Let b be a nonzero real number. Suppose f: R | vedclass

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

Question

(D) Given the differential equation $f^{\prime}(x) = \frac{f(x)}{b^2+x^2}$.
Separating variables,we get $\frac{f^{\prime}(x)}{f(x)} = \frac{1}{b^2+x^

Vedclass · Accepted Answer

$A, C$

Vedclass · Answer

(D) Given the differential equation $f^{\prime}(x) = \frac{f(x)}{b^2+x^2}$.
Separating variables,we get $\frac{f^{\prime}(x)}{f(x)} = \frac{1}{b^2+x^2}$.
Integrating both sides with respect to $x$,we have $\int \frac{f^{\prime}(x)}{f(x)} dx = \int \frac{1}{b^2+x^2} dx$.
This yields $\ln|f(x)| = \frac{1}{b} 	an^{-1}\left(\frac{x}{b}ight) + C$.
Using the initial condition $f(0)=1$,we get $\ln(1) = \frac{1}{b} 	an^{-1}(0) + C$,which implies $C=0$.
Thus,$f(x) = e^{\frac{1}{b} 	an^{-1}(\frac{x}{b})}$.
Now,consider $f(x)f(-x) = e^{\frac{1}{b} 	an^{-1}(\frac{x}{b})} \cdot e^{\frac{1}{b} 	an^{-1}(\frac{-x}{b})} = e^{\frac{1}{b} (	an^{-1}(\frac{x}{b}) - 	an^{-1}(\frac{x}{b}))} = e^0 = 1$. So,$(C)$ is true.
For $b>0$,$f^{\prime}(x) = \frac{f(x)}{b^2+x^2} > 0$ since $f(x) = e^{\dots} > 0$ and $b^2+x^2 > 0$. Thus,$f$ is an increasing function. So,$(A)$ is true.
Therefore,the correct statements are $(A)$ and $(C)$.

Similar Questions

The equation of the curve passing through $(3, 4)$ and satisfying the differential equation $y \left( \frac{dy}{dx} \right)^2 + (x - y) \frac{dy}{dx} - x = 0$ can be:

Verify that the given function $y = x \sin x$ is a solution of the differential equation $x y^{\prime} = y + x \sqrt{x^2 - y^2}$ (where $x \neq 0$ and $x > y$ or $x < -y$).

$A$ curve $y = f(x)$ passing through the point $\left(1, \frac{1}{\sqrt{e}}\right)$ satisfies the differential equation $\frac{dy}{dx} + x e^{-\frac{x^2}{2}} = 0.$ Then which of the following does not hold good?

The solution of $(1 + xy)y\,dx + (1 - xy)x\,dy = 0$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module