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AP EAMCET20225 Jul 2022Morning ShiftMathematicsApplication of DerivativesActual

If a, b>0 , then minimum value of y= b^2 a-x + a^2 x , 0 < x < a is

Options

  1. A(a+b)^2 a
  2. B(a+b)^2 b
  3. C(a-b)^2 a
  4. D(a-b)^2 b

Correct answer

A. (a+b)^2 a

Step-by-step solution

Given, y= b^2 a-x + a^2 x d y d x = -b^2(-1) (a-x)^2 + ( -a^2 x^2 ) d y d x = b^2 (a-x)^2 - a^2 x^2 d y d x =0, x= a^2 a b d^2 y d x^2 = 2 a^2 x^3 + 2 b^2 (a-x)^3 So, . d^2 y d x^2 |_ x= a^2 a+b >0 Therefore, y is minimum at x= a^2 a+b y_ = b^2 a- ( a^2 a+b ) + a^2 (a)^2 (a+b)= b^2(a+b) a^2+a b-a^2 +(a+b)=(a+b) [ b a +1 ]= (a+b)^2 a

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