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Consider a quadratic equation $a x^2+b x+c=0$, where $2 a+3 b+6 c=0$ and let $g(x)=a \frac{x^3}{3}+b \frac{x^2}{2}+c x$.
Statement 1: The quadratic equation has at least one root in the interval $(0,1)$.
Statement 2: The Rolle's theorem is applicable to function $g(x)$ on the interval $[0,1]$.
Options:
Statement 1: The quadratic equation has at least one root in the interval $(0,1)$.
Statement 2: The Rolle's theorem is applicable to function $g(x)$ on the interval $[0,1]$.
Solution:
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Verified Answer
The correct answer is:
Statement 1 is true, Statement 2 is true, , Statement 2 is a correct explanation for Statement 1.
Statement 1 is true, Statement 2 is true, , Statement 2 is a correct explanation for Statement 1.
Let $g(x)=\frac{a x^3}{3}+b \cdot \frac{x^2}{2}+c x$
$$
\mathrm{g}^{\prime}(x)=a x^2+b x+c
$$
Given: $a x^2+b x+c=0$ and $2 a+3 b+6 c=0$
Statement-2:
(i) $g(0)=0$ and $g(1)$
$$
\begin{aligned}
& =\frac{a}{3}+\frac{b}{2}+c=\frac{2 a+3 b+6 c}{6} \\
& =\frac{0}{6}=0 \\
& \Rightarrow g(0)=g(1)
\end{aligned}
$$
(ii) $g$ is continuous on $[0,1]$ and differentiable on $(0,1)$
$\therefore$ By Rolle's theorem $\exists k \in(0,1)$ such that $g^{\prime}(k)=0$
This holds the statement 2. Also, from statement-2, we can say $a x^2+b x+c=0$ has at least one root in $(0,1)$.
Thus statement-1 and 2 both are true and statement-2 is a correct explanation for statement-1.
$$
\mathrm{g}^{\prime}(x)=a x^2+b x+c
$$
Given: $a x^2+b x+c=0$ and $2 a+3 b+6 c=0$
Statement-2:
(i) $g(0)=0$ and $g(1)$
$$
\begin{aligned}
& =\frac{a}{3}+\frac{b}{2}+c=\frac{2 a+3 b+6 c}{6} \\
& =\frac{0}{6}=0 \\
& \Rightarrow g(0)=g(1)
\end{aligned}
$$
(ii) $g$ is continuous on $[0,1]$ and differentiable on $(0,1)$
$\therefore$ By Rolle's theorem $\exists k \in(0,1)$ such that $g^{\prime}(k)=0$
This holds the statement 2. Also, from statement-2, we can say $a x^2+b x+c=0$ has at least one root in $(0,1)$.
Thus statement-1 and 2 both are true and statement-2 is a correct explanation for statement-1.
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