Find the minimum value of $64 \sec x + 27 \csc x$ for $0 < x < \frac{\pi}{2}$.

If $p$ and $q$ are respectively the global maximum and global minimum of the function $f(x) = x^2 e^{2x}$ on the interval $[-2, 2]$,then $p e^{-4} + q e^4 =$

The curve $f(x) = e^x \sin x$ is defined in the interval $[0, 2 \pi]$. The value of $x$ for which the slope of the tangent drawn to the curve at $x$ is maximum,is

Let $f:[-1,1] \rightarrow R$ be defined as $f(x)=ax^{2}+bx+c$ for all $x \in[-1,1],$ where $a, b, c \in R$ such that $f(-1)=2, f^{\prime}(-1)=1$ and for $x \in(-1,1)$ the maximum value of $f^{\prime\prime}(x)$ is $\frac{1}{2}.$ If $f(x) \leq \alpha$ for all $x \in[-1,1],$ then the least value of $\alpha$ is equal to:

If $A = \{x \in R : \frac{\pi}{4} \leq x \leq \frac{\pi}{3}\}$ and $f(x) = \sin x - x$,then $f(A)$ is equal to

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