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Since as per Qs J is a positive no. We can try a pretty small no and check all options. Let say j=2 and it increase to 3 i) decrease from -3 to -8 ii) decrease from -2 to -4 iii) decrease from 1/4 to 1/9 Answer: E
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Last edited by SOURH7WK on 08 Aug 2012, 12:13, edited 1 time in total.

I \(y = 1 - x^2\) is the equation of a downward parabola, intercepting the x axis at -1 and 1. So, far away from the roots, the values of the function are decreasing as x increases. YES

II \(y=x-x^2\) is also the equation of a downward parabola, roots 0 and 1. Again, far away from the roots, the values of the function are decreasing as x increases. YES

III Definitely decreases, the larger the denominator, the smaller the value of the fraction. YES

Answer E
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Since as per Qs J is a positive no. We can try a pretty small no and check all options. Let say j=2 and it increase to 3 i) decrease from -3 to -8 ii) decrease from -2 to -4 iii) decrease from 1/4 to 1/9 Answer: C

The values of j are defined: 135 and 136!
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take a smaller number like 2 and 3 and solve. We get the ans to be E
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I. \(1-j^2\) Of course, a larger portion is taken from 1. Definitely the value will decrease. II. \(j - j^2\) I thought of smaller positive consecutive numbers such as 2 and 3. -2 vs. -6 shows a decrease. \(\frac{1}{j^2}\) The larger the denominater, the smaller the fraction.

Re: As j increases from 135 to 136, which of following must [#permalink]

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03 Mar 2013, 00:04

I. As j increases, j^2 will increase too. Therefore 1-j^2 will decrease. II. j-j^2 = j(1-j). As j increases, 1-j will decrease and be negative. Multiplied by the increased j, this gives us a more negative figure. Therefore j-j^2 will decrease too. III. Simple reciprocal. As j increases from 135 to 136, 1/j^2 is bound to decrease

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