Application of DerivativeHard
Question
Let f(x) = (1 + b2)x2 + 2bx + 1 and let m(b) be the minimum value of f(x). As b varies, the range of m(b) is
Options
A.[0, 1]
B.
C.
D.(0, 1]
Solution
Since coefficient of x2 is (+ ve)
⇒ m(b) =
⇒ m(b) = -
⇒ m(b) =
⇒ b2 ≥ 0
⇒ 1 + b2 ≥ 1
⇒ 0 <
≤ 1
⇒ m(b) ∈ (0, 1]
⇒ m(b) =
⇒ m(b) = -
⇒ m(b) =
⇒ b2 ≥ 0
⇒ 1 + b2 ≥ 1
⇒ 0 <
≤ 1⇒ m(b) ∈ (0, 1]
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