Let $f (x)$ and $g (x)$ be two continuous functions defined from $R \rightarrow R$, such that $f (x_1) > f (x_2)$ and $g (x_1) < g (x_2), \forall x_1 > x_2$ , then solution set of $f\,\left( {\,g({\alpha ^2} – 2\alpha )\,} \right) >f\,\left( {\,g(3\alpha – 4)\,} \right)$ is

Let $f$ be any function defined on $R$ and let it satisfy the condition

$|f( x )-f( y )| \leq\left|( x – y )^{2}\right|, \forall( x , y ) \in R$ If $f(0)=1,$ then

 $(i)$  $f (x)$ is continuous and defined for all real numbers

$(ii)$ $f '(-5) = 0 \,; \,f '(2)$ is not defined and $f '(4)  = 0$

$(iii)$ $(-5, 12)$ is a point which lies on the graph of $f (x)$

$(iv)$ $f ''(2)$ is undefined, but $f ''(x)$ is negative everywhere else.

$(v)$ the signs of  $f '(x)$ is given below

Possible graph of $y = f (x)$ is

In the mean value theorem, $f(b) – f(a) = (b – a)f'(c) $ if $a = 4$, $b = 9$ and $f(x) = \sqrt x $ then the value of $c$  is

From mean value theorem $f(b) – f(a) = $ $(b – a)f'({x_1});$   $a < {x_1} < b$ if $f(x) = {1 \over x}$, then ${x_1} = $