Question

Let $f\left(x\right)=\left[\frac{1}{cos\left\{x\right\}}\right]$ (where [.] and {.} are greatest integer function and fractional part function respectively) and g(x) = $k^2{x}^2$ – 11kx + 24. If g(f(x)) < 0, $\forall xϵR$, then number of integral values of k is

• A. 3
• B. 4
• C. 5
• D. 6

Answer 1

The correct option is B 4

$f\left(x\right)=1,\forall xϵR$
g(f(x)) = g(1) < 0
$\therefore k^2$ – 11k + 24 < 0
$\Rightarrow$ (k – 3) (k – 8) < 0
$\Rightarrow$ 3 < k < 8