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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

7 years ago

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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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we will get you the answer and detailed solution very quickly.We are all IITians and here to help you in your IIT JEE preparation. All the best.

Regards,

Naga Ramesh

IIT Kgp - 2005 batch

7 years ago

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