Question

Consider a quadratic equation $ax^2 + bx + c = 0,$ where $2a + 3b + 6c = 0$ and let $g\left(x\right)=a \frac{x^{3}}{3}+b \frac{x^{2}}{2}+cx.$ The quadratic equation has at least one root in the interval $(0,1).$ The Rolle?? 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)

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

Answer 2

Let $g\left(x\right)=\frac{ax^{3}}{3}+b.\frac{x^{2}}{2}+cx$ $g '\left(x\right)=ax^{2}+bx+c$ Given: $ax^{2} + bx+c=0$ and $2a+3 b + 6c=0$ 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?? theorem $\exists\,k\,\in\left(0, 1\right)$ such that $g' \left(k\right)=0$ This holds the statement 2. Also, from statement-2,we can say $ax^{2}+bx+c=0$ 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.