Let $\mathrm{ℝ}$ be the set of all real numbers.

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

Statement II : The function $f:[0,\infty )\to \mathrm{ℝ}$ defined by $f\left(x\right)=x^2$ is a one-one function.

Which of the above statements is(are) true?

Given,

$\mathrm{ℝ}$ be the set of all real numbers.

Now solving statement I : The function $f:\left(-\frac{\pi }{2},\frac{\pi }{2}\right)\to \mathrm{ℝ}$ defined by $f\left(x\right)=\sec x+\tan x$

Now differentiating the function with respect to $x$ we get,

$f^{\text{'}}\left(x\right)=\sec x\left(\sec x+\tan x\right)$

Now we know that $\sec x$ is always positive in given domain and $\sec x+\tan x$ will also be positive as $\tan x$ is negative in $\left(-\frac{\mathrm{\pi }}{2},0\right)$ but $\sec x\ge \tan x$ in $\left(-\frac{\mathrm{\pi }}{2},0\right)$

So, $f^{\text{'}}\left(x\right)\ge 0$ hence it is a one - one function.

Now solving statement II : The function $f:[0,\infty )\to \mathrm{ℝ}$ defined by

$f\left(x\right)=x^2$

Differentiating the function we get,

$f^{\text{'}}\left(x\right)=2x$

$\Rightarrow f^{\text{'}}\left(x\right)\ge 0 \forall x\in [0,\infty )$

Hence it is also a one-one function.

So both are true.