Question

The range of $k$ for which the inequality $k{\cos }^{2}x-k\cos x+1\ge 0\forall x$ is

• A. $k<\frac{-1}{2}$
• B. $k>4$
• C. $\frac{-1}{2}\le k\le 4$
• D. $\frac{-1}{2}\le k\le 2$

Answer 1

The correct option is A $k<\frac{-1}{2}$

Given $k{cos}^{2}x-kcosx+1\ge 0$ $\forall x\in (-\infty ,\infty )$

Put $cosx=t$

$\therefore k{t}^2-kt+1\ge 0$ $\forall t\in (-1,1)$

Now, Given expression must always be greater than zero for all values of t including extreme values of t.

Thus, put $t=-1$ in given expression, we get,

$k{(-1)}^2-k(-1)+1\ge 0$

$\therefore k\left(1\right)+k+1\ge 0$

$\therefore k+k+1\ge 0$

$\therefore 2k+1\ge 0$

$\therefore 2k\ge -1$

$\therefore k\ge \frac{-1}{2}$

$\therefore k\in (\frac{-1}{2},\infty )$

Similarly, put $t=1$ in the given expression, we get,

$k{\left(1\right)}^2-k\left(1\right)+1\ge 0$

$k-k+1\ge 0$

$1\ge 0$

This condition is always satisfied for all values of k.

$\therefore k\in (-\infty ,\infty )$

Thus, combining the solution is $k\in (\frac{-1}{2},\infty )$