Maxima and MinimaHard
Question
Let f(x) = (1 + b2)x2 + 2bx + 1 and m(b) is a minimum value of f(x). If b can assume different values, then range of m(b) is equal to-
Options
A.[0, 1]
B.(0, 1/2]
C.[1/2, 1]
D.(0, 1]
Solution
f(x) = (1 + b2) x2 + 2bx + 1
It is a quadratic expression with coefficient of x2
as (1 + b2) > 0
Whose minimum value is -
∴ m(b) = -
m(b) =
For range of m(b)
> 0 also b2 ≥ 0 ⇒ 1 + b2 ≥ 1
⇒
≤ 1⇒ range of m(b) is (0, 1]
It is a quadratic expression with coefficient of x2
as (1 + b2) > 0
Whose minimum value is -
∴ m(b) = -
m(b) =
For range of m(b)
> 0 also b2 ≥ 0 ⇒ 1 + b2 ≥ 1⇒
≤ 1⇒ range of m(b) is (0, 1] Create a free account to view solution
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