Trigonometric Identities Question 263
Question: The equation $ {{\sin }^{4}}x-(k+2){{\sin }^{2}}x-(k+3)=0 $ possesses a solution if
Options:
A) $ k>-3 $
B) $ k<-2 $
C) $ -3\le k\le -2 $
D) k is any positive integer
Show Answer
Answer:
Correct Answer: C
Solution:
We have, $ {{\sin }^{4}}x-(k+2){{\sin }^{2}}x-(k+3)=0 $
$ \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, $
$ \therefore ,0\le k+3\le 1 $ or $ -3\le k\le -2 $
BETA
