Question

The function $f\left(x\right)$ is defined by $f\left(x\right)=\{\begin{array}{c}{\log }_{(4x-3)}(x^2-2x+5),if\frac{3}{4}<x<1orx>1\\ 4ifx=1\end{array}$

• A. is continuous at $x=1$
• B. is discontinuous at $x=1$ since $f\left(1^{+}\right)$ does not exist through $f\left(1^{-}\right)$ exists
• C. is discontinuous at $x=1$ since $f\left(1^{-}\right)$ does not exist through $f\left(1^{+}\right)$ exists
• D. is discontinuous at since neither $f\left(1^{-}\right)$ nor $f\left(1^{+}\right)$ exists.

Answer 1

Answer: (B)

is discontinuous at $x=1$ since $f\left(1^{+}\right)$ does not exist through $f\left(1^{-}\right)$ exists

$\underset{x\to 1^{+}}{lim}f\left(x\right)=\underset{x\to 1}{lim}{\log }_{\left(4x-3\right)}\left(x^2-2x+5\right)$

$={\log }_1(1^2-2+5)$

$={\log }_{1}4$

$=\frac{1}{{\log }_{4}1}$

$=\frac{1}{0}$

$=\infty$

$\underset{x\to 1^{-}}{lim}f\left(x\right)=\underset{x\to 1}{lim}{\log }_{\left(4x-3\right)}\left(x^2-2x+5\right)$

$={\log }_1(1^2-2+5)$

$={\log }_{1}4$

$=\frac{1}{{\log }_{4}1}$

$=\frac{1}{0}$

$=\infty$

and $f\left(1\right)=4$

So, $f\left(x\right)$ is discontinuous at $x=1$ since $f\left(1^{+}\right)$ and $f\left(1^{-}\right)$ doesnot exist.