A) All real x
B) \[{{x}^{2}}\le 2\]
C) \[{{x}^{2}}\ge 2\]
D) None of these
Correct Answer: B
Solution :
On squaring the given relation \[\sin 2\theta ={{x}^{2}}-1\le 1\Rightarrow {{x}^{2}}\le 2\] or \[-\sqrt{2}\le x\le \sqrt{2}\] \[[\because \,\,\sin 2\theta \le 1]\] Now \[{{\sin }^{6}}\theta +{{\cos }^{6}}\theta \] \[={{({{\sin }^{2}}\theta +{{\cos }^{2}}\theta )}^{3}}-3{{\sin }^{2}}\theta {{\cos }^{2}}\theta ({{\sin }^{2}}\theta +{{\cos }^{2}}\theta )\] \[=1-3{{\sin }^{2}}\theta {{\cos }^{2}}\theta =1-\frac{3}{4}{{\sin }^{2}}2\theta \] \[=1-\frac{3}{4}{{({{x}^{2}}-1)}^{2}}=\frac{1}{4}\{4-3{{({{x}^{2}}-1)}^{2}}\}\] Thus the given result will hold true only when \[{{x}^{2}}\le 2\]and not for all real values of x.
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