Question

If the equation ${\sin }^{4}x-(k+2){\sin }^{2}x-(k+3)=0$ has a solution, then $k$ must lie in the interval

• A. $(-4,-2)$
• B. $[-3,2)$
• C. $(-4,-3)$
• D. $[-3,-2)$

Answer 1

The correct option is D $[-3,-2)$

$Si{n}^{4}x-(k+2)si{n}^{2}x-(k+3)=0$

Let $si{n}^{2}x=y$

$y^2-(k+2)y-k+3=0$

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

$y=\frac{k+2\pm \sqrt{k^2+4+4k+12}}{2}$

$y=\frac{k+2\pm \sqrt{k^2+8k+16}}{2}$

$y=\frac{(k+2)\pm k+4}{2}$

$y=\frac{k+2-k-4}{2};y=\frac{k+2+k+4}{2}$

$y=-1,y=k+3$

$si{n}^{2}x=-1$ $si{n}^{2}x=k+3$

Not possible $sinx=\sqrt{k+3}$

($\because$ It is given that

the ep. has sol)

$0\le sinx\le 1$

$sinx=0$

$0=k+3$

$k=-3$

Min value of $k=-3$

$sinx=1$

$1=k+3$

$-2=k$

Max value of $k=-2$

$\therefore kliesininterval[-3,-2)$