A) \[a>b>c\]
B) \[a<b<c\]
C) \[a>c<b\]
D) \[a<c<b\]
Correct Answer: D
Solution :
Let \[y=\frac{(x-a)(x-b)}{(x-c)}\] or \[y(x-c)={{x}^{2}}-(a+b)x+ab\] or \[{{x}^{2}}-(a+b+y)x+ab+cy=0\] \[\Delta ={{(a+b+y)}^{2}}-4(ab+cy)\] \[={{y}^{2}}+2y(a+b-2c)+{{(a-b)}^{2}}\] Since x is real and y assumes all real values, we must have \[\Delta \ge 0\]for all real values of y.The sign of a quadratic in y is same as of first term provided its discriminant \[{{B}^{2}}-4AC<0\] This will be so if \[4{{(a+b-2c)}^{2}}-4{{(a-b)}^{2}}<0\] or \[4(a+b-2c+a-b)(a+b-2c-a+b)<0\] or \[16(a-c)(b-c)<0\] or \[16(c-a)(c-b)=-\]ve \ c lies between a and b i.e., \[a<c<b\] .....(i) Where \[a<b\], but if \[b<a\]then the above condition will be \[b<c<a\]or \[a>c>b\] .....(ii) Hence from (i) and (ii) we observe that (d) is correct answer.
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