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Let the function f(x)=\left[x^2\left[\frac{1}{x^2}\right]\right], x \neq 0 is, ([x] represents the greatest integer \leq x )

 

 

Option: 1

 continuous \forall x \in R
 


Option: 2

 continuous at x=-1
 


Option: 3

 continuous at x=1
 


Option: 4

discontinuous at infinitely many points


Answers (1)

best_answer

Given,\mathrm{ f(x)=\left[x^2\left[\frac{1}{x^2}\right]\right], x \neq 0 }
Let x=1,-1
\mathrm{\therefore f(1)=f(-1)=1 }
Again \mathrm{x<-1\, then \, \, x^2>1 \Rightarrow 0<\frac{1}{x^2}<1 \Rightarrow f(x)=0 }
and \mathrm{x>1 then x^2>1 \Rightarrow 0<\frac{1}{x^2}<1 \Rightarrow f(x)=0 }
\mathrm{\therefore \quad f(x)= \begin{cases}1, & x= \pm 1 \\ 0, & |x|>1 \text { i.e. } x<-1, x>1\end{cases} }

\mathrm{\therefore } f(x) can't be continuous at x=1 & x=-1
Again, let \mathrm{\frac{1}{2}<x^2<1 \Rightarrow 1<\frac{1}{x^2}<2 }

\mathrm{\begin{aligned} & \Rightarrow\left[\frac{1}{x^2}\right]=1 \Rightarrow \frac{1}{2}<x^2\left[\frac{1}{x^2}\right]<1 \Rightarrow\left[x^2\left[\frac{1}{x^2}\right]\right]=0 \\ & \Rightarrow f(x)=0 \text { if } x \in\left(-1, \frac{1}{\sqrt{2}}\right) \cup\left(\frac{1}{\sqrt{2}}, 1\right) \end{aligned} }

Again, \mathrm{\frac{1}{3}<x^2<\frac{1}{2} \Rightarrow 2<\frac{1}{x^2}<3 }
\mathrm{\begin{aligned} & \Rightarrow\left[\frac{1}{x^2}\right]=2 \Rightarrow \frac{2}{3}<x^2\left[\frac{1}{x^2}\right]<1 \Rightarrow x^2\left[\frac{1}{x^2}\right]=0 \\ & \Rightarrow f(x)=0, \text { if } x \in\left(-\frac{1}{\sqrt{2}},-\frac{1}{\sqrt{3}}\right) \cup\left(\frac{1}{\sqrt{3}}, \frac{1}{\sqrt{2}}\right) \end{aligned} }
At
 \mathrm{x= \pm \frac{1}{\sqrt{2}} \Rightarrow x^2=\frac{1}{2} \& \frac{1}{x^2}=2 }
we have f(x)=1
Similarly, at different values of x, f(x) can be calculated.
\therefore f(x) is discontinuous at infinite number of points given
by \mathrm{x \in\left\{ \pm \frac{1}{\sqrt{n}}, n \in N\right\} }
Thus from above f(x) is also discontinuous at x=-1 as well as the infinite number of value of x.

Posted by

Kuldeep Maurya

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