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KVPY2017MathematicsApplication of Derivatives

Let a₁, a₂, a₃, a₄ be real numbers such that a₁²+a₂²+a₃²+a₄²=1 . Then the smallest possible value of the expression (a₁-a₂ )²+ (a₂-a₃ )²+ (a₃-a₄ )²+ (a₄-a₁ )² lies in the interval

Options

  1. A(0,1.5)
  2. B(-1.5,2.5)
  3. C(2.5,3)
  4. D(3,3.5)

Correct answer

B. (-1.5,2.5)

Step-by-step solution

a ₁²+ a ₂²+ a ₃²+ a ₄²=1 Smallest possible value of (a₁-a₂ )²+ (a₂-a₃ )²+ (a₃-a₄ )²+ (a₄-a₁ )²=0 if a₁=a₂=a₃=a₄= 1 2 (It should be bonus)

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