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`        Prove that if x is real, the expression (x-a)(x-c) / (x-b) is capable of assuming all values if a>b>c or a`
8 years ago

Ramesh V
70 Points
```										put Y = (x-a) (x-b) / (x-c)
so,its
x2 -(a+c+y)x + (ac+by) =0
now for roots to be real, its discriminant be positive
D>0 or (a+c+y)2 -4(ac+by) > 0
(a+c)2 + y2 +2y(a+c) -4ac -4by > 0
y2 + 2y(a+c-2b) + (a-c)2 > 0 ......(1)
here for any value of y(belongs to R) we have : y2  and  (a-c)2 always positive,
so for eqn (1) to be positive for all y(-ve or +ve) , 2(a+c-2b) shld be zero
i.e., a+c-2b = 0  or a,b,c in A.P series from where we can say that
either a<b<c  or c<b<a
--
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Regards,
Naga Ramesh
IIT Kgp - 2005 batch
```
8 years ago
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