If f(x) =
, then show that f(x) is
linear in r. Hence deduce that f(0) = 
where g(x) = (c 1 – x) (c 2 – x) (c 3 – x).
Text Solution
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Sol. Given f(x) =
…(1)
Operate C 2 → C 2 –C 1 ; C 3 → C 3 – C 2
f(x) = 
= x
+ 
⇒ f(x) is linear
Let f(x) = Px + Q
⇒ f(–a) = –aP + Q, f(–b) = –bP + Q
f(0) = 0.P + Q = Q
=
…(2)
From equation (1), we get.
f (–a) = 
= (c 1 – a) (c 2 – a) (c 3 – a)
Similarly f(–b) = (c 1 – b) (c 2 – b) (c 3 – b)
Given g(x) = (c 1 – x) (c 2 – x) (c 3 – x)
⇒ g(a) = (c 1 – a) (c 2 – a) (c 3 – a)
= f(–a)
Similarly g(b) = f(–b)
Putting these values in equation (2), we get
f(0) = 
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