Question

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

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

Answer 1

Answer: (C)

We have,${\sin }^{4}x-\left(k+2\right){\sin }^{2}x-\left(k+3\right)=0$
$\Rightarrow {\sin }^{2}x=\frac{\left(k+2\right)\pm \sqrt{{\left(k+2\right)}^2+4\left(k+3\right)}}{2}$
$=\frac{\left(k+2\right)\pm \left(k+4\right)}{2}$
$\Rightarrow {\sin }^{2}x=k+3$
$(\because {\sin }^{2}x=-1$ is not possible)
Since $0\le {\sin }^{2}x\le 1,$
$\therefore 0\le k+3\le 1or-3\le k\le -2$