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Let f(x)=(1+b^2 ) x^2 + 2bx + 1 and let m(b) be the minimum value of f(x). As b varies, the range of m(b) is: (A) [0, 1] (B) [0,1/2] (C) [1/2,1] (D) [0, 1] IIT JEE 2001 Please provide solution too.

5 years ago

Answers : (1)

Latika Leekha
askIITians Faculty
165 Points
							Hello student,
The given function is f(x) = (1+b2) x2 + 2bx + 1
We can write it as
f(x) = (1+b2) x2 + 2bx + 1
= 1+b2 { x2 + 2b/(1+b2)x + b2/(1+b2)2} – b2/(1+b2) + 1
= (1+b2) (x + b/(1+b2))2 + 1/(1+b2) ≥ 1/(1+b2)
Hence, m(b) = 1/(1+b2)
So, the range of m(b) = (0, 1].
I suppose you have mixed teh brackets in options (B) and (D), so correct answer would be (0, 1].
5 years ago
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