A) \[-3/2\le \alpha \le 1\]
B) \[0\le \alpha \le 1/2\]
C) \[-3/2\le \alpha \le 1/2\]
D) None
Correct Answer: C
Solution :
| \[{{\sin }^{4}}x+{{\cos }^{4}}x+\sin 2x+\alpha =0\] |
| \[\Rightarrow \,\,{{({{\sin }^{2}}x)}^{2}}+{{({{\cos }^{2}}x)}^{2}}+2{{\sin }^{2}}x{{\cos }^{2}}x\]\[-2{{\sin }^{2}}x{{\cos }^{2}}x+\sin 2x+\alpha =0\] |
| \[\Rightarrow \,\,{{\sin }^{2}}2x-2\sin 2x-2-2\alpha =0\] |
| \[\Rightarrow \,\,\sin 2x=\frac{2\pm \sqrt{{{(2)}^{2}}-4\times 1(-2-2\alpha )}}{2\times 1}\]\[=1\pm \sqrt{3+2\alpha }\] |
| If \[\sin 2x=1+\sqrt{3+2\alpha }>1.\] |
| It is not possible. |
| If \[\sin 2x=1-\sqrt{3+2\alpha }\] |
| \[\Rightarrow \,\,-1\le 1-\sqrt{3+2\alpha }\le 1\Rightarrow -\frac{3}{2}\le \alpha \le \frac{1}{2}\] |
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