If f(x) is differentiable on R,f(x) f^{prime} | vedclass

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

Question

(B) Given the differential equation $f(x) f^{\prime}(-x) - f(-x) f^{\prime}(x) = 0$. This can be rewritten as $\frac{d}{dx} [f(x) f(-x)] = f(x) f^{\pr

Vedclass · Accepted Answer

$9$

Vedclass · Answer

(B) Given the differential equation $f(x) f^{\prime}(-x) - f(-x) f^{\prime}(x) = 0$. This can be rewritten as $\frac{d}{dx} [f(x) f(-x)] = f(x) f^{\prime}(-x) (-1) + f^{\prime}(x) f(-x) = 0$. Wait,let us re-evaluate: $\frac{d}{dx} [f(x) f(-x)] = f^{\prime}(x) f(-x) + f(x) f^{\prime}(-x) (-1) = f^{\prime}(x) f(-x) - f(x) f^{\prime}(-x) = 0$. Since $f(x) f^{\prime}(-x) - f(-x) f^{\prime}(x) = 0$,we have $f^{\prime}(x) f(-x) - f(x) f^{\prime}(-x) = 0$. Thus,$\frac{d}{dx} [f(x) f(-x)] = 0$. This implies $f(x) f(-x) = c$,where $c$ is a constant. At $x = 0$,$f(0) f(0) = c \Rightarrow 3 	imes 3 = 9$,so $c = 9$. Thus,$f(x) f(-x) = 9$ for all $x \in R$. For $x = 3$,$f(3) f(-3) = 9 \Rightarrow 9 	imes f(-3) = 9 \Rightarrow f(-3) = 1$. Finally,$(1 + f(-3))^3 + 1 = (1 + 1)^3 + 1 = 2^3 + 1 = 8 + 1 = 9$.

If $f(x)$ is differentiable on $R$,$f(x) f^{\prime}(-x) - f(-x) f^{\prime}(x) = 0$,$f(0) = 3$,and $f(3) = 9$,then $(1 + f(-3))^3 + 1 = $

Similar Questions

If ${x^{2/3}} + {y^{2/3}} = {a^{2/3}}$,then $\frac{dy}{dx} = $

Find $\frac{dy}{dx}$ for the function $xy = e^{(x-y)}$.

If $xy \neq 0, x+y \neq 0$ and $x^m y^n=(x+y)^{m+n}$,where $m, n \notin N$,then $\frac{dy}{dx}$ is equal to

If $3^x y^x = x^{3y}$, then the value of $\frac{dy}{dx}$ at $x = 1$ is:

The curve $3y^2 = 2ax^2 + 6b$ passes through the point $P(3, -1)$ and the gradient of the curve at $P$ is $-1$. Then the values of $a$ and $b$ are:

Vedclass Test Series

Exam Paper Generator

Online Exam Module