The domain of definition of the function f(x) | vedclass

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

Question

(B) For the function $f(x) = \sqrt{1 + \log_{e}(1 - x)}$ to be defined, the expression inside the square root must be non-negative, and the argument o

Vedclass · Accepted Answer

$-\infty < x \leq \frac{e - 1}{e}$

Vedclass · Answer

(B) For the function $f(x) = \sqrt{1 + \log_{e}(1 - x)}$ to be defined, the expression inside the square root must be non-negative, and the argument of the logarithm must be positive.$1$. Condition for logarithm: $1 - x > 0 \implies x $2$. Condition for square root: $1 + \log_{e}(1 - x) \geq 0$.$\log_{e}(1 - x) \geq -1$.Exponentiating both sides with base $e$:$1 - x \geq e^{-1}$.$1 - x \geq \frac{1}{e}$.$x \leq 1 - \frac{1}{e}$.$x \leq \frac{e - 1}{e}$.Combining both conditions: $x Since $\frac{e - 1}{e} < 1$, the domain is $-\infty < x \leq \frac{e - 1}{e}$.

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

Similar Questions

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

Let $R-(\alpha, \beta)$ be the range of $f(x) = \frac{x+3}{(x-1)(x+2)}$. Then,the sum of the intercepts of the line $\alpha x + \beta y + 1 = 0$ on the coordinate axes is:

Domain of the function $f$,given by $f(x) = \frac{1}{\sqrt{(x - 2)(x - 5)}}$ is

If $[x]$ represents the greatest integer $\leq x$, then the range of the real-valued function $f(x) = \frac{1}{\sqrt{[x]^2+[x]-2}}$ is

The domain of the function $f(x) = \frac{1}{\sqrt{[x]^2 - [x] - 6}}$,where $[x]$ is the greatest integer function $\leq x$,is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module