A) \[0\le p\le 1\]
B) \[-1\le p\le 3\]
C) \[4\le p\le 6\]
D) \[p\ge 6\]
Correct Answer: B
Solution :
\[\because \,\,\,\sin \,x=\frac{p+3\pm \,\sqrt{{{\left( p+3 \right)}^{2}}-4\times 2(2p-2)}}{4}\] \[\Rightarrow \,\sin \,x=\frac{(p+3)\,\pm (p-5)}{4}\] \[\Rightarrow \,\sin x=\frac{p-1}{2},\,2\] \[\therefore \,\,\sin x=\frac{p-1}{2}(\sin x\ne 2)\] \[\therefore \,\,-1\le \sin \,x\le 1\] \[\Rightarrow \,-1\le \frac{p-1}{2}\le 1\] \[\Rightarrow \,-1\le p\le 3\]
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