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

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

Vedclass · Accepted Answer

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

Vedclass · Answer

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

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

Similar Questions

Let $y = \sqrt{\frac{(x + 1)(x - 3)}{(x - 2)}}$. Then,the set of all real values of $x$ for which $y$ takes real values is:

The domain of the function $f(x) = [\log_{10}(\frac{5x - x^2}{4})]^{1/2}$ is

Find the domain and the range of the real function $f$ defined by $f(x) = \sqrt{x-1}$.

Range of the function $f(x) = 3 + 2^x + 4^x$ is

If a relation $R$ on the set of natural numbers $N$ is defined by $x + 2y = 8$,then the domain of $R$ is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module