If a, b, c are real numbers such that a + b + | vedclass

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

Question

(A) Let $f(x) = ax^3 + bx^2 + cx$. Since $f(x)$ is a polynomial,it is continuous and differentiable on $\mathbb{R}$. We observe that $f(0) = a(0)^3 +

Vedclass · Accepted Answer

At least one root in $[0, 1]$

Vedclass · Answer

(A) Let $f(x) = ax^3 + bx^2 + cx$. Since $f(x)$ is a polynomial,it is continuous and differentiable on $\mathbb{R}$. We observe that $f(0) = a(0)^3 + b(0)^2 + c(0) = 0$. Given $a + b + c = 0$,we have $f(1) = a(1)^3 + b(1)^2 + c(1) = a + b + c = 0$. Since $f(0) = f(1) = 0$,by Rolle's Theorem,there exists at least one $\alpha \in (0, 1)$ such that $f'(\alpha) = 0$. Calculating the derivative,$f'(x) = 3ax^2 + 2bx + c$. Thus,$f'(\alpha) = 3a\alpha^2 + 2b\alpha + c = 0$ for some $\alpha \in (0, 1)$. Therefore,the quadratic equation $3ax^2 + 2bx + c = 0$ has at least one root in $[0, 1]$.

If $a, b, c$ are real numbers such that $a + b + c = 0$,then the quadratic equation $3ax^2 + 2bx + c = 0$ has

Similar Questions

Let $f:[1,3] \rightarrow R$ be continuous and differentiable in $(1,3)$ such that $f^{\prime}(x)=[f(x)]^2+4$ for all $x \in (1,3)$. Then:

If the function $f(x) = ax^3 + bx^2 + 11x - 6$ satisfies the conditions of Rolle's theorem for the interval $[1, 3]$ and $f'\left( 2 + \frac{1}{\sqrt{3}} \right) = 0$,then the values of $a$ and $b$ are respectively

In the Mean Value Theorem,$f(b) - f(a) = (b - a)f'(c)$. If $a = 4$,$b = 9$,and $f(x) = \sqrt{x}$,then the value of $c$ is:

If the function $f(x)=a x^3+b x^2+11 x-6$ satisfies the conditions of Rolle's theorem in $[1,3]$ and $f^{\prime}\left(2+\frac{1}{\sqrt{3}}\right)=0$,then $a+b=$

If a function $f$ is differentiable on $R$ such that $f^{\prime}(x) \leq 4$ for all $x \in R$; and if $f(2)=-6$ and $f(6)=8$,then the value of $f(4)$ belongs to the interval

Vedclass Test Series

Exam Paper Generator

Online Exam Module