- Home
- Standard 12
- Mathematics
Consider a quadratic equation $ax^2 + bx + c = 0,$ where $2a + 3b + 6c = 0$ and let $g(x) = a\frac{{{x^3}}}{3} + b\frac{{{x^2}}}{2} + cx.$
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 $[0, 1 ].$
Statement $1$ is false, Statement $2$ is true.
Statement $1$ is true, Statement $2$ is false.
Statement $1$ is true, Statement $2$ is true,Statement $2$ is not a correct explanation for Statement $1.$
Statement $1$ is true, Statement $2$ is true, Statement $2$ is a correct explanation for Statement $1.$
Solution
Let $g\left( x \right) = \frac{{a{x^3}}}{3} + b.\frac{{{x^2}}}{2} + cx$
$g'\left( x \right) = a{x^2} + bx + c$
given:$a{x^2} + bx + c = 0$ and $2a + 3b + 6c = 0$
Statement – $2$:
$(i) g(0) =0$ and $g(1)$
$ = \frac{a}{3} + \frac{b}{2} + c = \frac{{2a + 3b + 6c}}{6}$
$ = \frac{0}{6} = 0$
$ \Rightarrow g\left( 0 \right) = g\left( 1 \right)$
$(ii)$ $g$ is continuose on $[0,1]$ and differentiable on $(0,1)$
$\therefore $ By Rolle's theorem $\exists k \in \left( {0,1} \right)$such that $g'(k)=0$
This holds the statement $2$. Also, from statement – $2$, we can say $a{x^2} + bx + c = 0$ has at least one root in $(0,1)$
Thus, statement – $1$ and $2$ both are true and statement – $2$ is a correct explantion for statement – $1$.
