If the function f : R to R is defined by f(x) | vedclass

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

Question

(A) Given $f(x) = \log_a(x + \sqrt{x^2 + 1})$.Let $y = f(x)$,so $y = \log_a(x + \sqrt{x^2 + 1})$.By the definition of logarithm,$a^y = x + \sqrt{x^2 +

Vedclass · Accepted Answer

$\frac{a^x - a^{-x}}{2}$

Vedclass · Answer

(A) Given $f(x) = \log_a(x + \sqrt{x^2 + 1})$.Let $y = f(x)$,so $y = \log_a(x + \sqrt{x^2 + 1})$.By the definition of logarithm,$a^y = x + \sqrt{x^2 + 1}$.Now,consider $a^{-y} = \frac{1}{a^y} = \frac{1}{x + \sqrt{x^2 + 1}}$.Rationalizing the denominator: $a^{-y} = \frac{\sqrt{x^2 + 1} - x}{(\sqrt{x^2 + 1} + x)(\sqrt{x^2 + 1} - x)} = \frac{\sqrt{x^2 + 1} - x}{(x^2 + 1) - x^2} = \sqrt{x^2 + 1} - x$.We have two equations:$1) \sqrt{x^2 + 1} + x = a^y$$2) \sqrt{x^2 + 1} - x = a^{-y}$Subtracting equation $(2)$ from equation $(1)$:$(\sqrt{x^2 + 1} + x) - (\sqrt{x^2 + 1} - x) = a^y - a^{-y}$$2x = a^y - a^{-y}$$x = \frac{a^y - a^{-y}}{2}$.Since $x = f^{-1}(y)$,we have $f^{-1}(y) = \frac{a^y - a^{-y}}{2}$.Replacing $y$ with $x$,we get $f^{-1}(x) = \frac{a^x - a^{-x}}{2}$.

If the function $f : R \to R$ is defined by $f(x) = \log_a(x + \sqrt{x^2 + 1})$,where $(a > 0, a \neq 1)$,then $f^{-1}(x)$ is

Similar Questions

Let $f(x) = (x + 1)^2 - 1$ for $x \ge -1$. Then the set $S = \{ x : f(x) = f^{-1}(x) \}$ is

If $f: \mathbb{R} \to \mathbb{R}$ is defined by $f(x) = 3x - 4$,then ${f^{ - 1}}: \mathbb{R} \to \mathbb{R}$ is

$A$ condition for a function $y = f(x)$ to have an inverse is that it should be

The relation $R$ on the set of natural numbers is defined by $R = \{(a, b) : a = 2b\}$. Then ${R^{-1}}$ is:

Let $f : (4, 6) \to (6, 8)$ be a function defined by $f(x) = x + [\frac{x}{2}]$ (where $[.]$ denotes the greatest integer function),then $f^{-1}(x)$ is equal to

Vedclass Test Series

Exam Paper Generator

Online Exam Module