Differential EquationHard
Question
f(x) is a four degree polynomial with leading coefficient unity and f(1) = 1, f(2) = 4, f(3) = 9. If x = 1 is a point of extremum of f(x), then -
Options
A.f(4) = 16
B.f′(2) = 4
C.f″(x) = 2 has atleast one solution in (0,4)
D.f′(2) = 0
Solution
Let P(x) = f(x) - x2 = four degree polynomial.
∵ P(1) = P(2) = P(3) = 0
⇒ P(x) = (x - 1) (x - 2) (x - 3) (x - α)
⇒ f(x) = x2 + (x - α) (x - 1) (x - 2) (x - 3)
f′(x) = 2x + (x - α) (x - 1) (x - 2) (x - 3)
⇒ f′(1) = 2 + (1 - α) (1 - 2) (1 - 3) + 0 + 0 + 0 = 0
⇒ 2 + 2 - 2α = 0
α = 2
⇒ f(x) = x2 + (x - 1) (x - 2)2 (x - 3)
Now
∵
⇒ P″(c3) = 0 for same c3 ∈(c1, c2) ⊂ (1, 3)
⇒ f″(x) = 2 for same x ∈ (1,3) ⊂ (0, 4)
∵ P(1) = P(2) = P(3) = 0
⇒ P(x) = (x - 1) (x - 2) (x - 3) (x - α)
⇒ f(x) = x2 + (x - α) (x - 1) (x - 2) (x - 3)
f′(x) = 2x + (x - α) (x - 1) (x - 2) (x - 3)
⇒ f′(1) = 2 + (1 - α) (1 - 2) (1 - 3) + 0 + 0 + 0 = 0
⇒ 2 + 2 - 2α = 0
α = 2
⇒ f(x) = x2 + (x - 1) (x - 2)2 (x - 3)
Now
∵
⇒ P″(c3) = 0 for same c3 ∈(c1, c2) ⊂ (1, 3)
⇒ f″(x) = 2 for same x ∈ (1,3) ⊂ (0, 4)
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