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MHT CET20243 May 2024Evening ShiftMathematicsApplication of DerivativesActual

The minimum value of the function f (x)=2 x^3-15 x^2+36 x-48 on the set A = x x^2+20 9 x is

Options

  1. A-16
  2. B-7
  3. C16
  4. D7

Correct answer

A. -16

Step-by-step solution

aligned A & = x x^2+20 9 x & = x x^2-9 x+20 0 & = x (x-4)(x-5) 0 A & = 4,5 f (x) & =2 x^3-15 x^2+36 x-48 f ^ (x) & =6 x^2-30 x+36 & =6 (x^2-5 x+6 ) & =6(x-2)(x-3) 0 x (4,5) aligned f (x) is strictly increasing in the interval (4,5) . Minimum value of f (x) when x (4,5) is f(4)=-16

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