Let f : R to R be a twice differentiable func | vedclass

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

Question

(A) Given $f'(x) = f^2(x)$ and $f(0) = 0$. If $f(x)$ is not identically zero,then $\frac{f'(x)}{f^2(x)} = 1$. Integrating both sides,we get $-\frac{1}

Vedclass · Accepted Answer

$2f'(2) + 4f''(2)$

Vedclass · Answer

(A) Given $f'(x) = f^2(x)$ and $f(0) = 0$. If $f(x)$ is not identically zero,then $\frac{f'(x)}{f^2(x)} = 1$. Integrating both sides,we get $-\frac{1}{f(x)} = x + c$,which implies $f(x) = -\frac{1}{x+c}$. However,$f(0) = 0$ implies $-\frac{1}{c} = 0$,which is impossible. Thus,the only twice differentiable function satisfying $f(0) = 0$ and $f'(x) = f^2(x)$ is $f(x) = 0$ for all $x \in R$. If $f(x) = 0$,then $f'(x) = 0$ and $f''(x) = 0$. Therefore,$\lim_{x 	o 2} (f(x) + xf'(x) + x^2f''(x)) = 0 + 2(0) + 4(0) = 0$. Checking the options for $f(x) = 0$: $A: 2(0) + 4(0) = 0$. $B: 2(0) + 4(0) - 0 = 0$. $C: -1 + 0 + 0 = -1$. $D: 1 + 0 + 0 = 1$. Since $f(x) = 0$ is the solution,the expression evaluates to $0$. Option $A$ and $B$ both result in $0$. Given the structure,$A$ is the most simplified form.

Let $f : R \to R$ be a twice differentiable function satisfying $f(0) = f(1) = 0$ and $f'(x) = f^2(x)$ for all $x \in R$. Then $\lim_{x \to 2} (f(x) + xf'(x) + x^2f''(x))$ is equal to:

Similar Questions

If $x \ln(\ln x) - x^2 + y^2 = 4$ where $y > 0$,then $\frac{dy}{dx}$ at $x = e$ is equal to

Find $\frac{dy}{dx}$,if $y^{x}+x^{y}+x^{x}=a^{b}$.

If $\log (x+y)-2xy=0$,then $y^{\prime}(0)=$

If ${x^2} + {y^2} = t - \frac{1}{t}$ and ${x^4} + {y^4} = {t^2} + \frac{1}{t^2}$,then ${x^3}y\frac{dy}{dx} = $

If $\sin(xy) + \cos(xy) = 0$,then $\frac{dy}{dx} = $

Vedclass Test Series

Exam Paper Generator

Online Exam Module