A) \[a<b<c\]
B) \[a>b>c\]
C) c lies between a and b
D) \[a<c<b\]or \[a>b>c\]
Correct Answer: C
Solution :
(c): Let \[y=\frac{\left( x-a \right)\left( x-b \right)}{\left( x-c \right)}\] \[\Rightarrow {{x}^{2}}-\left( a+b+y \right)x+ab+cy=0\] Since x is real, then \[D\ge 0\] \[\Rightarrow {{\left( a+b+y \right)}^{2}}-4\left( ab+cy \right)\ge 0\] \[\Rightarrow {{y}^{2}}+2\left( a+b-2c \right)y+{{\left( a+b \right)}^{2}}-4ab\ge 0\] \[\Rightarrow {{y}^{2}}+2\left( a+b-2c \right)y+{{\left( a+b \right)}^{2}}\ge 0\]???(i) Since y takes all real values, (i) is possible if \[4{{\left( a+b-2c \right)}^{2}}-4{{\left( a-b \right)}^{2}}<0\] \[\Rightarrow \left( a+b-2c+a-b \right)\left( a+b-2c-a+b \right)<0\] \[\Rightarrow 4\left( a-c \right)\left( b-c \right)<0\Rightarrow \left( a-c \right)\left( b-c \right)<0\] This is possible if c lies between a and b, that is if \[a<c<b\]or\[a>c>b\].
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