The domain of the real valued function f(x) = | vedclass

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

Question

(B) For the function $f(x) = \log_{\sqrt{2}}(\sqrt{x^2+x} + \sqrt{x^2-x})$ to be defined,the argument of the logarithm must be positive and the square

Vedclass · Accepted Answer

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

Vedclass · Answer

(B) For the function $f(x) = \log_{\sqrt{2}}(\sqrt{x^2+x} + \sqrt{x^2-x})$ to be defined,the argument of the logarithm must be positive and the square roots must be defined.$1$. For $\sqrt{x^2+x}$ to be defined,$x^2+x \ge 0 \implies x(x+1) \ge 0$. This gives $x \in (-\infty, -1] \cup [0, \infty)$.$2$. For $\sqrt{x^2-x}$ to be defined,$x^2-x \ge 0 \implies x(x-1) \ge 0$. This gives $x \in (-\infty, 0] \cup [1, \infty)$.$3$. The intersection of these two conditions is $x \in (-\infty, -1] \cup [1, \infty) \cup \{0\}$.$4$. Additionally,the argument $\sqrt{x^2+x} + \sqrt{x^2-x} > 0$. At $x=0$,the expression becomes $\sqrt{0} + \sqrt{0} = 0$,and $\log_{\sqrt{2}}(0)$ is undefined.Thus,we exclude $x=0$.The domain is $(-\infty, -1] \cup [1, \infty)$.

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

Similar Questions

If $f(x) = \frac{2x-3}{(x-2)(x-3)}$ is a real-valued function,then the value that $f(x)$ does not take is:

The range of the function $f(x) = \frac{x^2}{x^2+1}$ is

Let $D = \{x \in R : f(x) = \sqrt{\frac{x-|x|}{x-[x]}} \text{ is defined} \}$ and $C$ be the range of the real function $g(x) = \frac{2x}{4+x^2}$. Then $D \cap C =$

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

If $f: R \rightarrow A$,defined by $f(x) = \cos x + \sqrt{3} \sin x - 1$,is an onto function,then $A =$

Vedclass Test Series

Exam Paper Generator

Online Exam Module