A) \[(b,c)\]
B) \[[b,c]\]
C) \[(-\infty ,b]\cup [c,\infty )\]
D) \[(-\infty ,b)\cup (c,\infty )\]
Correct Answer: C
Solution :
Let \[y=\frac{{{x}^{2}}-bc}{2x-b-c}\] \[\Rightarrow \,\,\,{{x}^{2}}-2yx+(b+c)y-bc=0\] \[\because \,\,x\,\in \,R,\,\,\,so \,4{{y}^{2}}-4\left( b+c \right)y+4bc\ge 0\] \[\Rightarrow \,\, x\le b\,\,or\,\,x\ge c~~~\,\,\,\,\left( \because \,\,b<c \right)\]You need to login to perform this action.
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