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Q.

Consider the quadratic polynomial  f(x)=x2px+q  where  f(x)=0 has prime roots If  p+q=11  &  a=p2+q2 then  f(a)=λ (a is an odd positive integer) Number of even  positive divisors of   is

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a

8

b

12

c

6

d

4

answer is B.

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Detailed Solution

[α+β=p,αβ=2  and p+q=11 α+β+αβ=11 α=11β1+β β=2, α=3 p=5,  q=6 a=61 f(61)=(61)26.61+6=3422   3422=2×29×59    

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