If $f: R \setminus \{0\} \rightarrow R$ is defined by $f(x) = x + \frac{1}{x}$,then the value of $(f(x))^2 =$

Let $f(x)$ and $g(x)$ be two functions given by $f(x) = \frac{2\sin(\pi x)}{x}$ and $g(x) = f(1 - x) + f(x)$. If $g(x) = k f(\frac{x}{2}) f(\frac{1 - x}{2})$,then the value of $k$ is

Let $f(x) = x^2, x \in R$. For any $A \subseteq R$,define $g(A) = \{x \in R : f(x) \in A\}$. If $S = [0, 4]$,then which one of the following statements is not true?

$[t]$ denotes the greatest integer function and $[t-m] = [t] - m$ when $m \in \mathbb{Z}$. If $k = 2[2x - 1] - 1$ and $3[2x - 2] + 1 = 2[2x - 1] - 1$,then the range of $f(x) = [k + 5x]$ is

If $a+\alpha=1, b+\beta=2$ and $af(x)+\alpha f\left(\frac{1}{x}\right)=bx+\frac{\beta}{x}$ for $x \neq 0$,then the value of the expression $\frac{f(x)+f\left(\frac{1}{x}\right)}{x+\frac{1}{x}}$ is ..... .

Let $f(x)$ and $g(x)$ be two continuous functions defined from $R \rightarrow R$,such that $f(x_1) > f(x_2)$ and $g(x_1) < g(x_2)$ for all $x_1 > x_2$. Then the solution set of $f(g(\alpha^2 - 2\alpha)) > f(g(3\alpha - 4))$ is