Question

The equation $\sin ^{4} x+\cos ^{4} x+\sin 2 x+\alpha=0$ is solvable for

• A. $-1 \frac{1}{2} \leq \alpha \leq \frac{1}{2}$
• B. $-3 \leq \alpha \leq 1$
• C. $-\frac{3}{2} \leq \alpha \leq \frac{1}{2}$
• D. $-1 \leq \alpha \leq 1$

Answer 1

Answer: (C)

$\sin ^{4} x+\cos ^{4} x+\sin 2 x+\alpha=0$
$\Rightarrow\left(\sin ^{2} x+\cos ^{2} x\right)^{2}-2 \sin ^{2} x \cos ^{2} x+\sin 2 x+\alpha=0$
$\Rightarrow \sin ^{2} 2 x-2 \sin 2 x-2-2 \alpha=0$
Let $\sin 2 x=y$.
Then the given equation becomes
$y^{2}-2 y-2(1+\alpha)=0$
where $-1, \leq y \leq 1,(\because-1 \leq \sin 2 x \leq 1)$
For real, discriminant $\geq 0 \Rightarrow 3+2 \alpha \geq 0$
$\Rightarrow \alpha \geq-\frac{3}{2}$
Also $-1 \leq y \leq 1$
$\Rightarrow-1 \leq 1-\sqrt{3+2 \alpha} \leq 1$
$\Rightarrow 3+2 \alpha \leq 4$
$\Rightarrow \alpha \leq \frac{1}{2}$.
Thus $-\frac{3}{2} \leq \alpha \leq \frac{1}{2}$