The range of the function f(x) = frac{1}{sqrt | vedclass

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

Question

(A) The function is defined as $f(x) = \frac{1}{\sqrt{x-[x]}}$.Domain of $f$:We know that $0 \leq x-[x] < 1$ for all $x \in \mathbb{R}$.Also,$x-[x] =

Vedclass · Accepted Answer

$(1, \infty)$

Vedclass · Answer

(A) The function is defined as $f(x) = \frac{1}{\sqrt{x-[x]}}$.Domain of $f$:We know that $0 \leq x-[x] Also,$x-[x] = 0$ when $x \in \mathbb{Z}$.Since the denominator cannot be zero,$x-[x] > 0$,which implies $x 
otin \mathbb{Z}$.Thus,the domain is $\mathbb{R} - \mathbb{Z}$.Range of $f$:For $x \in \mathbb{R} - \mathbb{Z}$,we have $0 Taking the square root,$0 Taking the reciprocal,$\frac{1}{\sqrt{x-[x]}} > 1$.Therefore,$f(x) \in (1, \infty)$.Hence,the range of $f$ is $(1, \infty)$.

The range of the function $f(x) = \frac{1}{\sqrt{x-[x]}}$ is

Similar Questions

If $f:[2, \infty) \rightarrow B$ defined by $f(x)=x^2-4x+5$ is a bijection, then $B$ is equal to

The domain of definition of the function $f(x)$ given by the equation $2^x + 2^y = 2$ is

The domain of the function $\log _{10}(x^2-5x+6)$ is

If the domain of the greatest integer function is the set of real numbers,then the range will be the set of

The range of the function $f(x) = (\sin x)^{\sin x}$ defined on $(0, \pi)$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module