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Question

Let f(x) = x2 -bx+c,b is odd positive integer, f(x) = 0 have two prime 

numbers as roots and b+c=35.Then the global minimum value of f(x) is



Your Answer
A

Your Answer
B

 

Correct Answer
C

 

Your Answer
D


Solution

The correct option is B

 


Let α,β  be the roots of x2bx+c=0,Then α+β=b 

one of the roots is '2'(Since a,b are primes and b is odd positive integer)

f(2) = 0 2b-c = 4 and b + c =35

 b = 13, c = 22

 Minimum value = f(132)=814 

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