Let f(x)=x2−px+q, p is an odd positive integer and the roots of the equation f(x)=0 are two distinct prime numbers, if p+q=35, then the value of f(10)(∑10r=1f(r))−878 equal to
∵ Roots of x2−px+q=0 are two distinct prime numbers.
Sum of roots = p = odd positive integer (given)
⇒ Both roots of the equation cannot be odd.
∴ One root must be 2(prime number)
and given p+q=35
Solving Eqn. (i) and (ii), we get
=−110 and f(10)=−8