If $f(a) = \log \left| \frac{1-a}{1+a} \right|$ for $a \neq \{-1, 1\}$,then the set of values of all $a$,for which $f\left( \frac{2a}{1+a^2} \right) > 0$ is

Suppose $f:[-2, 2] \to R$ is defined by $f(x) = \begin{cases} -1 & \text{for } -2 \le x \le 0 \\ x - 1 & \text{for } 0 < x \le 2 \end{cases}$,then find the set $\{ x \in (-2, 2) : x \le 0 \text{ and } f(|x|) = x \}$.

Let $f(x) = \begin{cases} x-1, & x \text{ is even} \\ 2x, & x \text{ is odd} \end{cases}$. If for some $a \in N, f(f(f(a))) = 21$,then $\lim_{x \rightarrow a^{-}} \left\{ \frac{|x|^3}{a} - \left[ \frac{x}{a} \right] \right\}$,where $[t]$ denotes the greatest integer less than or equal to $t$,is equal to:

Let $g(x) = ||x + 2| - 3|$. If $a$ denotes the number of relative minima,$b$ denotes the number of relative maxima,and $c$ denotes the product of the zeroes of $g(x)$,then the value of $(a + 2b - c)$ is:

Let $\alpha, \beta$ and $\gamma$ be three positive real numbers. Let $f(x) = \alpha x^5 + \beta x^3 + \gamma x, x \in \mathbb{R}$ and $g: \mathbb{R} \rightarrow \mathbb{R}$ be such that $g(f(x)) = x$ for all $x \in \mathbb{R}$. If $a_1, a_2, a_3, \dots, a_n$ are in arithmetic progression with mean zero,then the value of $f(g(\frac{1}{n} \sum_{i=1}^{n} f(a_i)))$ is equal to.

Let $f: R \rightarrow (0,1)$ be a continuous function. Then,which of the following function$(s)$ has(have) the value zero at some point in the interval $(0,1)$?