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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.
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 2001Please provide solution too.

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5 years ago Latika Leekha
165 Points
```							Hello student,The given function is f(x) = (1+b2) x2 + 2bx + 1We can write it asf(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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