If $f(x)=\sqrt{x}$ $(x \geq 0)$ and $g(x)=1+x^2$,then $(f \circ g)^{\prime}(1)=$

If $f$ and $g$ are increasing and decreasing functions respectively from $[0, \infty)$ to $[0, \infty)$,and $h(x) = f(g(x))$ with $h(0) = 0$,then $h(x) - h(1)$ is:

Give examples of two functions $f: N \rightarrow Z$ and $g: Z \rightarrow Z$ such that $g \circ f$ is injective but $g$ is not injective. (Hint: Consider $f(x) = x$ and $g(x) = |x|$)

For $x \in R$,two real-valued functions $f(x)$ and $g(x)$ are such that $g(x) = \sqrt{x} + 1$ and $(f \circ g)(x) = x + 3 - \sqrt{x}$. Then $f(0)$ is equal to:

$f(x) = \frac{1}{x}$ and $g(x) = \frac{1}{\sqrt{x}}$,then:

Let $f: R \rightarrow R$ be defined as $f(x) = 2x - 1$ and $g: R - \{1\} \rightarrow R$ be defined as $g(x) = \frac{x - 1/2}{x - 1}$. Then the composition function $f(g(x))$ is: