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

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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