If $f(x)=\left(\sin ^4 x+\cos ^4 x\right), 0< x<\frac{\pi}{2}$, then the function has minimum value at $x=$

$\lim _\limits{x \rightarrow 0} \frac{(1-\cos 2 x)(3+\cos x)}{x \tan 4 x}$ has the value

Let $a, b \in(a \neq 0)$. If the function $f$ is defined as

$$f(x)=\left\{\begin{array}{cc} \frac{2 x^2}{\mathrm{a}} & , 0 \leq x<1 \\ \mathrm{a} & , 1 \leq x<\sqrt{2} \\ \frac{2 \mathrm{~b}^2-4 b}{x} & , \sqrt{2} \leq x<\infty \end{array}\right.$$

is continuous in the interval $[0, \infty)$, then an ordered pair $(a, b)$ is

If the function $\mathrm{f}(x)=\left(\frac{5 x-8}{8-3 x}\right)^{\frac{3}{2 x-4}}$ if $x \neq 2$. $=\mathrm{k}$ if $x=2$. is continuous at $x=2$, then $\mathrm{k}=$