Let f: R rightarrow R be a continuous functio | vedclass

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

Question

(B) Given the equation: $f(x) + \int_{0}^{x} t f(t) dt + x^2 = 0$. Differentiating both sides with respect to $x$ using Leibniz's rule: $f'(x) + x f(x

Vedclass · Accepted Answer

$\lim_{x \rightarrow -\infty} f(x) = -2$

Vedclass · Answer

(B) Given the equation: $f(x) + \int_{0}^{x} t f(t) dt + x^2 = 0$. Differentiating both sides with respect to $x$ using Leibniz's rule: $f'(x) + x f(x) + 2x = 0$. Rearranging the terms: $f'(x) = -x(f(x) + 2)$. This is a first-order linear differential equation. Separating variables: $\frac{f'(x)}{f(x) + 2} = -x$. Integrating both sides: $\ln|f(x) + 2| = -\frac{x^2}{2} + C$. Thus,$f(x) + 2 = A e^{-x^2/2}$,or $f(x) = A e^{-x^2/2} - 2$. Using the initial condition from the original equation at $x = 0$: $f(0) + \int_{0}^{0} t f(t) dt + 0^2 = 0 \Rightarrow f(0) = 0$. Substituting $x = 0$ into $f(x) = A e^{-x^2/2} - 2$: $0 = A(1) - 2 \Rightarrow A = 2$. So,$f(x) = 2 e^{-x^2/2} - 2$. Now,evaluating the limits: $\lim_{x \rightarrow \infty} f(x) = \lim_{x \rightarrow \infty} (2 e^{-x^2/2} - 2) = 0 - 2 = -2$. $\lim_{x \rightarrow -\infty} f(x) = \lim_{x \rightarrow -\infty} (2 e^{-x^2/2} - 2) = 0 - 2 = -2$. Since $f(x) = 2(e^{-x^2/2} - 1)$,$f(x) = f(-x)$,so it is an even function. Comparing with the options,$\lim_{x \rightarrow -\infty} f(x) = -2$ is correct.

Let $f: R \rightarrow R$ be a continuous function satisfying $f(x) + \int_{0}^{x} t f(t) dt + x^2 = 0$ for all $x \in R$. Then:

Similar Questions

The solution of the differential equation $\left( {{e^{{x^2}}} + {e^{{y^2}}}} \right) y \frac{{dy}}{{dx}} + {e^{{x^2}}}(x{y^2} - x) = 0$ is

The integrating factor of the differential equation $x \frac{dy}{dx} + y \log x = x e^x \cdot x^{-1/2} \log x$ for $x > 0$ is:

Let the solution curve $y=y(x)$ of the differential equation $(1+e^{2x})(\frac{dy}{dx}+y)=1$ pass through the point $(0, \frac{\pi}{2})$. Then,$\lim_{x \rightarrow \infty} e^{x} y(x)$ is equal to.

Let $F:[3,5] \rightarrow R$ be a twice differentiable function on $(3,5)$ such that $F(x)=e^{-x} \int_{3}^{x} (3t^{2}+2t+4F^{\prime}(t)) \,dt$. If $F^{\prime}(4)=\frac{\alpha e^{\beta}-224}{(e^{\beta}-4)^{2}}$,then $\alpha+\beta$ is equal to $....$

The general solution of the differential equation $(x+2y^3) \frac{dy}{dx} - y = 0, y > 0$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module