Question

The function f(x) defined by $f\left(x\right)=\{\begin{array}{cc}\log 4x-3(x^2-2x+5),& \frac{3}{4}<x<1andx>1\\ 4,& x=1\end{array}$

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

Answer 1

The correct option is D is discontinuous at $x=1$ since neither $f\left(1^{+}\right)$ nor $f\left(1^{-}\right)$ exists.

We have $\underset{x\to 1^{-}}{lim}f\left(x\right)=\underset{h\to 0}{lim}f(1-h)$

$=\underset{h\to 0}{lim}\frac{log(4+g^2)}{log(1-4h)}=\infty$

and $\underset{x\to 1^{+}}{lim}f\left(x\right)=\underset{h\to 0}{lim}f(1+h)=\underset{h\to 0}{lim}\frac{log(4+h^2)}{log+4h}=\infty$

So, $f\left(1^{-}\right)$ and $f\left(1^{+}\right)$ does not exist.