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AP EAMCET201825 Apr 2018Morning ShiftMathematicsApplication of DerivativesActual

Let f x = x - a x - b - a + b 2 . If f x = 0 has both non-negative roots, then the minimum value of f x

Options

  1. A= a + b 4
  2. B≥ ( a + b ) 2 4
  3. C≥ - ( a + b ) 2 4
  4. D≤ - ( a + b ) 2 4

Correct answer

C. ≥ - ( a + b ) 2 4

Step-by-step solution

Given, f x = x - a x - b - a + b 2 ⇒ f x = x 2 - a + b x - a + b 2 Let α   &   β be the roots of f x . α ,   β > 0 where the minimum is obtained i.e. - b 2 a > 0 a + b 2 > 0 ⇒ a + b > 0     . . . i Now, x 2 - a + b x - a + b 2 = 0 ⇒ x = a + b ± a + b 2 - 4 a b + a + b 2 2 Roots are non-negative, so a + b ± a - b 2 + 2 a + b 2 ≥ 0 ⇒ a + b ± a - b 2 + 2 a + b ≥ 0 ⇒ a + b 2 ≥ a - b 2 + 2 a +

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