Question

Find the true set of values of $p$ for which the equation, $p\cdot 2^{{\cos }^{2}x}+p\cdot 2^{-{\cos }^{2}x}-2=0$ has real roots.

Answer 1

Let $y=2^{{\cos }^{2}x}\Rightarrow y+\frac{1}{y}=\frac{2}{p}$

$\because {\cos }^{2}x\ge 0\Rightarrow y\ge 1$

$\therefore y+\frac{1}{y}\ge 2\Rightarrow \frac{2}{p}\ge 2\Rightarrow p\le 1$

Also, $2^{{\cos }^{2}x}=y\le 2\Rightarrow \frac{2}{p}\le 2+\frac{1}{2}$

$\Rightarrow p\ge \frac{4}{5}\Rightarrow p\in \left[\frac{4}{5},1\right]$