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

The function $f(x) = \operatorname{sech}(x)$ on $R$ has the range

The domain of the function $y = \frac{1}{\sqrt{|x| - x}}$ is

If $f(x) = \frac{x + 2}{x^2 - 3x + 1}$, then the values of $x$ for which $f(x)$ is not defined are

Let $f: R \rightarrow R$ be defined by $f(x) = \frac{1}{x}$ for all $x \in R$. Then $f$ is:

If the domain of the function $f(x) = \log_7(1 - \log_4(x^2 - 9x + 18))$ is $(\alpha, \beta) \cup (\gamma, \delta)$,then $\alpha + \beta + \gamma + \delta$ is equal to

Vedclass Test Series

Exam Paper Generator

Online Exam Module