The domain of the definition of the function | vedclass

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

Question

(A) For the function $f(x) = \sqrt{\frac{4 - x^2}{[x] + 2}}$ to be defined,the expression inside the square root must be non-negative and the denomina

Vedclass · Accepted Answer

$(-\infty, -2) \cup [-1, 2]$

Vedclass · Answer

(A) For the function $f(x) = \sqrt{\frac{4 - x^2}{[x] + 2}}$ to be defined,the expression inside the square root must be non-negative and the denominator must not be zero.Case $1$: $\frac{4 - x^2}{[x] + 2} \geq 0$ and $[x] + 2 > 0$.This implies $4 - x^2 \geq 0$ and $[x] > -2$.$x^2 \leq 4 \Rightarrow x \in [-2, 2]$.$[x] > -2 \Rightarrow x \geq -1$.Intersection: $x \in [-1, 2]$.Case $2$: $\frac{4 - x^2}{[x] + 2} \geq 0$ and $[x] + 2 This implies $4 - x^2 \leq 0$ and $[x] $x^2 \geq 4 \Rightarrow x \in (-\infty, -2] \cup [2, \infty)$.$[x] Intersection: $x \in (-\infty, -2)$.Combining both cases,the domain is $x \in (-\infty, -2) \cup [-1, 2]$.

The domain of the definition of the function $f(x) = \sqrt{\frac{4 - x^2}{[x] + 2}}$ is (where $[.] \rightarrow \text{G.I.F.}$)

Similar Questions

The domain of the function $f(x) = \sqrt{x - x^2} + \sqrt{4 + x} + \sqrt{4 - x}$ is

If function $f : R \to S, f(x) = (\sin x - \sqrt{3} \cos x + 1)$ is onto,then $S$ is equal to

The domain of the function $f(x) = \log |\log x|$ is:

The number of integral values of $\lambda$ for which the function $f(x) = \sqrt{\ln(2\lambda \cos x + 5)}$ is defined for all $x \in R$ is:

The domain of the function $f(x) = \sin^{-1}\left[\log_4\left(\frac{x}{4}\right)\right] + \sqrt{17x - x^2 - 16}$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module