Consider the quadratic equation ax^2+bx+c=0,w | vedclass

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

Question

(D) Given $g(x) = \frac{ax^3}{3} + \frac{bx^2}{2} + cx$. Calculating $g(0)$ and $g(1)$: $g(0) = 0$. $g(1) = \frac{a}{3} + \frac{b}{2} + c = \frac{2a+3

Vedclass · Accepted Answer

Statement-$I$ is true,Statement-$II$ is true and Statement-$II$ is a correct explanation of Statement-$I$

Vedclass · Answer

(D) Given $g(x) = \frac{ax^3}{3} + \frac{bx^2}{2} + cx$. Calculating $g(0)$ and $g(1)$: $g(0) = 0$. $g(1) = \frac{a}{3} + \frac{b}{2} + c = \frac{2a+3b+6c}{6}$. Since $2a+3b+6c=0$,we have $g(1) = 0$. Since $g(x)$ is a polynomial,it is continuous on $[0,1]$ and differentiable on $(0,1)$. Since $g(0) = g(1) = 0$,by Rolle's theorem,there exists at least one $c_1 \in (0,1)$ such that $g'(c_1) = 0$. Note that $g'(x) = ax^2+bx+c$. Thus,$g'(c_1) = ac_1^2+bc_1+c = 0$. This implies that the quadratic equation $ax^2+bx+c=0$ has at least one root $c_1 \in (0,1)$. Therefore,Statement-$I$ is true. Statement-$II$ is also true because $g(x)$ satisfies all conditions of Rolle's theorem on $[0,1]$,and it is the correct explanation for Statement-$I$.

Consider the quadratic equation $ax^2+bx+c=0$,where $2a+3b+6c=0$ and let $g(x)=\frac{ax^3}{3}+\frac{bx^2}{2}+cx$.
Statement-$I$ : The given quadratic equation $ax^2+bx+c=0$ has at least one root in $(0,1)$.
Statement-$II$ : Rolle's theorem is applicable to $g(x)$ on $[0,1]$.
Then

Similar Questions

If Rolle's theorem holds for the function $f(x) = 2x^3 + ax^2 + bx$ in the interval $[-1, 1]$ for the point $c = \frac{1}{2}$,then the value of $2a + b$ is

The value of $c$ for the function $f(x) = \log x$ on $[1, e]$ if Lagrange's Mean Value Theorem $(LMVT)$ is applied,is

Suppose that $f$ is differentiable for all $x$ and that $f'(x) \le 2$ for all $x$. If $f(1) = 2$ and $f(4) = 8$,then $f(2)$ has the value equal to

Let $f(x)$ satisfy all the conditions of the Mean Value Theorem in $[0, 2]$. If $f(0) = 0$ and $|f'(x)| \le \frac{1}{2}$ for all $x$ in $[0, 2]$,then:

If $f(x)=|x-2|, x \in[0,4]$ then the Rolle's theorem cannot be applied to the function because

Vedclass Test Series

Exam Paper Generator

Online Exam Module

Consider the quadratic equation $ax^2+bx+c=0$,where $2a+3b+6c=0$ and let $g(x)=\frac{ax^3}{3}+\frac{bx^2}{2}+cx$.Statement-$I$ : The given quadratic equation $ax^2+bx+c=0$ has at least one root in $(0,1)$.Statement-$II$ : Rolle's theorem is applicable to $g(x)$ on $[0,1]$.Then