Let $R$ be the set of all real number

Statement I The function $f:\left(-\frac{\pi}{2}, \frac{\pi}{2}\right) \rightarrow R$ defined by $f(x)=\sec x+\tan x$ is one-one function.

Statement II The function $f:[0, \infty) \rightarrow R$ defined by $f(x)=x^2$ is a one-one function

Which of the above statements is (are) true?

Let $R$ be the set of all real numbers. Let $f: R \rightarrow R$ be a function defined by

$$ f(x)=\left\{\begin{array}{rcc} 2 x-5, & \text { if } & x<-3 \\ x+2, & \text { if } & -3 \leq x<5 \\ 3 x+1, & \text { if } & x \geq 5 \end{array}\right. $$

Match the following

$$ \begin{array}{llll} \hline & \text { List I } & & \text { List II } \\ \hline \text { A } & f(-5)+f(0)+f(-1)= & \text { I } & 16 \\ \hline \text { B } & f(f(5)+10 f(-3))= & \text { II } & 40 \\ \hline \text { C } & f(|f(-4)|)= & \text { III } & -32 \\ \hline \text { D } & f(f(f(1)))= & \text { IV } & -12 \\ \hline & & \text { V } & 19 \\ \hline \end{array} $$

The domain of the real valued function $f(x)=\frac{\sqrt{6 x^2+5 x-6}}{\sqrt{4-x}-\sqrt{x+4}}$ is

If $[x]$ represents the greatest integer $\leq x$, then the range of the real valued function $f(x)=\frac{1}{\sqrt{[x]^2+[x]-2}}$ is