Question

If the equation in $x$ given by $2{\left(\frac{1}{{\cos }^{-1}x}\right)}^{2\pi }-a+\frac{1}{2}{\left(2\frac{1}{{\cos }^{-1}x}\right)}^{\pi }$ $-a^2=0$ has only one real solution then exhaustive set of values of ' $a$ ' is

• A. $\left(-3,1\right)$
• B. $\left[-\infty ,-3\right]\cup \left[1,\infty \right]$
• C. $(-\infty ,-3)\cup (1,\infty )$
• D. $\left[-3,\infty \right]$

Answer 1

Answer: (B)

Let $\frac{\pi }{2^{co{s}^{-1}x}}=t\Rightarrow t\ge 2$
equation becomes $t^2-\left(a+\frac{1}{2}\right)t-a^2=0$
has one roots $2$ or greater than $2$ and other root less than $2,f\left(2\right)\le 0$
$\Rightarrow 4-\left(a+\frac{1}{2}\right)2-a^2\le 0$
$a^2+2a-3\ge 0$
$\left(a+3\right)\left(a-1\right)\ge 0$
$a\le -3$ or $a\ge 1$