The domain of the function f(x) = frac{1}{log | vedclass

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

Question

(D) For the function $f(x) = \frac{1}{\log_{10}(1-x)} + \sqrt{x+2}$ to be defined:$1$. The term inside the square root must be non-negative: $x + 2 \g

Vedclass · Accepted Answer

$[-2, 0) \cup (0, 1)$

Vedclass · Answer

(D) For the function $f(x) = \frac{1}{\log_{10}(1-x)} + \sqrt{x+2}$ to be defined:$1$. The term inside the square root must be non-negative: $x + 2 \geq 0 \implies x \geq -2$.$2$. The argument of the logarithm must be positive: $1 - x > 0 \implies x $3$. The denominator cannot be zero: $\log_{10}(1-x) 
eq 0 \implies 1 - x 
eq 10^0 \implies 1 - x 
eq 1 \implies x 
eq 0$.Combining these conditions: $x \geq -2$,$x Thus,the domain is $x \in [-2, 0) \cup (0, 1)$.

The domain of the function $f(x) = \frac{1}{\log_{10}(1-x)} + \sqrt{x+2}$ is

Similar Questions

The range of the real valued function $f(x) = \sqrt{\frac{x^2+2x+8}{x^2+2x+4}}$ is

Let $f(x) = \sqrt{\frac{x+1}{x+3}}$ and $g(x) = \sqrt{\frac{2-x}{x+3}}$ be two real-valued functions. Then the domain of $f/g$ is

The sum of the least positive integer and the greatest negative integer in the range of the function $f(x) = \frac{x^2-5x+7}{x^2-5x-7}$ is

The domain of the function $f(x) = \sqrt{x-1} + \sqrt{6-x}$ is...

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

Vedclass Test Series

Exam Paper Generator

Online Exam Module