Let f(x) = (1 + b^2)x^2 + 2bx + 1 and let m(b) be the… — JEE Main 2001 Mathematics PYQ

2001

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)

A.
$[0, 1]$
B.
$\left[0, \frac{1}{2}\right]$
C.
$\left[\frac{1}{2}, 1\right]$
D.
$(0, 1]$

Show answer and solution

Correct option: D

Solution

The minimum of f(x) = (1+b²)x² + 2bx + 1 occurs at x = -b/(1+b²), giving m(b) = 1 - b²/(1+b²) = 1/(1+b²).

As b varies over all reals, b² ∈ [0, ∞), so 1+b² ∈ [1, ∞), making m(b) = 1/(1+b²) ∈ (0, 1]. The value 1 is achieved at b=0, but m(b) → 0 as b → ∞ without ever reaching 0. Hence the range is (0, 1].

[2001]

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