Question

If $co{s}^{4}x+si{n}^{2}x-p=0,p\in R$ has real solution then

• A. $p\le 1$
• B. $\frac{3}{4}\le p\le 1$
• C. $p\ge \frac{3}{4}$
• D. none of these

Answer 1

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

${sin}^{2}x=1-{cos}^{2}x$
Substitute $t={cos}^{2}x$,
$t^2-t+1-p=0$
For real solution, $D=b^2-4ac\ge 0$
$1-4(1-p)\ge 0$
$1-4+4p\ge$
Hence, $p\ge \frac{3}{4}$
Also,
$p=co{s}^{4}x-co{s}^{2}x+1$
$p=(co{s}^{2}x-\frac{1}{2})+\frac{3}{4}$
The maximum value of $co{s}^{2}x=1$,
Hence, maximum value of $p={(1-\frac{1}{2})}^2+\frac{3}{4}$
Hence, maximum value of $p=1$
Hence, option B is correct.