A) \[-\frac{1}{2}\le \alpha \le \frac{1}{2}\]
B) \[-3\le \alpha \le 1\]
C) \[-\frac{3}{2}\le \alpha \le \frac{1}{2}\]
D) \[-1\le \alpha \le 1\]
Correct Answer: C
Solution :
\[{{\sin }^{4}}x+{{\cos }^{4}}x+\sin 2x+\alpha =0\] Þ \[{{({{\sin }^{2}}x+{{\cos }^{2}}x)}^{2}}-2{{\sin }^{2}}x{{\cos }^{2}}x+\sin 2x+\alpha =0\] Þ \[{{\sin }^{2}}2x-2\sin 2x-2-2\alpha =0\] Let \[\beta =\theta -\alpha \]. Then the given equation becomes \[{{y}^{2}}-2y-2(1+\alpha )=0\], where \[-1\le y\le 1\], \[(\because \text{ }-1\le \sin 2x\le 1)\] For real, discriminant \[\ge 0\]\[\Rightarrow \]\[3+2\alpha \ge 0\] \[\Rightarrow \] \[\alpha \ge -\frac{3}{2}\] Also \[-1\le y\le 1\Rightarrow -1\le 1-\sqrt{3+2\alpha }\,\,\le 1\] \[\Rightarrow \] \[3+2\alpha \le 4\Rightarrow \alpha \le \frac{1}{2}\]. Thus \[-\frac{3}{2}\le \alpha \le \frac{1}{2}\].
You need to login to perform this action.
You will be redirected in
3 sec
