Question

If ${\cos }^{4}x+{\sin }^{2}x-p=0,p\in R$ has real solutions, then

• A. $p\le 1$
• B. $\frac{3}{4}\le p\le 1$
• C. $p\ge \frac{3}{4}$
• D. None of the above.

Answer 1

The correct option is B $\frac{3}{4}\le p\le 1$

${\cos }^{4}x+{\sin }^{2}x-p=0$

$\therefore {\cos }^{2}x{\cos }^{2}x+{\sin }^{2}x-p=0$

$\therefore {\cos }^{2}x(1-{\sin }^{2}x)+{\sin }^{2}x-p=0$

$\therefore {\cos }^{2}x-{\sin }^{2}x{\cos }^{2}x+{\sin }^{2}x-p=0$

$\therefore 1-{\sin }^{2}x{\cos }^{2}x=p$

$\therefore 1-\frac{{\sin }^{2}2x}{4}=p$

$0\le {\sin }^{2}2x\le 1$

$\therefore \frac{3}{4}\le 1-\frac{{\sin }^{2}2x}{4}\le 1$

$\therefore \frac{3}{4}\le p\le 1$