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16 Sep 2014, 01:03
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16 Sep 2014, 01:03
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Official Solution:
Is integer \(x\) positive?
(1) \(x \gt x^3\).
\(x^3-x \lt 0\);
\(x(x-1)(x+1) < 0\).
Here is how to solve the above inequality easily. The "roots", in ascending order, are -1, 0, and 1, which gives us 4 ranges:
\(x < -1\);
\(-1 < x < 0\);
\(0 < x < 1\);
\(x > 1\).
Next, test an extreme value for \(x\): if \(x\) is some large enough number, say 10, then all three multiples will be positive, giving a positive result for the whole expression. So when \(x > 1\), the expression is positive. Now the trick: as in the 4th range, the expression is positive, then in the 3rd it'll be negative, in the 2nd it'll be positive, and finally in the 1st, it'll be negative: \(\text{(- + - +)}\). So, the ranges when the expression is negative are: \(x < -1\) and \(0 < x < 1\).
Since we are told that \(x\) is an integer and there is no integer in the range \(0 < x < 1\), then \(x < -1\), so \(x\) is a negative integer. Sufficient.
P.S. You can apply this technique to any inequality of the form \((ax - b)(cx - d)(ex - f)... > 0\) or \( < 0\). If one of the factors is in the form \((b - ax)\) instead of \((ax - b)\), simply rewrite it as \(-(ax - b)\) and adjust the inequality sign accordingly. For example, consider the inequality \((4 - x)(2 + x) > 0\). Rewrite it as \(-(x - 4)(2 + x) > 0\). Multiply by -1 and flip the sign: \((x - 4)(2 + x) < 0\). The "roots", in ascending order, are -2 and 4, giving us three ranges: \(x < -2\), \(-2 < x < 4\), and \(x > 4\). If \(x\) is some large enough number, say 100, both factors will be positive, yielding a positive result for the entire expression. Thus, when \(x > 4\), the expression is positive. Now, apply the alternating sign trick: as the expression is positive in the 3rd range, it will be negative in the 2nd range, and positive in the 1st range: \(\text{(+ - +)}\). Therefore, the ranges where the expression is negative are: \(-2 < x < 4\).
(2) \(x \lt x^2\).
\(x^2-x \gt 0\);
\(x(x-1) > 0\).
Using the same technique as above, we find the "roots" of the inequality \(x(x-1) \gt 0\) to be 0 and 1, giving us three ranges: \(x < 0\), \(0 < x < 1\), and \(x > 1\). If \(x\) is some large enough number, say 100, both factors will be positive, yielding a positive result for the entire expression. Thus, when \(x > 1\), the expression is positive. Now, apply the alternating sign trick: as the expression is positive in the 3rd range, it will be negative in the 2nd range, and positive in the 1st range: \(\text{(+ - +)}\). Therefore, the ranges where the expression is positive are: \(x < 0\) and \(x > 1\). So, integer \(x\) could be either negative or positive. Not sufficient.
Answer: A
Is integer \(x\) positive?
(1) \(x \gt x^3\).
\(x^3-x \lt 0\);
\(x(x-1)(x+1) < 0\).
Here is how to solve the above inequality easily. The "roots", in ascending order, are -1, 0, and 1, which gives us 4 ranges:
\(x < -1\);
\(-1 < x < 0\);
\(0 < x < 1\);
\(x > 1\).
Next, test an extreme value for \(x\): if \(x\) is some large enough number, say 10, then all three multiples will be positive, giving a positive result for the whole expression. So when \(x > 1\), the expression is positive. Now the trick: as in the 4th range, the expression is positive, then in the 3rd it'll be negative, in the 2nd it'll be positive, and finally in the 1st, it'll be negative: \(\text{(- + - +)}\). So, the ranges when the expression is negative are: \(x < -1\) and \(0 < x < 1\).
Since we are told that \(x\) is an integer and there is no integer in the range \(0 < x < 1\), then \(x < -1\), so \(x\) is a negative integer. Sufficient.
