The maximum value of $\frac{\log x}{x}$ is

For the function $f(x) = x \cos \frac{1}{x}, \quad x \geq 1$,consider the following statements:
$(A)$ For at least one $x$ in the interval $[1, \infty), f(x+2)-f(x) < 2$
$(B)$ $\lim _{x \rightarrow \infty} f^{\prime}(x) = 1$
$(C)$ For all $x$ in the interval $[1, \infty), f(x+2)-f(x) > 2$
$(D)$ $f^{\prime}(x)$ is strictly decreasing in the interval $[1, \infty)$
Which of the following combinations of statements is correct?

Find the absolute maximum of $x^{40}-x^{20}$ on the interval $[0,1]$.

If $m$ is the minimum value of $k$ for which the function $f(x) = x\sqrt{kx - x^2}$ is increasing in the interval $[0, 3]$ and $M$ is the maximum value of $f$ in $[0, 3]$ when $k = m$,then the ordered pair $(m, M)$ is equal to

If the local maximum value of the function $f(x)=\left(\frac{\sqrt{3 e}}{2 \sin x}\right)^{\sin ^2 x}, \quad x \in\left(0, \frac{\pi}{2}\right)$,is $\frac{k}{e}$,then $\left(\frac{ k }{ e }\right)^8+\frac{ k ^8}{ e ^5}+ k ^8$ is equal to

Let $P(x) = x^4 + ax^3 + bx^2 + cx + d$ be such that $x = 0$ is the only real root of $P'(x) = 0$. If $P(-1) < P(1)$,then in the interval $[-1, 1]$: