Question

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

• A. $−1\frac{1}{2}≤Î\pm ≤\frac{1}{2}$
• B. $−3≤Î\pm ≤1$
• C. $−\frac{3}{2}≤Î\pm ≤\frac{1}{2}$
• D. $−1≤Î\pm ≤1$

Answer 1

Answer: (C)

${\sin }^{4}x+{\cos }^{4}x+\sin 2x+Î\pm =0$
$⇒{\left({\sin }^{2}x+{\cos }^{2}x\right)}^2−2{\sin }^{2}x{\cos }^{2}x+\sin 2x+Î\pm =0$
$⇒{\sin }^{2}2x−2\sin 2x−2−2Î\pm =0$
Let $\sin 2x=y$.
Then the given equation becomes
$y^2−2y−2(1+Î\pm )=0$
where $−1,≤y≤1,(âˆ\mu −1≤\sin 2x≤1)$
For real, discriminant $â‰\yen 0⇒3+2Î\pm â‰\yen 0$
$⇒Î\pm â‰\yen −\frac{3}{2}$
Also $−1≤y≤1$
$⇒−1≤1−\sqrt{3+2Î\pm }≤1$
$⇒3+2Î\pm ≤4$
$⇒Î\pm ≤\frac{1}{2}$.
Thus $−\frac{3}{2}≤Î\pm ≤\frac{1}{2}$