Let $f: R \rightarrow R$ be defined by $f(x) = 2x + 6$,which is a bijective mapping. Then $f^{-1}(x)$ is given by:

If a function $f:(-1,1) \rightarrow B(\subseteq R)$ is defined as $f(x)=x+x^2+x^3+\ldots \infty$, then in order to have the inverse function of $f$, $B$ is equal to

$f: R \rightarrow R, f(x) = 3x + 2$ and $g: R \rightarrow R, g(x) = 6x + 5$. Find the value of $(g \circ f^{-1})(10)$.

If the function $f:[1, \infty) \to [1, \infty)$ is defined by $f(x) = 2^{x(x - 1)}$,then $f^{-1}(x)$ is

Let $f(x)=(x+1)^2-1$ for $x \geq -1$.
Statement-$1$: $S=\{x:f(x)=f^{-1}(x)\}=\{0, -1\}$
Statement-$2$: $f$ is a bijection.

Let $f: R \rightarrow R$ and $g: R \rightarrow R$ be differentiable functions such that $(f \circ g)(x) = x$. If $f(x) = 2x + \cos x + \sin^2 x$, then the value of $\sum_{n=1}^{99} g(1 + (2n - 1) \pi)$ is