If f(x) is a function such that f^{prime}(x)= | vedclass

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

Question

(B) Given $f^{\prime}(x) = \sqrt{f^2(x)-1}$.We can write this as $\frac{f^{\prime}(x)}{\sqrt{f^2(x)-1}} = 1$.Integrating both sides with respect to $x

Vedclass · Accepted Answer

$\frac{e^2+1}{2 e}$

Vedclass · Answer

(B) Given $f^{\prime}(x) = \sqrt{f^2(x)-1}$.We can write this as $\frac{f^{\prime}(x)}{\sqrt{f^2(x)-1}} = 1$.Integrating both sides with respect to $x$,we get $\int \frac{f^{\prime}(x)}{\sqrt{f^2(x)-1}} dx = \int 1 dx$.Using the standard integral $\int \frac{1}{\sqrt{t^2-1}} dt = \log |t + \sqrt{t^2-1}| + C$,we have $\log |f(x) + \sqrt{f^2(x)-1}| = x + C$.Given $f(0) = 1$,substituting $x=0$ gives $\log |f(0) + \sqrt{f^2(0)-1}| = 0 + C$.$\log |1 + \sqrt{1-1}| = C \Rightarrow \log(1) = C \Rightarrow C = 0$.So,$\log |f(x) + \sqrt{f^2(x)-1}| = x$.At $x=1$,$\log |f(1) + \sqrt{f^2(1)-1}| = 1$.Taking the exponential of both sides,$f(1) + \sqrt{f^2(1)-1} = e^1 = e$.$\sqrt{f^2(1)-1} = e - f(1)$.Squaring both sides,$f^2(1) - 1 = e^2 + f^2(1) - 2ef(1)$.$-1 = e^2 - 2ef(1) \Rightarrow 2ef(1) = e^2 + 1$.Thus,$f(1) = \frac{e^2+1}{2e}$.

If $f(x)$ is a function such that $f^{\prime}(x)=\sqrt{f^2(x)-1}$ and $f(0)=1$,then $f(1)=$

Similar Questions

The equation of a curve passing through $(1,0)$ and having the slope of the tangent at any point $(x, y)$ of the curve as $\frac{y-1}{x^2+x}$ is

The curve passing through the point $(1,2)$ given that the slope of the tangent at any point $(x, y)$ is $\frac{3x}{y}$ represents

The solution of $\frac{dy}{dx} + 1 = e^{x+y}$ is

If the differential equation $\frac{dy}{dx} + \frac{x}{y} = \frac{a}{y}$,where $a$ is a constant,represents a family of circles,then the radius of the circle is ......

The particular solution of the differential equation $\cos \left(\frac{dy}{dx}\right) = 0.5$ with the condition $y = 1$ at $x = 0$ is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module