For the function $f(x) = x^{40} - x^{20}$,find the absolute minimum value in the interval $[0, 1]$.

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

Divide $10$ into two parts such that the sum of twice the first part and the square of the second part is minimum. The two parts are:

Divide $20$ into two parts such that the product of one part and the cube of the other is maximum. The two parts are

Let $S=(-1, \infty)$ and $f: S \rightarrow R$ be defined as $f(x)=\int_{-1}^x (e^t-1)^{11}(2t-1)^5(t-2)^7(t-3)^{12}(2t-10)^{61} dt$. Let $p$ be the sum of the squares of the values of $x$ where $f(x)$ attains local maxima on $S$,and $q$ be the sum of the values of $x$ where $f(x)$ attains local minima on $S$. Then,the value of $p^2+2q$ is

If $f(x) = 3x + \frac{12}{x}$ is continuous on $R - \{0\}$ and $M$ is its local maximum value,then $\lim_{x \rightarrow M} f(x) = $