Question

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

• A. $2$
• B. $\frac{7}{3}$
• C. $\frac{24}{11}$
• D. None of these
  

Answer 1

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

The correct options are
A $2$
B $\frac{7}{3}$
C $\frac{24}{11}$

$f\left(x\right)=\frac{1}{x^2+7x+66}$

$f\left(x\right)=\frac{1}{x^2-11x-6x+66}$

$=\frac{1}{x(x-11)-6(x-11)}$

$y=f\left(x\right)=\frac{1}{(x-6)(x-11)}=f\left(t\right)=\frac{1}{(t-6)(t-11)}$

let say $t=\frac{2}{x-2}$

we observe that ${}^{\prime }{t}^{\prime }$ is $n^{\prime }t$ derived for $x=2$ and ${}^{\prime }{y}^{\prime }$ is $n^{\prime }t$ derived for $t=6,t=11$

$\therefore 6\frac{2}{x-2}\Rightarrow 3x-6=1$

$\Rightarrow 3x=7$

$\Rightarrow x=\frac{7}{3}$

$11=\frac{2}{x-2}\Rightarrow 11x-22=2$

$\Rightarrow x=\frac{24}{11}$

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

$\therefore f\left(\frac{2}{x-2}\right)$ is continuous for $x\in R-\{2,\frac{7}{3},\frac{24}{11}\}$

Continuous for $x\in R-\{2,\frac{7}{3},\frac{24}{11}\}$