The domain and range of f(x) = frac{1}{sqrt{| | vedclass

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

Question

(C) For the domain $A$,we require $|x| - x^2 > 0$. Since $|x|^2 = x^2$,this is $|x| - |x|^2 > 0$,which implies $|x|(1 - |x|) > 0$. This holds when $0

Vedclass · Accepted Answer

$(-1, 0) \cup (0, 1) \cup [1, \infty)$

Vedclass · Answer

(C) For the domain $A$,we require $|x| - x^2 > 0$. Since $|x|^2 = x^2$,this is $|x| - |x|^2 > 0$,which implies $|x|(1 - |x|) > 0$. This holds when $0 For the range $B$,let $y = \frac{1}{\sqrt{|x| - x^2}}$. As $x 	o 0$,$|x| - x^2 	o 0^+$,so $y 	o \infty$. As $|x| 	o 1$,$|x| - x^2 	o 0^+$,so $y 	o \infty$. The maximum value of $|x| - x^2$ occurs at $|x| = 1/2$,giving $1/2 - 1/4 = 1/4$. The minimum value of the denominator is $0$ (exclusive) and the maximum value is $1/4$. Thus,$\sqrt{|x| - x^2} \in (0, 1/2]$. Therefore,$y \in [2, \infty)$,so $B = [2, \infty)$. Finally,$A \cup B = ((-1, 0) \cup (0, 1)) \cup [2, \infty)$.

The domain and range of $f(x) = \frac{1}{\sqrt{|x| - x^2}}$ are $A$ and $B$ respectively. Then $A \cup B =$

Similar Questions

Let the domain of the function $f(x) = \log_3 \log_5 (7 - \log_2 (x^2 - 10 x + 85)) + \sin^{-1} ( | \frac{3 x - 7}{17 - x} | )$ be $(\alpha, \beta]$. Then $\alpha + \beta$ is equal to :

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

The range of $f(x) = \frac{x^2 + 34x - 71}{x^2 + 2x - 7}$ is

Let $f(x) = \frac{1}{7 - \sin 5x}$ be a function defined on $R$. Then the range of the function $f(x)$ is equal to:

Is it true that $x = e^{\log x}$ for all real $x$?

Vedclass Test Series

Exam Paper Generator

Online Exam Module