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 $[0, 1].$

$A$ value of $c$ for which the conclusion of the Mean Value Theorem holds for the function $f(x) = \log_e x$ on the interval $[1, 3]$ is

If $27a + 9b + 3c + d = 0$,then the equation $4ax^3 + 3bx^2 + 2cx + d = 0$ has at least one root between which of the following?

Let $f, g:[-1,2] \rightarrow R$ be continuous functions which are twice differentiable on the interval $(-1,2)$. Let the values of $f$ and $g$ at the points $-1, 0$ and $2$ be as given in the following table:

$x$	$x=-1, 0, 2$
$f(x)$	$3, 6, 0$
$g(x)$	$0, 1, -1$

In each of the intervals $(-1,0)$ and $(0,2)$ the function $(f-3g)^{\prime \prime}$ never vanishes. Then the correct statement$(s)$ is(are):
$(A)$ $f^{\prime}(x)-3g^{\prime}(x)=0$ has exactly three solutions in $(-1,0) \cup (0,2)$
$(B)$ $f^{\prime}(x)-3g^{\prime}(x)=0$ has exactly one solution in $(-1,0)$
$(C)$ $f^{\prime}(x)-3g^{\prime}(x)=0$ has exactly one solution in $(0,2)$
$(D)$ $f^{\prime}(x)-3g^{\prime}(x)=0$ has exactly two solutions in $(-1,0)$ and exactly two solutions in $(0,2)$

Consider $f(x) = |1 - x|$ for $1 \le x \le 2$ and $g(x) = f(x) + b \sin(\frac{\pi}{2}x)$ for $1 \le x \le 2$. Which of the following is correct?

For all real values of $a_{0}, a_{1}, a_{2}, a_{3}$ satisfying $a_{0}+\frac{a_{1}}{2}+\frac{a_{2}}{3}+\frac{a_{3}}{4}=0$, the equation $a_{0}+a_{1} x+a_{2} x^{2}+a_{3} x^{3}=0$ has a real root in the interval