Q48 / 90 · JEE Main 2012 (19 May Online) · Mathematics

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 $[0,1]$.

  • 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.
  • Correct answer
    D
    Statement 1 is true, Statement 2 is true, , Statement 2 is a correct explanation for Statement 1.

Solution

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.

Hence, the correct answer is option: 'D'.

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