If f(x) = (frac{3}{5})^x + (frac{4}{5})^x - 1 | vedclass

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

Question

(B) Given $f(x) = (\frac{3}{5})^x + (\frac{4}{5})^x - 1$. Setting $f(x) = 0$,we get $(\frac{3}{5})^x + (\frac{4}{5})^x = 1$. Dividing by $5^x$ is equi

Vedclass · Accepted Answer

one solution

Vedclass · Answer

(B) Given $f(x) = (\frac{3}{5})^x + (\frac{4}{5})^x - 1$. Setting $f(x) = 0$,we get $(\frac{3}{5})^x + (\frac{4}{5})^x = 1$. Dividing by $5^x$ is equivalent to solving $3^x + 4^x = 5^x$. Divide both sides by $5^x$: $(\frac{3}{5})^x + (\frac{4}{5})^x = 1$. Let $g(x) = (\frac{3}{5})^x + (\frac{4}{5})^x$. Since both $(\frac{3}{5})^x$ and $(\frac{4}{5})^x$ are strictly decreasing functions for $x \in R$,their sum $g(x)$ is also a strictly decreasing function. $A$ strictly decreasing function can intersect the horizontal line $y = 1$ at most once. By inspection,for $x = 2$,we have $(\frac{3}{5})^2 + (\frac{4}{5})^2 = \frac{9}{25} + \frac{16}{25} = \frac{25}{25} = 1$. Thus,$x = 2$ is the unique solution.

If $f(x) = (\frac{3}{5})^x + (\frac{4}{5})^x - 1$,$x \in R$,then the equation $f(x) = 0$ has

Similar Questions

The number of functions $f: \{1, 2, \ldots, 100\} \rightarrow \{0, 1\}$ that assign $1$ to exactly one of the positive integers less than or equal to $98$ is equal to $\qquad$.

Let $A = \{x : x \in R, x \text{ is not a positive integer}\}$. Define $f: A \rightarrow R$ as $f(x) = \frac{2x}{x-1}$. Then $f$ is:

The function $f: R \rightarrow R$ defined by $f(x)=\frac{x}{\sqrt{1+x^2}}$ is

Let $A = \{x, y, z, u\}$ and $B = \{a, b\}$. $A$ function $f: A \rightarrow B$ is selected randomly. The probability that the function is an onto function is

Let $f: R \to R$ be defined as $f(x) = x^3$. Then $f$ is . . . . . . .

Vedclass Test Series

Exam Paper Generator

Online Exam Module