If the function $f(x) = x^3 - 3(a - 2)x^2 + 3ax + 7$,for some $a \in R$,is increasing in $(0, 1]$ and decreasing in $[1, 5)$,then a root of the equation $\frac{f(x) - 14}{(x - 1)^2} = 0$ $(x \neq 1)$ is

The function $f(x) = x^3 + ax^2 + bx + c$ where $a^2 \leq 3b$ has:

$f(x) = \int\limits_0^x t(t - 1)(t - 2) dt$ takes on its minimum value when:

Let $f(x)$ be a polynomial of degree $4$ having extreme values at $x = 1$ and $x = 2$. If $\mathop {\lim }\limits_{x \to 0} \left( {\frac{{f(x)}}{{{x^2}}} + 1} \right) = 3$,then $f(-1)$ is equal to

Let $f(x) = x^3 + px + 1$ and consider the following three statements:
$(i)$ For $p \geqslant 0$,$f(x) = 0$ has one negative root and $f(x)$ is monotonic.
$(ii)$ For $-1 < p < 0$,$f(x) = 0$ has one negative root and $f(x)$ is non-monotonic.
$(iii)$ For $p < -3/\sqrt[3]{4}$,$f(x) = 0$ has three real and distinct roots.
Which of the following is correct?

The maximum value of $f(x) = (x + 1)^{\frac{1}{3}} - (x - 1)^{\frac{1}{3}}$ for $x \in [0, 1]$ is ....