The domain of f(x) = sin log left( frac{sqrt{ | vedclass

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

Question

(A) For $f(x) = \sin \log \left( \frac{\sqrt{4-x^2}}{1-x} \right)$ to be defined,the argument of the logarithm must be strictly positive. First,the sq

Vedclass · Accepted Answer

$(-2, 1)$

Vedclass · Answer

(A) For $f(x) = \sin \log \left( \frac{\sqrt{4-x^2}}{1-x} ight)$ to be defined,the argument of the logarithm must be strictly positive. First,the square root $\sqrt{4-x^2}$ is defined when $4-x^2 \geq 0$,which implies $x \in [-2, 2]$. Second,the expression inside the logarithm must be positive: $\frac{\sqrt{4-x^2}}{1-x} > 0$. Since $\sqrt{4-x^2} \geq 0$,we must have $1-x > 0$ (i.e.,$x Thus,$x Combining $x \in [-2, 2]$ with $x < 1$ and $x 
eq \pm 2$,we get $x \in (-2, 1)$.

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

Similar Questions

If $f:[2,3] \rightarrow R$ is defined by $f(x)=x^3+3x-2$,then the range of $f(x)$ is contained in the interval

The range of the function $f(x) = \frac{1}{\sqrt{x-[x]}}$ is

The domain of $f(x) = \frac{\log_{(x+1)}(x-2)}{x^2 - (2x + 3)}$ for $x \in R$ is

The domain of the real-valued function $f(x) = \frac{\sqrt{\log_{10}\left(\frac{x}{x-2}\right)}}{\sqrt{[x]^2-5[x]+6}}$ is (where $[x]$ denotes the greatest integer function):

If a real valued function $f:[-1,2] \rightarrow B$ defined by $f(x) = \begin{cases} 1-x, & -1 \leq x \leq 1 \\ x-1, & 1 < x \leq 2 \end{cases}$ is a surjection,then $B=$

Vedclass Test Series

Exam Paper Generator

Online Exam Module