Let $f: R - \left\{-\frac{1}{2}\right\} \rightarrow R$ be defined by $f(x) = \frac{x-2}{2x+1}$. If $\alpha$ and $\beta$ satisfy the equation $f(f(x)) = -x$,then $4(\alpha^2 + \beta^2) = $

Let $f(x)=\log (\sin x), 0 < x < \pi$ and $g(x)=\sin ^{-1}(e^{-x}), x \geq 0$. If $\alpha$ is a positive real number such that $a=(f \circ g)^{\prime}(\alpha)$ and $b=(f \circ g)(\alpha)$,then

If $f(x) = \sin^2 x + \sin^2(x + \frac{\pi}{3}) + \cos x \cos(x + \frac{\pi}{3})$ and $g(\frac{5}{4}) = 1$,then $(g \circ f)(x) = $

If $f(x)=\frac{3x+4}{5x-7}$ and $g(x)=\frac{7x+4}{5x-3}$,then $f(g(x))=$

If $g(x) = x^2 + x - 2$ and $\frac{1}{2}g(f(x)) = 2x^2 - 5x + 2$,then $f(x)$ is

If $f(x) = \begin{cases} \sin x, & x \neq n\pi, n \in \mathbb{Z} \\ 0, & \text{otherwise} \end{cases}$ and $g(x) = \begin{cases} x^2 + 1, & x \neq 0, 2 \\ 4, & x = 0 \\ 5, & x = 2 \end{cases}$,then $\lim_{x \to 0} g(f(x)) = $