The domain of the function f(x) = sqrt{log_el | vedclass

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

Question

(C) For the function $f(x) = \sqrt{\log_e\left(\frac{1}{(x-2)^2}\right)} + \sin^{-1}(x^2-2)$ to be defined:$1$. The term inside the square root must b

Vedclass · Accepted Answer

$[1, \sqrt{3}]$

Vedclass · Answer

(C) For the function $f(x) = \sqrt{\log_e\left(\frac{1}{(x-2)^2}ight)} + \sin^{-1}(x^2-2)$ to be defined:$1$. The term inside the square root must be non-negative: $\log_e\left(\frac{1}{(x-2)^2}ight) \ge 0$.This implies $\frac{1}{(x-2)^2} \ge e^0 = 1$,so $(x-2)^2 \le 1$,which means $|x-2| \le 1$,giving $1 \le x \le 3$. Since $x 
eq 2$,the domain is $[1, 2) \cup (2, 3]$.$2$. The term inside $\sin^{-1}$ must be in $[-1, 1]$: $-1 \le x^2-2 \le 1$.This implies $1 \le x^2 \le 3$,so $x \in [-\sqrt{3}, -1] \cup [1, \sqrt{3}]$.$3$. Taking the intersection of both conditions: $[1, 2) \cup (2, 3] \cap ([-\sqrt{3}, -1] \cup [1, \sqrt{3}])$.This results in $[1, \sqrt{3}] \setminus \{2\}$. Since $2$ is not in $[1, \sqrt{3}]$,the domain is $[1, \sqrt{3}]$.

The domain of the function $f(x) = \sqrt{\log_e\left(\frac{1}{x^2-4x+4}\right)} + \sin^{-1}(x^2-2)$ is

Similar Questions

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

Domain of the function $f(x) = \frac{x - 3}{(x - 1)\sqrt{x^2 - 4}}$ is

The domain of the real-valued function $f(x) = \sqrt{\frac{2x^2 - 7x + 5}{3x^2 - 5x - 2}}$ is

Let $R$ be the set of real numbers and $f: R \rightarrow R$ be given by $f(x) = \sqrt{|x|} - \log(1 + |x|)$. We now make the following assertions:
$I.$ There exists a real number $A$ such that $f(x) \leq A$ for all $x$.
$II.$ There exists a real number $B$ such that $f(x) \geq B$ for all $x$.

For the function $f(x) = (1 + \frac{1}{x})^x$,the domain of $f(x)$ is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module