FunctionHard
Question
Let f(x) = (1 + b2)x2 + 2bx + 1 and let m (b) the minimum value of f(x) As b varies, the renge of m (b) is
Options
A.[0, 1]
B.
C.
D.(0,1]
Solution
Given, f(x) = (1 + b2)x2 + 2bx + 1
= (1 + b2)
m (b) = minimum value of f(x)
is positive and
m (b) varies from 1 to 0, so range = (0,1]
= (1 + b2)
m (b) = minimum value of f(x)
is positive and m (b) varies from 1 to 0, so range = (0,1]
Create a free account to view solution
View Solution FreeMore Function Questions
If f : R → R, f(x) = x2 + 3, then pre- image of 2 under f is -...Let function f : R → R be defined by f(x) = 2x + sin x for x ∈ R Then, f is...f(x) = is not defined for-...f(x) = |x - 1|, f: R+ → R, g(x) = ex, g : [- 1, ∞) → R. If the function fog (x) is defined, then its d...Let f(x) = |x - 1|, then...