Question

The equation ${\sin }^{6}x+{\cos }^{6}x=a^2$ has real solutions if-

• A. $aϵ(-1,1)$
• B. $aϵ[-1,-\frac{1}{2}]$
• C. $aϵ(-\frac{1}{2},\frac{1}{2})$
• D. $aϵ[\frac{1}{2},1]$.

Answer 1

Answer: (B), (D)

The correct options are
B $aϵ[-1,-\frac{1}{2}]$
D $aϵ[\frac{1}{2},1]$.

${\sin }^6\theta +{\cos }^6\theta =1-\frac{3}{4}{\sin }^{2}2\theta$
$\Rightarrow \frac{1}{4}\le {\sin }^6\theta +{\cos }^6\theta \le 1$
$\Rightarrow \frac{1}{4}\le a^2\le 1,aϵ[-1-\frac{1}{2}]\cup [\frac{1}{2},1]$
Hence Options B and D are correct