Question

If $2^{\frac{2\pi }{{\sin }^{-1}x}}-2(a+2)2^{\frac{\pi }{{\sin }^{-1}x}}+8a<0$ for at least one real $x$, then

• A. $\frac{1}{8}\le a<2$
• B. $a<2$
• C. $a\in R-\left\{2\right\}$
• D. $a\in (0,\frac{1}{8}]\cup [2,\infty )$

Answer 1

Answer: (D)

$a\in (0,\frac{1}{8}]\cup [2,\infty )$

Let $2^{\frac{\pi }{{\sin }^{-1}x}}=t$
then

$-\frac{\pi }{2}\le {\sin }^{-1}x\le \frac{\pi }{2}$

Therefore

$t\in (0,2^{-2}]\cup [2^2,\infty )$

Now
$2^{\frac{2\pi }{{\sin }^{-1}x}}-2(a+2)2^{\frac{\pi }{si{n}^{-1}x}}+8a<0$
$\Rightarrow t^2-2(a+2)t+8a<0\Rightarrow t=\frac{2(a+2)\pm \sqrt{{4(a+2)}^2-32a}}{2}=(a+2)\pm (a-2)=2a,4$

$\because t\in (0,2^{-2}]\cup [2^2,\infty )$

$\Rightarrow 2a\in (0,2^{-2}]\cup [2^2,\infty )$

$\Rightarrow a\in (0,2^{-3}]\cup [2,\infty )$

Hence, option D.