If [x] denotes the greatest integer leq x,the | 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 denominator

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.$1$. Denominator condition: $[x] + 2 
eq 0 \Rightarrow [x] 
eq -2$. This implies $x 
otin [-2, -1)$.$2$. Inequality condition: $\frac{4-x^2}{[x]+2} \geq 0$.Case $I$: If $[x] + 2 > 0$,then $[x] > -2$,which means $x \geq -1$. Then $4 - x^2 \geq 0$ $\Rightarrow x^2 \leq 4$ $\Rightarrow x \in [-2, 2]$. Combining $x \geq -1$ and $x \in [-2, 2]$,we get $x \in [-1, 2]$.Case $II$: If $[x] + 2 Then $4 - x^2 \leq 0$ $\Rightarrow x^2 \geq 4$ $\Rightarrow x \in (-\infty, -2] \cup [2, \infty)$. Combining $x Combining both cases,the domain is $(-\infty, -2) \cup [-1, 2]$.

If $[x]$ denotes the greatest integer $\leq x$,then the domain of the function $f(x)=\sqrt{\frac{4-x^2}{[x]+2}}$ is

Similar Questions

Let $R-(\alpha, \beta)$ be the range of $f(x) = \frac{x+3}{(x-1)(x+2)}$. Then,the sum of the intercepts of the line $\alpha x + \beta y + 1 = 0$ on the coordinate axes is:

If $n$ is an integer,the domain of the function $\sqrt{\sin 2x}$ is

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

The function is defined as $f(x) = (1 + \frac{1}{x})^x$. What is the range of the function $f(x)$?

Domain of $f(x) = \frac{x}{1-|x|}$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module