The domain of the function f(x) = sqrt{log_{0 | vedclass

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

Question

(D) We have $f(x) = \sqrt{\log_{0.5} x!}$.For $f(x)$ to be defined,the expression inside the square root must be non-negative:$\log_{0.5} x! \geq 0$.S

Vedclass · Accepted Answer

$\{0, 1\}$

Vedclass · Answer

(D) We have $f(x) = \sqrt{\log_{0.5} x!}$.For $f(x)$ to be defined,the expression inside the square root must be non-negative:$\log_{0.5} x! \geq 0$.Since the base of the logarithm is $0.5$ (which is between $0$ and $1$),the inequality sign reverses when we remove the logarithm:$x! \leq (0.5)^0$.$x! \leq 1$.The factorial $x!$ is defined for non-negative integers $x$. We check the values:For $x = 0$,$0! = 1 \leq 1$ (True).For $x = 1$,$1! = 1 \leq 1$ (True).For $x = 2$,$2! = 2 
ot\leq 1$ (False).Thus,the domain is $\{0, 1\}$.

The domain of the function $f(x) = \sqrt{\log_{0.5} x!}$ is

Similar Questions

Let $a > 1$ be a constant. If $f: A \rightarrow A$ and $(x, y) \in f$ satisfy $a^x + a^y = a$,then $A =$

Domain of $y = \sqrt{\log _{10} \frac{3x - x^2}{2}}$ is

Domain of the function $f$,given by $f(x) = \frac{1}{\sqrt{(x - 2)(x - 5)}}$ is

If the equation $\frac{1}{x} + \frac{1}{x - 1} + \frac{1}{x - 2} = 3x^3$ has $k$ real roots,then $k$ is equal to -

The range of the function $f(x)=\sin [x]$,where $-\frac{\pi}{4} < x < \frac{\pi}{4}$ and $[x]$ denotes the greatest integer $\leq x$,is

Vedclass Test Series

Exam Paper Generator

Online Exam Module