Question

Let the function $f\left(x\right)=(\frac{|x{|}^3}{a}-{\left[\frac{x}{a}\right]}^3),a>0.$ Then
(where $[.]$ denotes the greatest integer function)

• A. $f\left(x\right)$ is discontinuous at $x=a$
• B. $f\left(x\right)$ is continuous at $x=a$
• C. $\underset{x\to a^{-}}{lim}f\left(x\right)=a^2$
• D. $\underset{x\to a^{+}}{lim}f\left(x\right)=a^2-1$

Answer 1

Answer: (A), (C), (D)

$\underset{x\to a^{+}}{lim}f\left(x\right)=a^2-1$

Given : $f\left(x\right)=(\frac{|x{|}^3}{a}-{\left[\frac{x}{a}\right]}^3),a>0$

$R.H.L.=\underset{x\to a^{+}}{lim}\{\frac{|x{|}^3}{a}-{\left[\frac{x}{a}\right]}^3\}$
$=\underset{h\to 0}{lim}\{\frac{|a+h{|}^3}{a}-{\left[\frac{a+h}{a}\right]}^3\}$
$=\frac{a^3}{a}-1=a^2-1$
$L.H.L.=\underset{x\to a^{-}}{lim}\{\frac{|x{|}^3}{a}-{\left[\frac{x}{a}\right]}^3\}$
$=\underset{h\to 0}{lim}\{\frac{|a-h{|}^3}{a}-{\left[\frac{a-h}{a}\right]}^3\}$
$=\frac{a^3}{a}-0=a^2$

Clearly, $L.H.L.\ne R.H.L.$
$\therefore$ $f\left(x\right)$ is discontinuous at $x=a$