If [x] denotes the greatest integer leq x, th | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) The expression $x-[x]$ represents the fractional part of $x$, denoted as $\{x\}$.Since $x$ is a real number, the fractional part $\{x\}$ lies in t

Vedclass · Accepted Answer

$(1, \infty)$

Vedclass · Answer

(C) The expression $x-[x]$ represents the fractional part of $x$, denoted as $\{x\}$.Since $x$ is a real number, the fractional part $\{x\}$ lies in the interval $[0, 1)$.However, the denominator $\sqrt{x-[x]}$ cannot be zero, so $x-[x] 
eq 0$.Therefore, $x-[x] \in (0, 1)$.As $x-[x]$ approaches $0$ from the right, $\frac{1}{\sqrt{x-[x]}}$ approaches $\infty$.As $x-[x]$ approaches $1$ from the left, $\frac{1}{\sqrt{x-[x]}}$ approaches $1$.Thus, the range of $f(x)$ is $(1, \infty)$.

If $[x]$ denotes the greatest integer $\leq x$, then the range of the real-valued function $f(x) = \frac{1}{\sqrt{x-[x]}}$ is

Similar Questions

Let $f(x) = \cos(\pi(|x| + 2[x]))$,where $[.]$ represents the greatest integer function. Then:

The domain of the real valued function $f(x) = \frac{1}{\sqrt{\log_{0.5}(2x - 3)}} + \sqrt{4 - 9x^2}$ is:

The largest possible set of real numbers which can be the domain of $f(x) = \sqrt{1 - \frac{1}{x}}$ is

If $[x]^2-5[x]+6=0$,where $[\cdot]$ denotes the greatest integer function,then $x$ belongs to:

If $e^x + e^{f(x)} = e$, then the domain of $f(x)$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module