Question

Let $f\left(x\right)=\{\begin{array}{c}\frac{2^x+2^{3-x}-6}{\sqrt{2^{-2}}-2^{1-x}};x>2\\ \\ \frac{x^2-4}{x-\sqrt{3x-2}};x<2\end{array}$

then which of the following is correct?

• A. $f\left(2\right)=8\Rightarrow f$ is continuous at $x=2$
• B. $f\left(2\right)=16\Rightarrow f$ is continuous at $x=2$
• C. $f\left(2^{-}\right)\ne f\left(2^{+}\right)\Rightarrow f$ is discontinuous
• D. $f$ has removable discontinuity at $x=2$

Answer 1

The correct option is C $f\left(2^{-}\right)\ne f\left(2^{+}\right)\Rightarrow f$ is discontinuous

$LHL={lim}_{x\to 2^{-}}f\left(x\right)$

$=\underset{x\to 2^{-}}{lim}\frac{x^2-4}{x-\sqrt{3x-2}}$

$=\underset{x\to 2^{-}}{lim}\frac{(x^2-4)(x+\sqrt{3x-2})}{x^2-3x+2}$

$=\underset{x\to 2^{-}}{lim}\frac{(x-2)(x+2)(x+\sqrt{3x-2})}{(x-2\left)\right(x-1)}$

$=\underset{x\to 2^{-}}{lim}\frac{(x+2)(x+\sqrt{3x-2})}{(x-1)}=16$

$RHL={lim}_{x\to 2^{+}}f\left(x\right)$

$=\underset{x\to 2^{+}}{lim}\frac{2^x+2^{3-x}-6}{\sqrt{2^{-2}}-2^{1-x}}\left(\frac{0}{0}\right)$

$=\underset{x\to 2^{+}}{lim}\frac{2^x\log 2-2^{3-x}\log 2}{2^{1-x}\log 2}$

$={lim}_{x\to 2^{+}}\frac{2^x-2^{3-x}}{2^{1-x}}=8$

Since $LHL\ne RHL$

Hence, $f\left(x\right)$ is discontinuous at $x=2$