Domain of the function $f(x) = \sqrt {2 - {{\sec }^{ - 1}}x} $ is
$\left( { - \infty , - 1} \right] \cup \left[ {1,\infty } \right)$
$\left( { - \infty , - 1} \right] \cup \left[ {\sec 1,\infty } \right)$
$\left( { - \infty ,\sec 2} \right] \cup \left[ {1,\infty } \right)$
$\left( { - \infty ,\sec 2} \right] \cup \left[ {\sec 1,\infty } \right)$
Statement $-1$ : The equation $x\, log\, x = 2 - x$ is satisfied by at least one value of $x$ lying between $1$ and $2$
Statement $-2$ : The function $f(x) = x\, log\, x$ is an increasing function in $[1, 2]$ and $g (x) = 2 -x$ is a decreasing function in $[ 1 , 2]$ and the graphs represented by these functions intersect at a point in $[ 1 , 2]$
If $f(x)$ satisfies the relation $f\left( {\frac{{5x - 3y}}{2}} \right) = \frac{{5f(x) - 3f(y)}}{2}\forall x,y\, \in \,R$ and $f(0)=1, f'(0)=2$ then the period of $sin(f(x))$ is
Range of the function , $f (x) = cot ^{-1}$ $\left( {{{\log }_{4/5}}\,\,(5\,{x^2}\,\, - \,\,8\,x\,\, + \,\,4)\,} \right)$ is :
The range of values of the function $f\left( x \right) = \frac{1}{{2 - 3\sin x}}$ is
The value of $b$ and $c$ for which the identity $f(x + 1) - f(x) = 8x + 3$ is satisfied, where $f(x) = b{x^2} + cx + d$, are