, then $f\left(\frac{2}{x-2}\right)$ is discontinuous at $x=$

• A. $2$
• B. $\frac{7}{3}$
• C. $\frac{24}{11}$
• D. $6,11$

Answer 1

Answer: (A), (B), (C)

$\frac{7}{3}$
$\frac{24}{11}$
$2$

$f\left(x\right)=\frac{1}{x^2-17x+66}$

$f\left(\frac{2}{x-2}\right)=\frac{1}{(\frac{2}{x-2}{)}^2-17(\frac{2}{x-2})+66}$

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

$g\left(x\right)$ is discontinuous at $x=2$

$g\left(x\right)=\frac{x^2-4x+4}{4-34(x-2)+66(x-2{)}^2}$

$g\left(x\right)$ is discontinuous when

$4-34(x-2)+66(x-2{)}^2$

$4-34x+68+66(x^2-4x+4)=0$

$66x^2-298x+336=0$

$x=\frac{7}{3},\frac{24}{11}$

So $f\left(\frac{2}{x-2}\right)$ is discontinuous at $x=2,\frac{7}{3},\frac{24}{11}$