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

The function $f(x) = x^{25}(1 - x)^{75}$ for $x \in [0, 1]$ attains its maximum at $x = \dots$

Let a function $f(x)$ be defined as $f(x) = \begin{cases} \cos^{-1}(\mu) + x^2, & 0 < x < 1 \\ 4x, & x \geqslant 1 \end{cases}$. The function $f(x)$ can have a local minimum at $x = 1$ if the value of $\mu$ lies in the interval:

Let the set of all positive values of $\lambda$,for which the point of local minimum of the function $f(x) = 1 + x(\lambda^2 - x^2)$ satisfies $\frac{x^2+x+2}{x^2+5x+6} < 0$,be $(\alpha, \beta)$. Then $\alpha^2 + \beta^2$ is equal to:

If $f(x) = \int\limits_0^x {{e^{\frac{{ - {t^2}}}{2}}}} \left( {1 - {t^2}} \right)\,dt$,then $f(x)$ is minimum at $x = \dots$

The sum of two non-zero numbers is $4$. The minimum value of the sum of their reciprocals is