$f: R \rightarrow R$ and $g:[0, \infty) \rightarrow R$ are defined by $f(x)=x^2$ and $g(x)=\sqrt{x}$. Which one of the following is not true?

If $f(x)$ and $g(x)$ are two functions such that $g(x) = x - \frac{1}{x}$ and $(f \circ g)(x) = x^3 - \frac{1}{x^3}$,then $f'(1)$ is equal to:

If $f(x)=e^x$ and $g(x)=\ln(x)$ for all $x \in [1, \infty)$,then $f \circ g$ is . . . . . .

Let $f, g :(1, \infty) \rightarrow \mathbb{R}$ be defined as $f(x) = \frac{2x+3}{5x+2}$ and $g(x) = \frac{2-3x}{1-x}$. If the range of the function $f \circ g : [2, 4] \rightarrow \mathbb{R}$ is $[\alpha, \beta]$,then $\frac{1}{\beta-\alpha}$ is equal to

Let $S, T, U$ be three non-void sets and $f: S \rightarrow T, g: T \rightarrow U$ and the composed mapping $g \circ f: S \rightarrow U$ be defined. If $g \circ f$ is an injective mapping, then:

If $f(x) = \begin{cases} \sin x, & x \ne n\pi, n \in \mathbb{Z} \\ 2, & \text{otherwise} \end{cases}$ and $g(x) = \begin{cases} x^2 + 1, & x \ne 0, 2 \\ 4, & x = 0 \\ 5, & x = 2 \end{cases}$,then $\lim_{x \to 0} g(f(x))$ is