Question

The equation ${\cos }^{4}x-{\sin }^{4}x+\cos 2x+{\alpha }^2+\alpha =0,$ will have at least one solution if:

• A. $-2\le \alpha \le 2$
• B. $-3\le \alpha \le 1$
• C. $-2\le \alpha \le 1$
• D. $-1\le \alpha \le 2$

Answer 1

The correct option is C $-2\le \alpha \le 1$

Given equation can be written as,
$(co{s}^{2}x-si{n}^{2}x)(co{s}^{2}x+si{n}^{2}x)+cos2x+{\alpha }^2+\alpha =0$
$\Rightarrow 2cos2x+{\alpha }^2+\alpha =0$
this equation have at least one solution only if,
$-2\le {\alpha }^2+\alpha \le 2$
$i.e.{\alpha }^2+\alpha \le 2....\left(1\right)$ and ${\alpha }^2+\alpha \ge -2.....\left(2\right)$
solving (1)
$(\alpha -1)(\alpha +2)\le 0\Rightarrow -2\le \alpha \le 1$
now solving (2)
${\alpha }^2+\alpha +2\ge 0$, This is true for every real value of $\alpha$
combining both, final answer is
$-2\le \alpha \le 1$
Hence, option 'C' is correct.