Question

lf $f\left(x\right)=\{\begin{array}{c}\frac{a^{2\left[x\right]+\left\{x\right\}}-1}{2\left[x\right]+\left\{x\right\}};x\ne 0\\ \log a;x=0\end{array}$ where $[.]$ and $\{.\}$ denote integral and fractional part respectively, then

• A. $f\left(x\right)$ is continuous at $x=0$
• B. $f\left(x\right)$ is discontinuous at $x=0$
• C. $f\left(x\right)$ is continuous $\forall x\in R$
• D. $f\left(x\right)$ is differentiable at $x=0$

Answer 1

The correct option is B $f\left(x\right)$ is discontinuous at $x=0$

Given definition of $f\left(x\right)$ can be written as
$f\left(x\right)=\{\begin{array}{c}\frac{a^{\left[x\right]+x}}{\left[x\right]+x};x\ne 0\\ \log a;x=0\end{array}$ ($\because \left\{x\right\}=x-\left[x\right]$)

To check continuity of $f\left(x\right)$ at $x=0$
$f\left(0^{+}\right)=f\left(0^{-}\right)=f\left(0\right)$
$f\left(0^{+}\right)={lim}_{h\to 0}\frac{a^{h+\left[h\right]}}{\left[h\right]+h}$
$={lim}_{h\to 0}\frac{a^h-1}{h}=\log \left(a\right)$
$f\left(0^{-}\right)={lim}_{h\to 0}\frac{a^{[-h]-h}}{[-h]-h}$
$={lim}_{h\to 0}\frac{a^{-1-h}-1}{-1-h}$
$=1-\frac{1}{a}$
$f\left(0^{+}\right)\ne f\left(0^{-}\right)$
So, $f\left(x\right)$ is discontinuous at $x=0$