Let f(x) = [x] ( [x + 1] ), where [.] denotes the greatest intege… - Vidyadip

MCQ

Let $f(x) = [x]\sin \left( {\frac{\pi }{{[x + 1]}}} \right)$, where $[.]$ denotes the greatest integer function. The domain of $f$ is ….and the points of discontinuity of $f$ in the domain are

A
$\left\{ {x \in R|x \in [ - 1,\;0)} \right\},\;\;I - \{ 0\} $
B
$\left\{ {x \in R|x \notin [1,\;0)} \right\},\;\;I - \{ 0\} $
✓
$\left\{ {x \in R|x \notin [ - 1,\;0)} \right\},\;\;I - \{ 0\} $
D
None of these

✓

Answer

Correct option: C.

$\left\{ {x \in R|x \notin [ - 1,\;0)} \right\},\;\;I - \{ 0\} $

c
(c) Note that $[x + 1] = 0$, if $0 \le x + 1 < 1$
i.e., $[x + 1] = 0$ if $ - 1 \le x < 0.$
Thus domain of f is $R - [ - 1,\,\,0) = \left\{ {x \in R|x \notin {\rm{[--1,}}\,{\rm{0)}}} \right\}$
We have $\sin \,\left( {\frac{\pi }{{[x + 1]}}} \right)$ is continuous at all points of $R - [ - 1,\,\,0)$ and $[x]$ is continuous on $R - I,$ where I denotes the set of integers.
Thus the points where $f$ can possibly be discontinuous are ………..,$ -3, -2, -1, 0, 1, 2,$ …… But for $0 \le x < 1$, $[x] = 0$ and $\sin \left( {\frac{\pi }{{[x + 1]}}} \right)$ is defined.
Therefore $f(x) = 0$ for $0 \le x < 0$. Also $f(x)$ is not defined on $ - 1 \le x < 0$.
Therefore continuity of $f$ at $O$ means continuity of $f$ from right at $O$. Since $f$ is continuous from right at $O$, $f$ is continuous at $O$. Hence $f$ set of points of discontinuities of $f$ is $I - \{ O\} $

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