Question

To the equation $2^{2\pi /{\cos }^{-1}x}-(a+\frac{1}{2})2^{\pi /{\cos }^{-1}x}-a^2=0$ has only one real root, then

• A. $1\le a\le 3$
• B. $a\ge 1$
• C. $a\le -3$
• D. $a\ge 3$

Answer 1

Answer: (B), (C)

The correct options are
B $a\ge 1$
C $a\le -3$

Let $2^{\frac{\pi }{{cos}^{-1}x}}=t$
Then $0\le {cos}^{-1}x\le \pi \Rightarrow \frac{1}{\pi }\le \frac{1}{{cos}^{-1}x}\le \infty \Rightarrow 1\le \frac{\pi }{{cos}^{-1}x}\le \infty \Rightarrow 2\le t\le \infty$ ...(1)
$2^{2\pi /{\cos }^{-1}x}-(a+\frac{1}{2})2^{\pi /{\cos }^{-1}x}-a^2=0$
$\Rightarrow t^2-(a+\frac{1}{2})t-a^2=0\Rightarrow t=\frac{(a+\frac{1}{2})\pm \sqrt{{(a+\frac{1}{2})}^2+4a^2}}{2}$
From (1)
$2\le \frac{(a+\frac{1}{2})\pm \sqrt{{(a+\frac{1}{2})}^2+4a^2}}{2}\le \infty$

$\Rightarrow 4\le (a+\frac{1}{2})\pm \sqrt{{(a+\frac{1}{2})}^2+4a^2}\le \infty \Rightarrow \pm \sqrt{{(a+\frac{1}{2})}^2+4a^2}\ge 4-(a+\frac{1}{2})=\frac{7}{2}-a$

$\Rightarrow {(a+\frac{1}{2})}^2+4a^2\ge {(\frac{7}{2}-a)}^2\Rightarrow a^2+\frac{1}{4}+a+4a^2\ge \frac{49}{4}+a^2-7a$

$\Rightarrow 4a^2+8a-12\ge 0\Rightarrow a^2+2a-3\ge 0$
$\Rightarrow (a-1)(a+3)\ge 0\Rightarrow a\le -3$ or $a\ge 1$