A) \[k>-3\]
B) \[k<-2\]
C) \[-3\le k\le -2\]
D) k is any positive integer
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 \[-3\le k\le -2\]
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