If the function $f(x) = (\frac{1}{x})^{2x}$ for $x > 0$ attains the maximum value at $x = \frac{1}{e}$,then:

Let $f, g$ and $h$ be real-valued functions defined on the interval $[0,1]$ by $f(x)=e^{x^2}+e^{-x^2}$,$g(x)=x e^{x^2}+e^{-x^2}$ and $h(x)=x^2 e^{x^2}+e^{-x^2}$. If $a, b$ and $c$ denote,respectively,the absolute maximum of $f, g$ and $h$ on $[0,1]$,then

The condition for the function $f(x) = x^3 + px^2 + qx + r$ $(x \in R)$ to have no extreme value is:

If $f(x) = 7e^{\sin^2 x} - 7e^{\cos^2 x} + 2$,then $\sqrt{7f_{\min} + f_{\max}}$ is equal to:

If $ax + \frac{b}{x} \ge c$ for all positive $x$,where $a, b > 0$,then:

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