P.S. You can apply this technique to any inequality of the form \((ax - b)(cx - d)(ex - f)... > 0\) or \( < 0\). If one of the factors is in the form \((b - ax)\) instead of \((ax - b)\), simply rewrite it as \(-(ax - b)\) and adjust the inequality sign accordingly. For example, consider the inequality \((4 - x)(2 + x) > 0\). Rewrite it as \(-(x - 4)(2 + x) > 0\). Multiply by -1 and flip the sign: \((x - 4)(2 + x) < 0\). The "roots", in ascending order, are -2 and 4, giving us three ranges: \(x < -2\), \(-2 < x < 4\), and \(x > 4\). If \(x\) is some large enough number, say 100, both factors will be positive, yielding a positive result for the entire expression. Thus, when \(x > 4\), the expression is positive. Now, apply the alternating sign trick: as the expression is positive in the 3rd range, it will be negative in the 2nd range, and positive in the 1st range: \(\text{(+ - +)}\). Therefore, the ranges where the expression is negative are: \(-2 < x < 4\).
(2) \(x \lt x^2\).
\(x^2-x \gt 0\);
\(x(x-1) > 0\).
Using the same technique as above, we find the "roots" of the inequality \(x(x-1) \gt 0\) to be 0 and 1, giving us three ranges: \(x < 0\), \(0 < x < 1\), and \(x > 1\). If \(x\) is some large enough number, say 100, both factors will be positive, yielding a positive result for the entire expression. Thus, when \(x > 1\), the expression is positive. Now, apply the alternating sign trick: as the expression is positive in the 3rd range, it will be negative in the 2nd range, and positive in the 1st range: \(\text{(+ - +)}\). Therefore, the ranges where the expression is positive are: \(x < 0\) and \(x > 1\). So, integer \(x\) could be either negative or positive. Not sufficient.
Answer: A
General Discussion
16 Feb 2017, 23:33
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Solution provided by Bunuel is self-explanatory. But I saw few posters were having issue in understanding how to get to \(x<−1\) or \(0<x<1\) from \(x(x−1)(x+1)<0\)
To break any equality of this type - First, we will find the trigger point. Trigger points are the point on which equality will give output = 0 .
So in this equation, we have 3 i.e. \(x=0\) , \(x=1\) and \(x=-1\)
Now we will mark these points on number line.
___-1____0_____1____
Here we are having 4 regions
1. \(x <-1\)
2. \(-1<x<0\)
3. \(0<x<1\)
4. \(x>1\)
Second step , test one value from each region and see if you get the value < 0
\(x(x−1)(x+1)<0\)
1. \(x <-1\) ---> Put x = -2 ; -6 < 0 ; So this range is valid .
2. -\(1<x<0\) ---> Put x = -1/2 ; Some +ve < 0 ;So this range is not valid.
3. \(0<x<1\) ---> Put x = 1/2 ; Some -ve value < 0 ; So this range is valid .
4.\(x>1\) ---> Put x =2 ; SOme +ve value < 0 ; So this range is not valid.
So from here we get 2 ranges \(x <-1\) or \(0<x<1\)
Same way try for Statement 2 .
Please +1 kudos If this post helps.
To break any equality of this type - First, we will find the trigger point. Trigger points are the point on which equality will give output = 0 .
So in this equation, we have 3 i.e. \(x=0\) , \(x=1\) and \(x=-1\)
Now we will mark these points on number line.
___-1____0_____1____
Here we are having 4 regions
1. \(x <-1\)
2. \(-1<x<0\)
3. \(0<x<1\)
4. \(x>1\)
Second step , test one value from each region and see if you get the value < 0
\(x(x−1)(x+1)<0\)
1. \(x <-1\) ---> Put x = -2 ; -6 < 0 ; So this range is valid .
2. -\(1<x<0\) ---> Put x = -1/2 ; Some +ve < 0 ;So this range is not valid.
3. \(0<x<1\) ---> Put x = 1/2 ; Some -ve value < 0 ; So this range is valid .
4.\(x>1\) ---> Put x =2 ; SOme +ve value < 0 ; So this range is not valid.
So from here we get 2 ranges \(x <-1\) or \(0<x<1\)
Same way try for Statement 2 .
Please +1 kudos If this post helps.
