If ${e^x} = y + \sqrt {1 + {y^2}} $,then $y =$

Let $A = \{1, 2, 3\}$ and $B = \{1, 3, 5\}$. $A$ relation $R: A \to B$ is defined by $R = \{(1, 3), (1, 5), (2, 1)\}$. Then ${R^{-1}}$ is defined by:

Consider $f: \{1, 2, 3\} \rightarrow \{a, b, c\}$ and $g: \{a, b, c\} \rightarrow \{\text{apple, ball, cat}\}$ defined as $f(1)=a, f(2)=b, f(3)=c$ and $g(a)=\text{apple}, g(b)=\text{ball}, g(c)=\text{cat}$. Show that $f, g$ and $g \circ f$ are invertible. Find $f^{-1}, g^{-1}$ and $(g \circ f)^{-1}$ and show that $(g \circ f)^{-1} = f^{-1} \circ g^{-1}$.

Let $g(x)$ be the inverse of an invertible function $f(x)$ which is differentiable at $x = c$. Then $g'(f(c))$ equals:

If $g$ is the inverse of $f$ and $f^{\prime}(x)=\frac{1}{1+x^3}$,then $g^{\prime}(x)$ is

If $[\cdot]$ denotes the greatest integer function and if $f:(5,10) \rightarrow(7,12)$ is a function defined by $f(x)=x+2\left[\frac{x}{5}\right]$,then