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

For the function $f(x) = \left[ \frac{1}{[x]} \right]$, where $[x]$ denotes the greatest integer less than or equal to $x$, which of the following statements is true?

If the domain of the function $f(x) = \log_e(4x^2 + 11x + 6) + \sin^{-1}(4x + 3) + \cos^{-1}\left(\frac{10x + 6}{3}\right)$ is $(\alpha, \beta]$,then $36|\alpha + \beta|$ is equal to:

The range of the real valued function $f(x) = \frac{x^2 + 2x - 15}{2x^2 + 13x + 15}$ is

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

Find the domain and range of the following real function:
$f(x) = -|x|$

Vedclass Test Series

Exam Paper Generator

Online Exam Module