Is xy < x^2*y^2? 1) xy>0 2) x+y=1
Solution:
First simplify to xy < (xy)^2 using law of exponents With statement (1), you know that either both x and y are negative or both x and y are positive. Otherwise their product could not be positive. However, even within this space the answer to the inequality is ambiguous since for -1 < x < 1 and -1 < y < 1, the inequality does not hold, but for two negative numbers or two positive numbers greater than 1 or less than -1, it does hold.
With statement (2) you know that either x and y are both greater than zero and less than one such that their sum equals 1 (e.g. - 1/3 and 2/3), or you know that they are two numbers (one positive and one negative) where the positive number has an absolute value 1 greater than the negative number. This statement too is ambiguous since as in the example given the product of 1/3 and 2/3 is greater than their product squared but the product of 8 and -7 is less than their product squared (-56 < (-56)^2). Take the two together and you see that a non-negative product of two numbers whose sum equals 1 only allows for 0 < x < 1 and 0 < y < 1, and in this case, the inequality is always false. Answer is C.
If x is positive, is x>3? a) (x-1)^2 > 4 b) (x-2)^2 > 9
Solution:
(1). Consider (x-1)^2 > 4 which will mean (x-1)^2-4>0 ((x-1)+2)((x-1)-2)>0 [using the identity a^2-b^2 =(a+b)(a-b)] Simplifying, (x+1)(x-3)>0 now for a product of two terms {(x+1),(x+3)} to be positive either both x+1 and x-3 to be positive which is possible only when x>3 or both x+1 and x-3 to be negative which is possible only when x<-1. so we can safely conclude that (x-1)^2 > 4 will mean x>3 or x<-1
(2). (x-2)^2 > 9 =>(x-2+3)(x-2-3)>0 upon solving which we get x>5 or x<-1 since x is positive , x>5 So b is sufficient to answer.
Answer is D.
Solution:
First simplify to xy < (xy)^2 using law of exponents With statement (1), you know that either both x and y are negative or both x and y are positive. Otherwise their product could not be positive. However, even within this space the answer to the inequality is ambiguous since for -1 < x < 1 and -1 < y < 1, the inequality does not hold, but for two negative numbers or two positive numbers greater than 1 or less than -1, it does hold.
With statement (2) you know that either x and y are both greater than zero and less than one such that their sum equals 1 (e.g. - 1/3 and 2/3), or you know that they are two numbers (one positive and one negative) where the positive number has an absolute value 1 greater than the negative number. This statement too is ambiguous since as in the example given the product of 1/3 and 2/3 is greater than their product squared but the product of 8 and -7 is less than their product squared (-56 < (-56)^2). Take the two together and you see that a non-negative product of two numbers whose sum equals 1 only allows for 0 < x < 1 and 0 < y < 1, and in this case, the inequality is always false. Answer is C.
If x is positive, is x>3? a) (x-1)^2 > 4 b) (x-2)^2 > 9
Solution:
(1). Consider (x-1)^2 > 4 which will mean (x-1)^2-4>0 ((x-1)+2)((x-1)-2)>0 [using the identity a^2-b^2 =(a+b)(a-b)] Simplifying, (x+1)(x-3)>0 now for a product of two terms {(x+1),(x+3)} to be positive either both x+1 and x-3 to be positive which is possible only when x>3 or both x+1 and x-3 to be negative which is possible only when x<-1. so we can safely conclude that (x-1)^2 > 4 will mean x>3 or x<-1
(2). (x-2)^2 > 9 =>(x-2+3)(x-2-3)>0 upon solving which we get x>5 or x<-1 since x is positive , x>5 So b is sufficient to answer.
Answer is D.
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