The absolute maximum value of the function $f(x)=2 x^3-9 x^2+12 x+1$ on the interval $[0,2]$ is:

If the distance $s$ covered in time $t$ by a particle moving on a straight line is given by $s = t^5 - 40t^3 + 30t^2 + 80t - 250$,then its minimum acceleration 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?

The function $f(x) = x^2 \log x$ in the interval $(1, e)$ has

If $a, b > 0$,then the minimum value of $y = \frac{b^2}{a-x} + \frac{a^2}{x}$ for $0 < x < a$ is

If the function $f(x) = 2x^3 - 9ax^2 + 12a^2x + 1$,where $a > 0$,attains its maximum and minimum at $p$ and $q$ respectively such that $p^2 = q$,then $a$ equals: