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27 Jul 2026, 09:38
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Given y = (x - 6)(x - 5)(x - 4)(x - 3) and 'x' is a positive integer. If y > 0 then which of following must be true?
(A) x < 3 or x > 6
(B) x < 6
(C) 3 < x < 0
(D) x > 3
(E) None of these
(A) x < 3 or x > 6
(B) x < 6
(C) 3 < x < 0
(D) x > 3
(E) None of these
27 Jul 2026, 09:59
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Given y = (x - 6)(x - 5)(x - 4)(x - 3) and 'x' is a positive integer. If y > 0 then which of following must be true?
(A) x < 3 or x > 6
(B) x < 6
(C) 3 < x < 0
(D) x > 3
(E) None of these
We have:
\((x - 6)(x - 5)(x - 4)(x - 3) > 0\)
The transition points, in ascending order, are \(x = 3\), \(x = 4\), \(x = 5\), and \(x = 6\) (these are the values of x at which the expression changes sign). This gives us five ranges:
\(x < 3\)
\(3 < x < 4\)
\(4 < x < 5\)
\(5 < x < 6\)
\(x > 6\)
Next, test an extreme value for x: if x is a large enough number, say 100, then all four factors will be positive, resulting in a positive value for the whole expression. Therefore, when \(x > 6\), the expression is positive.
Now, here’s the trick: since the expression is positive in the 5th range, it will be negative in the 4th range, positive again in the 3rd range, negative again in the 2nd range, and positive again in the 1st range, following the pattern: +, -, +, -, +.
Thus, the expression is positive for \(x < 3\), \(4 < x < 5\), and \(x > 6\).
However, since \(x\) is an integer, there are no integer values between 4 and 5. Therefore, \(y > 0\) only when \(x < 3\) or \(x > 6\).
Answer: A.
Second method:
We have:
\((x - 6)(x - 5)(x - 4)(x - 3) > 0\)
Since x is an integer, test the possible ranges separated by the zeros 3, 4, 5, and 6.
For x < 3, all four factors are negative, so y > 0.
For x = 3, 4, 5, or 6, y = 0.
There are no integers strictly between 3 and 4, 4 and 5, or 5 and 6.
For x > 6, all four factors are positive, so y > 0.
Therefore, y > 0 only when x < 3 or x > 6.
Answer: A.
(A) x < 3 or x > 6
(B) x < 6
(C) 3 < x < 0
(D) x > 3
(E) None of these
We have:
\((x - 6)(x - 5)(x - 4)(x - 3) > 0\)
The transition points, in ascending order, are \(x = 3\), \(x = 4\), \(x = 5\), and \(x = 6\) (these are the values of x at which the expression changes sign). This gives us five ranges:
\(x < 3\)
\(3 < x < 4\)
\(4 < x < 5\)
\(5 < x < 6\)
\(x > 6\)
Next, test an extreme value for x: if x is a large enough number, say 100, then all four factors will be positive, resulting in a positive value for the whole expression. Therefore, when \(x > 6\), the expression is positive.
Now, here’s the trick: since the expression is positive in the 5th range, it will be negative in the 4th range, positive again in the 3rd range, negative again in the 2nd range, and positive again in the 1st range, following the pattern: +, -, +, -, +.
Thus, the expression is positive for \(x < 3\), \(4 < x < 5\), and \(x > 6\).
However, since \(x\) is an integer, there are no integer values between 4 and 5. Therefore, \(y > 0\) only when \(x < 3\) or \(x > 6\).
Answer: A.
Second method:
We have:
\((x - 6)(x - 5)(x - 4)(x - 3) > 0\)
Since x is an integer, test the possible ranges separated by the zeros 3, 4, 5, and 6.
For x < 3, all four factors are negative, so y > 0.
For x = 3, 4, 5, or 6, y = 0.
There are no integers strictly between 3 and 4, 4 and 5, or 5 and 6.
For x > 6, all four factors are positive, so y > 0.
Therefore, y > 0 only when x < 3 or x > 6.
Answer: A.
27 Jul 2026, 10:10
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Why did you take x=3, 4, 5, 6? You also could take 1,2,3,4 or any other numbers. Why those particular numbers?
Bunuel
