Question

Let $f\left(\right.x\left.\right)=\left[\frac{1}{c o s \left\{\right. x \left.\right\}}\right]$ and $g\left(\right.x\left.\right)=2x^{2}-3x\left(\right.k+1\left.\right)+k\left(\right.3k+1\left.\right)$ (where $\left[\right.x\left]\right.\&\left\{\right.x\left.\right\}$ denote the greatest integer and fractional part function respectively) are two real valued function. If $g\left(\right.f\left(\right.x\left.\right)\left.\right) < 0,$ $\forall x\in R$ , then the number of integral value of $k$ is :

Answer 1

$\because f(x)=\left[\frac{1}{\cos \{x\}}\right]=1 $
$\therefore g(f(x))=g(1) < 0 $
$\Rightarrow 2-3(k+1)+k(3 k+1)<0 $
$3 k^2-2 k-1<0$
$\Rightarrow k \in\left(-\frac{1}{3}, 1\right)$
$\therefore$ The integral value of $k$ is $0$