The value $C$ of the Lagrange's mean value theorem for the function $f(x)=x(x-1)(x-2)$ in the interval $[0, 1/2]$ is

If $2a + 3b + 6c = 0$,then at least one root of the equation $ax^2 + bx + c = 0$ lies in which interval?

If $f(x) = x^{3}$ and $g(x) = x^{3} - 4x$ in the interval $[-2, 2]$,consider the following statements:
$(a)$ $f(x)$ and $g(x)$ satisfy the Mean Value Theorem.
$(b)$ $f(x)$ and $g(x)$ both satisfy Rolle's Theorem.
$(c)$ Only $g(x)$ satisfies Rolle's Theorem.
Which of these statements is correct?

Let $f$ and $g$ be twice differentiable even functions on $(-2, 2)$ such that $f(\frac{1}{4}) = 0, f(\frac{1}{2}) = 0, f(1) = 1$ and $g(\frac{3}{4}) = 0, g(1) = 2$. Then,the minimum number of solutions of $f(x)g''(x) + f'(x)g'(x) = 0$ in $(-2, 2)$ is equal to

For the Mean Value Theorem $f(b) - f(a) = (b - a) f'(x_1)$ where $a < x_1 < b$,if $f(x) = 1/x$,then $x_1 = ?$

If $2a + 3b + 6c = 0$,then at least one root of the equation $ax^2 + bx + c = 0$ lies in the interval: