The domain of the function f(x) = frac{1}{sqr | vedclass

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

Question

(A) For the function $f(x) = \frac{1}{\sqrt{[x]^2 - [x] - 6}}$ to be defined,the expression inside the square root must be strictly positive: $[x]^2 -

Vedclass · Accepted Answer

$(-\infty, -2) \cup [4, \infty)$

Vedclass · Answer

(A) For the function $f(x) = \frac{1}{\sqrt{[x]^2 - [x] - 6}}$ to be defined,the expression inside the square root must be strictly positive: $[x]^2 - [x] - 6 > 0$ Factoring the quadratic expression: $([x] - 3)([x] + 2) > 0$ This inequality holds when: $[x] > 3$ or $[x] If $[x] > 3$,then the smallest integer value for $[x]$ is $4$,which implies $x \geq 4$,or $x \in [4, \infty)$. If $[x] Combining these,the domain is $x \in (-\infty, -2) \cup [4, \infty)$.

The domain of the function $f(x) = \frac{1}{\sqrt{[x]^2 - [x] - 6}}$,where $[x]$ is the greatest integer function $\leq x$,is:

Similar Questions

The domain of the real-valued function $f(x) = \sqrt[3]{\frac{x-2}{2x^2-7x+5}} + \log(x^2-x-2)$ is

If the domain of the function $f(x) = \sqrt{\ln(m\sin x + 4)}$ is $R$,then the number of possible integral values of $m$ is:

If the function $f: R-\{-1, 1\} \rightarrow A$ defined by $f(x) = \frac{x^2}{1-x^2}$ is surjective,then $A$ is equal to

If $f(x) = \frac{2x-3}{(x-2)(x-3)}$ is a real-valued function, then the value that $f(x)$ does not take is:

The range of the function $f(x) = \begin{cases} 4x - 1, & x > 3 \\ x^2 - 2, & -2 \leq x \leq 3 \\ 3x + 4, & x < -2 \end{cases}$ is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module