The maximum value of the function $f(x)=2x^3-15x^2+36x-48$ on the set $A=\{x \mid x^2+20 \leq 9x\}$ is

If the absolute maximum value of the function $f(x) = (x^2 - 2x + 7) e^{(4x^3 - 12x^2 - 180x + 31)}$ in the interval $[-3, 0]$ is $f(\alpha)$,then:

If $4+3x-7x^2$ attains its maximum value $M$ at $x=\alpha$ and $5x^2-2x+1$ attains its minimum value $m$ at $x=\beta$, then $\frac{28(M-\alpha)}{5(m+\beta)}=$

The points on the curve $y = 12x - x^3$ at which the gradient is zero are

The function $S(x) = \int\limits_0^x {\sin \left( {\frac{{\pi {t^2}}}{2}} \right)\,dt} $ has two critical points in the interval $[1, 2.4]$. One of the critical points is a local minimum and the other is a local maximum. The local minimum occurs at $x =$

The vertices of a triangle are $(0, 0)$,$(x, \cos x)$,and $(\sin^3 x, 0)$ where $0 < x < \frac{\pi}{2}$. The maximum area for such a triangle in sq. units is: