The set of all $a \in R$ for which the equation $x|x-1|+|x+2|+a=0$ has exactly one real root is:

Let $S$ be a finite set. Then a non-identity function $f: S \rightarrow S$ can be

$f: R \rightarrow R$ is a function defined by $f(x) = \frac{1}{e^x + 2e^{-x}}$. Assertion $(A): f(c) = \frac{1}{3}$ for some values of $c \in R$. Reason $(R): 0 < f(x) \leq \frac{1}{2\sqrt{2}}$ for all $x \in R$. Then which of the following options is correct?

Consider a function $f: R \to R$ such that $f(x + a) = \frac{1}{2} + \sqrt{f(x) - f^2(x)}$,where $a$ is a real constant. Then $f(x)$ must be

If $S$ is a set of polynomials $P(x)$ of degree $\le 2$ such that $P(0) = 0$,$P(1) = 1$,and $P'(x) > 0$ for all $x \in (0, 1)$,then which of the following describes $S$?

Let $g(x) = 1 + x - [x]$ and $f(x) = \begin{cases} -1, & x < 0 \\ 0, & x = 0 \\ 1, & x > 0 \end{cases}$,where $[x]$ denotes the greatest integer less than or equal to $x$. Then for all $x$,$f(g(x)) = $