If f is a real-valued function from A onto B | vedclass

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

Question

(A) Given $f(x) = \frac{1}{\sqrt{|x - |x||}}$.For the function to be defined, the expression inside the square root must be strictly positive: $|x - |

Vedclass · Accepted Answer

$\phi$

Vedclass · Answer

(A) Given $f(x) = \frac{1}{\sqrt{|x - |x||}}$.For the function to be defined, the expression inside the square root must be strictly positive: $|x - |x|| > 0$.If $x \ge 0$, then $|x| = x$, so $|x - x| = 0$, which is not $> 0$.If $x  0$.Thus, the domain $A = (-\infty, 0)$.For $x \in (-\infty, 0)$, $f(x) = \frac{1}{\sqrt{-2x}}$. As $x$ ranges from $(-\infty, 0)$, $-2x$ ranges from $(0, \infty)$, and $\sqrt{-2x}$ ranges from $(0, \infty)$.Therefore, the range $B = (0, \infty)$.Finally, $A \cap B = (-\infty, 0) \cap (0, \infty) = \phi$.

If $f$ is a real-valued function from $A$ onto $B$ defined by $f(x) = \frac{1}{\sqrt{|x - |x||}}$, then $A \cap B = $

Similar Questions

The natural domain of the real valued function defined by $f(x) = \sqrt{x^2 - 1} + \sqrt{x^2 + 1}$ is

Domain of the real valued function $f(x) = \frac{x+2}{9-x^{2}}$ is

The domain of $f(x) = \log \left[(2.5)^{3-x^2} - (0.4)^{x+9}\right]$ is

If the domain of the function $f(x) = \frac{[x]}{1+x^2}$,where $[x]$ is the greatest integer $\leq x$,is $(2, 6)$,then its range is

The domain of the function $f(x) = \sqrt{\frac{x-7}{9-x}}$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module