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

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

The domain of the function $f(x) = \exp (\sqrt {5x - 3 - 2{x^2}} )$ is

If $f: R-\{2\} \rightarrow R$ is a function defined by $f(x)=\frac{x^2-4}{x-2}$,then the range is

The range of the function $f(x) = \sqrt{3-x} + \sqrt{2+x}$ is

The domain of $\sqrt{|x|-x}$ is

What is the range of values for $\frac{x}{x^2 + 4}$ for all real values of $x$?

Vedclass Test Series

Exam Paper Generator

Online Exam Module