Let f : R to R be defined by f(x) = ln(x + sq | vedclass

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

Question

(B) Given $f(x) = \ln(x + \sqrt{x^2 + 1})$.Let $y = \ln(x + \sqrt{x^2 + 1})$.Then $e^y = x + \sqrt{x^2 + 1}$.Also,$e^{-y} = \frac{1}{x + \sqrt{x^2 + 1

Vedclass · Accepted Answer

$2$

Vedclass · Answer

(B) Given $f(x) = \ln(x + \sqrt{x^2 + 1})$.Let $y = \ln(x + \sqrt{x^2 + 1})$.Then $e^y = x + \sqrt{x^2 + 1}$.Also,$e^{-y} = \frac{1}{x + \sqrt{x^2 + 1}} = \sqrt{x^2 + 1} - x$.Subtracting the two equations: $e^y - e^{-y} = 2x$,which implies $x = \frac{e^y - e^{-y}}{2} = \sinh(y)$.Thus,$f^{-1}(x) = \sinh(x) = \frac{e^x - e^{-x}}{2}$.We need to find the number of solutions to $|f^{-1}(x)| = e^{-|x|}$,which is $|\sinh(x)| = e^{-|x|}$.Since both sides are even functions,we can analyze for $x \ge 0$: $\sinh(x) = e^{-x}$.At $x = 0$,$\sinh(0) = 0$ and $e^0 = 1$,so $0 
eq 1$.As $x$ increases,$\sinh(x)$ is strictly increasing from $0$ to $\infty$,and $e^{-x}$ is strictly decreasing from $1$ to $0$.Therefore,there is exactly one intersection point for $x > 0$.By symmetry,there is also exactly one intersection point for $x Thus,there are $2$ solutions in total.

Let $f : R \to R$ be defined by $f(x) = \ln(x + \sqrt{x^2 + 1})$. Then the number of solutions of $|f^{-1}(x)| = e^{-|x|}$ is

Similar Questions

If $g$ is the inverse of a function $f$ and $f'(x) = \frac{1}{1 + x^5}$,then $g'(x)$ is equal to:

$f: R \rightarrow R, f(x) = 3x + 2$ and $g: R \rightarrow R, g(x) = 6x + 5$. Find the value of $(g \circ f^{-1})(10)$.

$f: (-\infty, 0] \rightarrow [0, \infty)$ is defined as $f(x) = x^2$. The domain and range of its inverse are

If $f(x) = \frac{2x - 3}{3x - 4}$,$x \neq \frac{4}{3}$,then the value of $f^{-1}(x)$ is

Let $f$ and $g$ be two differentiable functions satisfying $g^{\prime}(5)=\frac{3}{4}$, $g(5)=6$ and $g=f^{-1}$. Then $f^{\prime}(6)$ is equal to

Vedclass Test Series

Exam Paper Generator

Online Exam Module