On the interval $[0,1]$,the function $f(x) = x^{25}(1-x)^{75}$ takes its maximum value at the point

If $y = a \log |x| + b x^2 + x$ has its extremum values at $x = -1$ and $x = 2$,then

Two particles move in the same straight line starting at the same moment from the same point in the same direction. The first moves with constant velocity $u$ and the second starts from rest with constant acceleration $f$. Then,

${a_1, a_2, ....., a_n, .....}$ is a progression where $a_n = \frac{n^2}{n^3 + 200}$. The largest term of this progression is

Let a function $f(x)$ be continuous in an interval $[a, b]$. Let $\delta > 0$ be a very small real number. Let $c \in (a, b)$ be such that $f(c - \delta) < f(c)$ and $f(c + \delta) < f(c)$ for every $\delta > 0$. Let $(f(\alpha - \delta) - f(\alpha))(f(\alpha + \delta) - f(\alpha)) < 0$ for all $\alpha \in (a, b)$ and $\alpha \neq c$. Then:

The function $f(x) = 2x^3 - 21x^2 + 36x + 7$ has a local maximum at $x = $ ........