Maxima and MinimaHard

Question

If maximum value ofy=ax+bx2+1  is 20, then range of a2 + 80b is-

Options

A.

[1600,∞)

B.

[-∞,1600)

C.

{1600}

D.

(–∞,∞) – {1600}

Solution

ax+bx2+1≤20∀x∈R
20x2 – ax + (20 – b)≥20∀x∈R
a2 – 80(20 – b) = 0
a2 + 80b = 1600

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