If 27a + 9b + 3c + d = 0,then the equation 4a | vedclass

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

Question

(C) Let $f(x) = ax^4 + bx^3 + cx^2 + dx + e$. Then,$f'(x) = 4ax^3 + 3bx^2 + 2cx + d$. We are given that $f'(x) = 0$ has at least one root. Consider th

Vedclass · Accepted Answer

$0$ and $3$

Vedclass · Answer

(C) Let $f(x) = ax^4 + bx^3 + cx^2 + dx + e$. Then,$f'(x) = 4ax^3 + 3bx^2 + 2cx + d$. We are given that $f'(x) = 0$ has at least one root. Consider the values of $f(x)$ at $x=0$ and $x=3$: $f(0) = a(0)^4 + b(0)^3 + c(0)^2 + d(0) + e = e$. $f(3) = a(3)^4 + b(3)^3 + c(3)^2 + d(3) + e = 81a + 27b + 9c + 3d + e$. We can rewrite $f(3)$ as $f(3) = 3(27a + 9b + 3c + d) + e$. Since it is given that $27a + 9b + 3c + d = 0$,we have $f(3) = 3(0) + e = e$. Thus,$f(0) = f(3) = e$. Since $f(x)$ is a polynomial,it is continuous on $[0, 3]$ and differentiable on $(0, 3)$. By Rolle's Theorem,there exists at least one $c \in (0, 3)$ such that $f'(c) = 0$. Therefore,the equation $4ax^3 + 3bx^2 + 2cx + d = 0$ has at least one root in the interval $(0, 3)$.

If $27a + 9b + 3c + d = 0$,then the equation $4ax^3 + 3bx^2 + 2cx + d = 0$ has at least one root between which of the following?

Similar Questions

If $f(x)$ satisfies the conditions of Rolle's theorem in $[1, 2]$ and $f(x)$ is continuous in $[1, 2]$,then $\int_1^2 f'(x) dx$ is equal to

If $2a + 3b + 6c = 0$,then at least one root of the equation $ax^2 + bx + c = 0$ lies in the interval:

Consider the function $f(x) = e^{-2x} \sin 2x$ over the interval $(0, \pi/2)$. $A$ real number $c \in (0, \pi/2)$,as guaranteed by Rolle's theorem,such that $f'(c) = 0$ is

$A$ value of $c$ according to the Lagrange's mean value theorem for $f(x)=(x-1)(x-2)(x-3)$ in $[0,4]$ is

In the interval $[0, 1]$,Lagrange's Mean Value Theorem is $NOT$ applicable to which of the following functions?

Vedclass Test Series

Exam Paper Generator

Online Exam Module