I posted this problem before this .I have satisfied explaination given by markus-scheuer sir and siminore sir . I also found here .

I have read the Wikipedia posts for continuous function and bounded function and I have concluded that a function should be defined on given range for satisfying conditions of continuity and bounded function .

I posted my answer at GateOverflow(a site old Q/A of GATE ) , but they not accepted my answer and they have given reason answer key is correct , but not my explanation .

Let $f(x)=x^{-(1/3)}$ and $A$ denote the area of region bounded by $f(x)$ and the $X-$axis, when $x$ varies from $-1$ to $1$. Which of the following statements is/are TRUE?

  1. $f$ is continuous in $[-1, 1]$
  2. $f$ is not bounded in $[-1, 1]$
  3. $A$ is nonzero and finite

Given answer key of this question $:$

  1. False
  2. True
  3. True

In my opinion all given statement should be false .

My doubts are $:$

  1. If a function is not continuous then is it possible for bounded function in given range ? What about function in given problem .
  2. Is it possible bounded area zero or infinite ? What about statement $(3)$ of problem ?
  • $\begingroup$ Sorry but I think this question would do better if it was rewritten. Getting rid of everything above 'original problem is', and rewrite the question with $\LaTeX$. $\endgroup$ – user233746 Oct 24 '15 at 13:18
  • $\begingroup$ Done , please suggest me , if anything more . $\endgroup$ – 1 0 Oct 24 '15 at 13:23
  • $\begingroup$ The leading section is necessary because of the links. It is not clear how this question differs from the previous version, what are the new problems or misunderstandings. The given answer was comprehensive, with answers for alternative interpretations of the question. As it is, this is a duplicate question. $\endgroup$ – Dr. Lutz Lehmann Oct 24 '15 at 13:23
  • 1
    $\begingroup$ @Siminore: The continuity of $f(x)=\sin(1/x)$ depends on the domain of $f$. If the domain of $f$ is specified e.g. as $(0,\infty)$, the function is continuous at each point of its domain and therefore continuous. $\endgroup$ – Markus Scheuer Oct 24 '15 at 22:20
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    $\begingroup$ @MarkusScheuer sir , yes . I support this. $\endgroup$ – 1 0 Oct 25 '15 at 6:11

1) A non continuous function can be either bounded (e.g. step function, see $g(x)$ below) and unbounded (e.g. the function in the question).

$$g(x)=\begin{cases} 1& x>0\\-1 & x\le 0 \end{cases} $$ 2) If $f$ is not bounded, how does it possible that the area bounded between $f$ and the X-axis is finite? Very roughly, area is height times width, $f(x)$ is the height here, and the width is a segment of $x$. The segment is finite, but since $f(x)$ is unbounded it is not, and therefore the area cannot be finite.

In addition, if the length of the segment (the width) is not zero and the function (the height) is not zero everywhere, the area cannot be zero as a product of two non zero numbers.


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