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because the terms are in AP x2 = x1 d ; x3= x1 2d and so on till xn = x1 (n-1)d; No subsitute this ino the equation and solve.. we know 2A is not an integer, therefore A is not an integer. if A is an integer and A=B in an integer, B must also be an integer. And if A , B , C are all integers, f(x) must be an integer. There may be a more amthematical way of approaching this, but Im not sure how..

q7 - take f(x) = 3x^3 (or any function that satisfies the equation) and solve q11. In this case for f(x) to be an integer, C must be an integer. b) Now if C is an integer we need x(Ax B) [let this funtion be g] to be an integer as well. so for g to be an integer, Ax B must be an integer.

But, that can happen only when (x,x1,x2)=(1,1,1) Will you please help me with 22 ? If A B is an integer and we know A is an integer, then B must also be an ineger.  so ans is (d) q18) It is clear by looking that the equation will be satisfied at (0,1). if |x| |y| = 2; then to satisfy the eq of the circle, we need x^2 (2-x)^2 = 1 = we find this has no real solution.

the answer comes different from the 3 equations that are witten.. Now if we can show |x| |y| = 2 (d) is our answer else (b) must be answer. if A B is an integer, it is not necessary for Ax B to be an integer.

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