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Question: Answered & Verified by Expert
Let fx=x-ax-b-a+b2. If fx=0 has both non-negative roots, then the minimum value of fx
MathematicsApplication of DerivativesAP EAMCETAP EAMCET 2018 (25 Apr Shift 1)
Options:
  • A =a+b4
  • B (a+b)24
  • C -(a+b)24
  • D -(a+b)24
Solution:
1700 Upvotes Verified Answer
The correct answer is: -(a+b)24

Given,

fx=x-ax-b-a+b2

fx=x2-a+bx-a+b2

Let α & β be the roots of fx.

α, β>0 where the minimum is obtained i.e. -b2a>0

a+b2>0a+b>0  ...i

Now, 

x2-a+bx-a+b2=0

x=a+b±a+b2-4ab+a+b22

Roots are non-negative, so

a+b±a-b2+2a+b20

a+b±a-b2+2a+b0

a+b2a-b2+2a+b

a+b2ab  ...ii

By i & ii, we have

0<a+b2ab

Minimum value=-D4a

 =-a-b2+2a+b4

   =-a-b2+4ab4

   =-a+b24

Therefore, fx-a+b24.

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