The domain of the function f(x) = frac{sin^{- | vedclass

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

Question

(B) For the function $f(x) = \frac{\sin^{-1}(3 - x)}{\ln(|x| - 2)}$ to be defined,the following conditions must be met:$1$. The argument of $\sin^{-1}

Vedclass · Accepted Answer

$(2, 3) \cup (3, 4]$

Vedclass · Answer

(B) For the function $f(x) = \frac{\sin^{-1}(3 - x)}{\ln(|x| - 2)}$ to be defined,the following conditions must be met:$1$. The argument of $\sin^{-1}$ must be in $[-1, 1]$:$-1 \le 3 - x \le 1$$-4 \le -x \le -2$$2 \le x \le 4$So,the domain of the numerator is $[2, 4]$.$2$. The argument of $\ln$ must be positive,and the denominator cannot be zero:$|x| - 2 > 0 \implies |x| > 2 \implies x \in (-\infty, -2) \cup (2, \infty)$.Also,$\ln(|x| - 2) 
eq 0 \implies |x| - 2 
eq 1 \implies |x| 
eq 3 \implies x 
eq 3, x 
eq -3$.Combining these,the domain of the denominator is $(-\infty, -3) \cup (-3, -2) \cup (2, 3) \cup (3, \infty)$.$3$. The domain of $f(x)$ is the intersection of these sets:$[2, 4] \cap ((-\infty, -3) \cup (-3, -2) \cup (2, 3) \cup (3, \infty)) = (2, 3) \cup (3, 4]$.Thus,the correct option is $B$.

The domain of the function $f(x) = \frac{\sin^{-1}(3 - x)}{\ln(|x| - 2)}$ is

Similar Questions

The domain of the real valued function $f(x) = \frac{\sqrt{\log _{0.5}(x-3)}}{\sqrt{x-1}}$ is

Let $A = \{9, 10, 11, 12, 13\}$ and let $f: A \rightarrow N$ be defined by $f(n) = \text{the highest prime factor of } n$. Find the range of $f$.

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

The domain of the real-valued function $f(x) = \frac{3}{4-x^2} + \log_{10}(x^3-x)$ is

The domain of the function $f(x) = \frac{1}{\sqrt{x+|x|}}$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module