The domain of the function f(x) = log_{3+x}(x | vedclass

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

Question

(C) For the function $f(x) = \log_{3+x}(x^2 - 1)$ to be defined,the following conditions must be satisfied:$1$. The argument of the logarithm must be

Vedclass · Accepted Answer

$(-3, -2) \cup (-2, -1) \cup (1, \infty)$

Vedclass · Answer

(C) For the function $f(x) = \log_{3+x}(x^2 - 1)$ to be defined,the following conditions must be satisfied:$1$. The argument of the logarithm must be positive: $x^2 - 1 > 0 \implies x^2 > 1 \implies x \in (-\infty, -1) \cup (1, \infty)$.$2$. The base of the logarithm must be positive and not equal to $1$: $3 + x > 0 \implies x > -3$ and $3 + x 
eq 1 \implies x 
eq -2$.Combining these conditions:$x > -3$ and $x 
eq -2$ and $(x  1)$.Intersection of these sets:$(-3, -2) \cup (-2, -1) \cup (1, \infty)$.

The domain of the function $f(x) = \log_{3+x}(x^2 - 1)$ is

Similar Questions

If $[x]^2-5[x]+6=0$,where $[\cdot]$ denotes the greatest integer function,then $x$ belongs to:

If ${ }^{n} C_{r}$ denotes the number of combinations of $n$ distinct things taken $r$ at a time,then the domain of the function $g(x)={ }^{(16-x)} C_{(2 x-1)}$ is

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

The domain of the real valued function $f(x) = \sqrt{2+x} + \sqrt{3-x}$ 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