Question

The equation ${\cos }^4\theta +{\sin }^2\theta +\lambda =0$ admits of real solution for $\theta$ if

• A. $\lambda \ge -1$
• B. $\lambda \le -1$
• C. $-\frac{3}{2}\le \lambda \le -\frac{1}{2}$
• D. $-1\le \lambda \le -\frac{3}{4}$

Answer 1

The correct option is D $-1\le \lambda \le -\frac{3}{4}$

${\cos }^4\theta +{\sin }^2\theta +\lambda =0$

$\Rightarrow {\cos }^4\theta +1-{\cos }^2\theta +\lambda =0$

$\Rightarrow \lambda =-1-({\cos }^4\theta -{\cos }^2\theta )=-\frac{3}{4}-{({\cos }^2\theta -\frac{1}{2})}^2$

We know $0\le {\cos }^2\theta \le 1$

Hence for given equation to have solution $-3/4-1/4\le \lambda \le -1\Rightarrow -1\le \lambda \le -3/4$