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Let $A$ denote the set of all real numbers $x$ such that $x^3-[x]^3=\left(x-[x]^3\right)$, where $[x]$ is the greatest integer less than or equal to $x$. Then,
$A$ is a discrete set of a least two points
$A$ contains an interval, but is not an interval
$A$ is an interval, but a proper subset of $(-\infty, \infty)$
$A=(-\infty, \infty)$
Solution
(b)
Given, $x^3-[x]^3=(x-[x])^3$
$(x-\mid x])\left(x^2+[x]^2+x[x]\right)=(x-[x])^3$
$x-[x]=0$ or $x[x]=0$
$\{x\}=0$ or $x[x]=0$
$x \in Z$ or $x \in[0,1)$
Hence, solution in $x \in(0, 1 ) \cup Z$.
Similar Questions
Let $\quad E_1=\left\{x \in R : x \neq 1\right.$ and $\left.\frac{x}{x-1}>0\right\}$ and $\quad E_2=\left\{x \in E_1: \sin ^{-1}\left(\log _e\left(\frac{x}{x-1}\right)\right)\right.$ is a real number $\}$.
(Here, the inverse trigonometric function $\sin ^{-1} x$ assumes values in $\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$ )
Let $f : E _1 \rightarrow R$ be the function defined by $f(x)=\log _c\left(\frac{x}{x-1}\right)$ and $g: E_2 \rightarrow R$ be the function defined by $g(x)=\sin ^{-1}\left(\log _e\left(\frac{x}{x-1}\right)\right)$
$LIST I$ | $LIST II$ |
$P$ The range of $f$ is | $1$ $\left(-\infty, \frac{1}{1- e }\right] \cup\left[\frac{ e }{ e -1}, \infty\right)$ |
$Q$ The range of $g$ contains | $2$ $(0,1)$ |
$R$ The domain of $f$ contains | $3$ $\left[-\frac{1}{2}, \frac{1}{2}\right]$ |
$S$ The domain of $g$ is | $4$ $(-\infty, 0) \cup(0, \infty)$ |
$5$ $\left(-\infty, \frac{ e }{ e -1}\right]$ | |
$6$ $(-\infty, 0) \cup\left(\frac{1}{2}, \frac{ e }{ e -1}\right]$ |
The correct option is: