As j increases from 135 to 136, which of the following must decrease?

Question

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

SOURH7WK · Answer

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

EvaJager · Answer

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

EvaJager · Answer

The values of j are defined: 135 and 136!

metallicafan · Answer

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

Thank you!

rajathpanta · Answer

take a smaller number like 2 and 3 and solve. We get the ans to be E

premnath · Answer

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

mbaiseasy · Answer

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)

GyanOne · Answer

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.

shanmugamgsn · Answer

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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As j increases from 135 to 136, which of the following must decrease?

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As j increases from 135 to 136, which of the following must decrease?

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