The graph of function $f$ contains the point $P (1, 2)$ and $Q(s, r)$. The equation of the secant line through $P$ and $Q$ is $y = \left( {\frac{{{s^2} + 2s - 3}}{{s - 1}}} \right)$ $x - 1 - s$. The value of $f ‘ (1)$, is

If $f:R \to R$ and $g:R \to R$ are given by $f(x) = \;|x|$ and $g(x) = \;|x|$ for each $x \in R$, then $\{ x \in R\;:g(f(x)) \le f(g(x))\} = $

The domain of the function $f(x) =\frac{{\,\cot^{-1} \,x}}{{\sqrt {{x^2}\,\, - \,\,\left[ {{x^2}} \right]} }}$ , where $[x]$ denotes the greatest integer not greater than $x$, is :

The range of the function,

$\mathrm{f}(\mathrm{x})=\log _{\sqrt{5}}(3+\cos \left(\frac{3 \pi}{4}+\mathrm{x}\right)+\cos \left(\frac{\pi}{4}+\mathrm{x}\right)+\cos \left(\frac{\pi}{4}-\mathrm{x}\right)$

$-\cos \left(\frac{3 \pi}{4}-\mathrm{x}\right))$ is :

Let $f (x) = a^x (a > 0)$ be written as $f( x) = f_1( x) + f_2( x)$ , where $f_1( x)$ is an even function and $f_2( x)$ is an odd function. Then $f_1( x + y) + f_1( x - y )$ equals

For $x\,\, \in \,R\,,x\, \ne \,0,$ let ${f_0}(x) = \frac{1}{{1 - x}}$ and ${f_{n + 1}}(x) = {f_0}({f_n}(x)),$ $n\, = 0,1,2,....$  Then the value of ${f_{100}}(3) + {f_1}\left( {\frac{2}{3}} \right) + {f_2}\left( {\frac{3}{2}} \right)$ is equal to