Statement 1: ax>bxSince it is an inequality and we don't the value of x, we cannot cancel x just like that. Instead, let's take all the variables on one side.
Thus, ax-bx>0
This implies that x(a-b)>0
Again, since its an inequality, we can either have x>0 and a>b (i.e. both elements are positive) or x<0 and a<b (i.e. both elements are negative). Since, we cannot conclusively say whether a<b or a>b,
this statement is insufficient.Statement 2: \(x*a^2 + x*b^2 < 0\)
This implies [m]x*(a^2 + b^2) < 0
Now, [m](a^2 + b^2) will be either 0 (if a=b=0) or positive. Sum of squares of two numbers can never be negative.
Thus, for statement 2 to be true, we will necessarily have x<0. However, this statement doesn't give us any knowledge of whether a<b or a>b.
Thus, insufficientCombining 1& 2, we definitely know that x<0. Thus, according to statement 1, a<b. This means that the question will be false and 1& 2 together are sufficient to answer the question.
Hence, answer is C
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