Question

Consider a quadratic equation $a{x}^2+bx+c=0,$ where $2a+3b+6c=0$ and let $g\left(x\right)=a\frac{x^3}{3}+b\frac{x^2}{2}+cx.$
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 $[0,1].$

• A. Statement 1 is false, Statement 2 is true
• B. Statement 1 is true, Statement 2 is false
• C. Statement 1 is true. Statement 2 is true. Statement 2 is not a correct explanation for Statement 1
• D. Statement 1 is true. Statement 2 is true,, Statement 2 is a correct explanation for Statement 1

Answer 1

Answer: (D)

Let $g\left(x\right)=\frac{a{x}^3}{3}+b.\frac{x^2}{2}+cx$
$g^{\prime }\left(x\right)=a{x}^2+bx+c$
Given: $a{x}^2+bx+c=0$ and $2a+3b+6c=0$ Statement-2:
(i) $g(0)=0$ and $g(l)$
$=\frac{a}{3}+\frac{b}{2}+c=\frac{2a+3b+6c}{6}$
$=\frac{0}{0}=0$
$\Rightarrow g\left(0\right)=g\left(1\right)$
(«) g is continuous on $[0,1]$ and differentiable on $(0,1)$
$\therefore$ By Rolle’s theorem $\exists k\in \left(0,1\right)$ such that $g^{\prime }\left(k\right)=0$
This holds the statement 2. Also, from statement-2,we can say $a{x}^2+bx+c=0$ has at least one root in $\left(0,1\right).$
Thus statement-1 and 2 both are true and statement-2 is a correct explanation for statement-1.