Consider a quadratic equation ax^2 + bx + c = | vedclass

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

Question

(D) Let $g(x) = \frac{ax^3}{3} + \frac{bx^2}{2} + cx.$Then $g'(x) = ax^2 + bx + c.$We are given $2a + 3b + 6c = 0.$Statement $2:$$g(0) = 0.$$g(1) = \f

Vedclass · Accepted Answer

Statement $1$ is true,Statement $2$ is true,Statement $2$ is a correct explanation for Statement $1.$

Vedclass · Answer

(D) Let $g(x) = \frac{ax^3}{3} + \frac{bx^2}{2} + cx.$Then $g'(x) = ax^2 + bx + c.$We are given $2a + 3b + 6c = 0.$Statement $2:$$g(0) = 0.$$g(1) = \frac{a}{3} + \frac{b}{2} + c = \frac{2a + 3b + 6c}{6} = \frac{0}{6} = 0.$Since $g(0) = g(1) = 0$ and $g(x)$ is a polynomial,it is continuous on $[0, 1]$ and differentiable on $(0, 1).$By Rolle's theorem,there exists at least one $k \in (0, 1)$ such that $g'(k) = 0.$Thus,Statement $2$ is true.Statement $1:$Since $g'(k) = ak^2 + bk + c = 0$ for some $k \in (0, 1),$ the quadratic equation $ax^2 + bx + c = 0$ has at least one root in $(0, 1).$Thus,Statement $1$ is true and Statement $2$ is the correct explanation for Statement $1.$

Similar Questions

Verify the Mean Value Theorem for the function $f(x) = x^{3} - 5x^{2} - 3x$ in the interval $[1, 3]$. Find all $c \in (1, 3)$ such that $f^{\prime}(c) = \frac{f(3) - f(1)}{3 - 1}$.

For all twice differentiable functions $f: \mathbb{R} \rightarrow \mathbb{R},$ with $f(0)=f(1)=f^{\prime}(0)=0,$ which of the following is true?

The function $f(x) = x^3 - 6x^2 + ax + b$ satisfies the conditions of Rolle's theorem in $[1, 3]$. Then the values of $a$ and $b$ are respectively

If $f$ is a differentiable function such that $f(2x + 1) = f(1 - 2x)$ for all $x \in R$,then the minimum number of roots of the equation $f'(x) = 0$ in $x \in (-5, 10)$,given that $f(2) = f(5) = f(10)$,is:

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

Vedclass Test Series

Exam Paper Generator

Online Exam Module