Which statement among the following is true? (i) The function f ( x ) = x | x | is strictly increasing on R…

$\left(i\right)$ Given that $f\left(x\right) = x\left|x\right|$

Differentiate on both sides

$f^{\text{'}}\left(x\right) = x\frac{\left|x\right|}{x} + \left|x\right|$

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

$f^{\text{'}}\left(x\right) > 0 \forall x \in R - \left\{0\right\}$

So $f\left(x\right)$ is strictly increasing function.

So it's true.

$\left(ii\right)$ $f\left(x\right) = {\log }_{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$4$}\right.}\left(x\right)$

Differentiating on both sides

$f^{\text{'}}\left(x\right) = \frac{1}{x\log \left({\displaystyle \raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$4$}\right.}\right)} = -\frac{1}{x\log \left(4\right)}$

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

So it's false.

$\left(iii\right)$ A one-one function can be either increasing or decreasing but can not be always increasing function.

$\left(iv\right) f\left(x\right) = x^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}$

Differentiating on both sides

$f\text{'}\left(x\right) = \frac{1}{3}{x}^{\raisebox{1ex}{$-2$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}$

$\Rightarrow f\text{'}\left(x\right) = \frac{1}{3{\left[x^{{\displaystyle \raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}}\right]}^2} > 0 \forall x \in R$

So it's always increasing function .