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SOURH7WK
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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Ouch! I didn't notice it was a function. :(

Thank you!
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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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It can be solved using values

(1) \(1- j^2\) : Obviously if j increases, value of \(J^2\) will increase. This will in turn decrease the value of \(1- j^2\)

(2) \(J-J^2\) = \(J*(1-J)\)

when J=135, we get -135*134
when J=136, we get -136*135
Therefore, from the values we can easily see it is decreasing.

(3) \(1/J^2\) : as the denominator decrease, so the result will decrease.

As in all three cases, values decrease the answer is 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.

Answer: I,II, III or (E)
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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

E it is.
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As j increases from 135 to 136, which of the following must decrease?

I. \(1 - j^2\)

II. \(j - j^2\)

III. \(\frac{1}{j^2}\)

(A) I only
(B) II only
(C) I and II
(D) I and III
(E) I, II and III

Source: Gmat Hacks 1800



Assume j increases from 2 to 4
I. \(1 - j^2\)
1- \(2^2\) = 1-4 = -3
1- \(3^2\) = 1-9 = -8
So this is true

II. \(j - j^2\)
2- \(2^2\) = 2-4 = -2
3- \(3^2\) = 3-9 = -6
So this is true

III. \(\frac{1}{j^2}\)
\(\frac{1}{2^2}\) = 0.25
\(\frac{1}{4^2}\) = 0.0XX
So this is true

Therefor , E :)
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