Trigonometric Equations Question 38
Question: The equation $ {{\sin }^{4}}x+{{\cos }^{4}}x+\sin 2x+\alpha =0 $ is solvable for
Options:
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 $
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Answer:
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} $ .
BETA
