j_shreyans wrote:Hi Guru ,
Pls correct me if i am wrong and suggest some technique to solve this kind of questions.
Is (x-2)^2 > x^2?
(1) x^2>x
(2) 1/x>0
(x-2)^2>x^2
if we solve the above will get x<1 this is our main target right?
statement 1 - X^2>x by solving this will get 1<x so x must be positive greater than 1 right?
So can this statement be sufficient by saying that x can not less than 1?
Your solution assumes that x is positive.
If x is negative, then dividing each side of x² > x by x requires that the direction of the inequality flip from > to <.
A safer approach is to subtract x from each side, as follows:
x² - x > 0
x(x-1) > 0.
The CRITICAL POINTS are the values of x such that x(x-1) = 0:
x=0 and x=1.
To determine the ranges where x(x-1) > 0, test one value to the left and right of each critical point.
x<0:
If x=-1, then x(x-1) = (-1)(-2) = 2, which is greater than 0.
Thus, x<0 is a valid range.
0<x<1:
If x=1/2, then x(x-1) = (1/2)(-1/2) = -1/4, which is NOT greater than 0.
Thus, 0<x<1 is NOT a valid range.
x>1:
If x=2, then x(x-1) = (2)(1) = 2, which is greater than 0.
Thus, x>1 is a valid range.
Result:
The valid ranges are x<0 and x>1.
Since it's possible that x<1 or that x>1, INSUFFICIENT.
Statement 2 - 1/x>0
if we multiply both side by x so will get 1>0 right?
Now what next should i do?
Pls help me because i really need a help on this kind of questions.
Here, algebra makes the situation unnecessarily confusing.
Rather than use algebra, I suggest the following line of reasoning:
1/x > 0
1/x = positive
(positive)/x = positive.
Since (positive)/(positive) = positive, the equation above holds true only if x>0.
Since it's possible that x<1 or that x>1, INSUFFICIENT.
Statements combined:
The only range that satisfies both statements is x>1.
SUFFICIENT.
The correct answer is
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