Question

Let f(x) = $x^2$ -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

• A.
• B.
• C.
• D.

Answer 1

The correct option is C

Let $\alpha ,\beta$ be the roots of $x^2-bx+c=0$,Then $\alpha +\beta =b$

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

$\therefore$ f(2) = 0 $\Rightarrow$ 2b-c = 4 and b + c =35

$\therefore$ b = 13, c = 22

Minimum value = $f\left(\frac{13}{2}\right)=-\frac{81}{4}$