If the function $$f(x)=\left(\frac{1}{x}\right)^{2 x} ; x>0$$ attains the maximum value at $$x=\frac{1}{\mathrm{e}}$$ then :

Let $$f(x)=\frac{1}{7-\sin 5 x}$$ be a function defined on $$\mathbf{R}$$. Then the range of the function $$f(x)$$ is equal to :

The function $$f(x)=\frac{x^2+2 x-15}{x^2-4 x+9}, x \in \mathbb{R}$$ is

Let $$f, g: \mathbf{R} \rightarrow \mathbf{R}$$ be defined as :

$$f(x)=|x-1| \text { and } g(x)= \begin{cases}\mathrm{e}^x, & x \geq 0 \\ x+1, & x \leq 0 .\end{cases}$$

Then the function $$f(g(x))$$ is