Let a 1 be a constant. If f: A rightarrow A | vedclass

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

Question

(C) Given $a^x + a^y = a$. Since $f: A \rightarrow A$,the domain and range are both $A$. For $y$ to be defined,we must have $a^y > 0$,which implies $a

Vedclass · Accepted Answer

$(-\infty, 1)$

Vedclass · Answer

(C) Given $a^x + a^y = a$. Since $f: A \rightarrow A$,the domain and range are both $A$. For $y$ to be defined,we must have $a^y > 0$,which implies $a - a^x > 0$,so $a^x Since $a > 1$,taking $\log_a$ on both sides gives $x Thus,the domain is $x \in (-\infty, 1)$. Since the range must also be $A$,we have $y = \log_a(a - a^x)$. As $x \rightarrow -\infty$,$a^x \rightarrow 0$,so $y \rightarrow \log_a(a) = 1$. As $x \rightarrow 1^-$,$a^x \rightarrow a^-$,so $a - a^x \rightarrow 0^+$,which means $y \rightarrow -\infty$. Thus,the range is $(-\infty, 1)$. Therefore,$A = (-\infty, 1)$.

Let $a > 1$ be a constant. If $f: A \rightarrow A$ and $(x, y) \in f$ satisfy $a^x + a^y = a$,then $A =$

Similar Questions

Range of the function $\frac{1}{2 - \sin 3x}$ is

If the domain of the function $f(x) = \sqrt{\ln(m\sin x + 4)}$ is $R$,then the number of possible integral values of $m$ is:

Find the range of the function $f(x)$ defined by:
$f(x) = \begin{cases} 2x-3, & x < -1 \\ 1-x^2, & -1 \leq x \leq 1 \\ 3x^2+2, & x > 1 \end{cases}$

Let $[x]$ denote the greatest integer not more than $x$. If $A$ and $B$ are the domains of the functions $f(x)=\frac{x-[x]}{\sqrt{|x|-x}}$ and $g(x)=\frac{x-[x]}{\sqrt{|x|+x}}$ respectively,then

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