A function f(x) is defined as f(x) = x + x2 for an integer x. If a is

IMO B

Stmt 1: f(a) + f(a+1) is a prime number

=>a + a^2 + (a+1) + (a+1)^2
=> a(a+1) + (a+1)(1+a+1)
(a+1)(a+1+a+1) = (a+1)(2a+2) = 2(a+1)^2

Since this is a prime number & 2 is the only even prime number, ==> 2(a+1)^2 = 2
==> (a+1)^2 = 1 ==> a+1 = +1 ,-1

a = 0 , -2

INSUFFICIENT

Stmt 2: The greatest common divisor of f(3) & f(4) is f(a)*

f(3) = 3+9 =12
f(4) = 4+16 =20

GCD for f(3) & f(4) = 4


Given: f(a)*2=4
OR f(a)=2

SUFFICIENT

The ans is B .

f(x) = x + x^2 for an integer x. If a is an integer, is f(a) a prime number?
i.e
f(a) = a + a^2 = a (1+a) Now f (a) Can only be prime if a = 1 or a = -2

So potentially we need to find that whether a has a unique value of either 1 and -2 or other values from the options .

(1) f(a) + f(a+1) is a prime number

a = 0 and -2 it will satisfy the equation .

Hence Statement 1 is Not sufficient .

(2) The greatest common divisor of f(3) & f(4) is f(a)*2.

f(3) = 12 , f(4) = 20 . GCD of 12 and 20 is 4 . Now f(a)*2. = 4 => f(a) = 2 . SO a = 1 or -2

We can definitely say that value of a is from 1 or -2 .
Hence F(a) = Prime .

So statement 2 is sufficient .
Ans B

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