Question

If $2^{\frac{2\pi }{{\sin }^{-1}x}}-2(a+2)2^{\frac{\pi }{{\sin }^{-1}x}}+8a<0$ for atleast 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

The correct option is D $a\in [0,\frac{1}{8})\cup (2,\infty )$

$2^{2\pi /{\sin }^{-1}x}-2(a+2)2^{\pi /{\sin }^{-1}x}+8a<0$

$(2^{\pi /{\sin }^{-1}x}-4)(2^{\pi /{\sin }^{-1}x}-2a)<0$

Now, $2^{\pi /{\sin }^{-1}x}\in (0,\frac{1}{4}]\cup [4,\infty )$

for $2^{\pi /{\sin }^{-1}x}\in (0,\frac{1}{4}]\text{.We have}$

$(2^{\pi /{\sin }^{-1}x}-4)<0$
$\therefore 2^{2\pi /{\sin }^{-1}x}-2a>0$

or$0\le a<\frac{1}{8}$
Similarly, for $2^{\pi /{\sin }^{-1}x}\in [4,\infty ),a>2$
So $a\in [0,\frac{1}{8})\cup (2,\infty )$