The minimum value of $\frac{\log x}{x}$ in the interval $(2, \infty)$ is

The maximum value of $f(x) = x^2 \log x$ on the interval $[1, e]$ is:

Observe the statements given below :
Assertion $(A)$ : $f(x)=x e^{-x}$ has the maximum at $x=1$
Reason $(R)$ : $f^{\prime}(1)=0$ and $f^{\prime \prime}(1) < 0$
Which of the following is correct?

$A(1,15), B(3,-12), C(6,12)$ are three consecutive turning points of a continuous curve $y=f(x)$. If $f(x)=0$ only for $x=\alpha$ and $x=\beta$,then $|\beta-\alpha| < $

$A$ particle moving in a straight line starts from rest and the acceleration at any time $t$ is $a - kt^2$, where $a$ and $k$ are positive constants. The maximum velocity attained by the particle 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: