Question

The equation ${\sin }^{4}x-(k+2){\sin }^{2}x-(k+3)=0$ possesses a solution if

• A. $k>-3$
• B. $k<-2$
• C. $-3\le k\le -2$
• D. $k$ is any positive integer

Answer 1

The correct option is C $-3\le k\le -2$

We have, ${\sin }^{4}x-(k+2){\sin }^{2}x-(k+3)=0$

We have a formula for solving quadratic equation $a{x}^2+bx+c=0$ is

$x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

$\Rightarrow {\sin }^{2}x=\frac{(k+2)\pm \sqrt{{(k+2)}^2+4(k+3)}}{2}=\frac{(k+2)\pm (k+4)}{2}$

$\Rightarrow {\sin }^{2}x=k+3(\because {\sin }^{2}x=-1$ is not possible $)$

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

$\Rightarrow 0\le k+3\le 1$

$\Rightarrow -3\le k\le -2$