If x is the reciprocal of a positive integer, then the

BTGmoderatorLU wrote:Source: Manhattan GMAT

If x is the reciprocal of a positive integer, then the maximum value of x^y, where y=-x^2, is achieved when x is the reciprocal of

A. 1
B. 2
C. 3
D. 4
E. 5

Let a be the positive integer that x is reciprocal of, i.e., x = 1/a. We have x^y = x^(-x^2) = 1/(x^(x^2)). In terms of a, we have:

1/{(1/a)^[(1/a)^2]}

a^[1/a^2]

If a = 1, then a^[1/a^2] = 1.

If a = 2, then a^[1/a^2] = 2^(1/4).

If a = 3, then a^[1/a^2] = 3^(1/9).

If a = 4, then a^[1/a^2] = 4^(1/16).

If a = 5, then a^[1/a^2] = 5^(1/25).

We can eliminate a = 1, since when a is any of the other 4 values, a^[1/a^2] is greater than 1. Now, let’s do a pairwise comparison between the other 4 choices.

2^(1/4) vs. 3^(1/9)

Let’s raise both by 36th power:

[2^(1/4)]^36 = 2^9 = 512

[3^(1/9)]^36 = 3^4 = 81

We see that 2^(1/4) is greater. Let’s compare it with 4^(1/16).

2^(1/4) vs. 4^(1/16)

Let’s raise both by 16th power:

[2^(1/4)]^16 = 2^4 = 16

[4^(1/16)]^16 = 4^1 = 4

We see that 2^(1/4) is still greater. Let’s compare it with 5^(1/25).

2^(1/4) vs. 5^(1/25)

Let’s raise both by 100th power:

[2^(1/4)]^100 = 2^25 = (2^3)^8 * 2^1 = 8^8 * 2

[5^(1/25)]^100 = 5^4

We see that 2^(1/4) is still greater. Therefore, 2^(1/4) is the greatest.

Answer: B

BTGmoderatorLU wrote:Source: Manhattan GMAT

If x is the reciprocal of a positive integer, then the maximum value of x^y, where y=-x^2, is achieved when x is the reciprocal of

A. 1
B. 2
C. 3
D. 4
E. 5

The OA is B.

We have to maximize that value of x^y = x^(–x^2) = 1 / x^(x^2)

Say the positive integer is a, thus, x = 1/a

1 / x^(x^2) = 1 / (1/a)^{(1/a)^2} = a^(1/a^2)

So, We have to maximize that value of a^(1/a^2)

Let's try with options:

A. a = 1: a^(1/a^2) = 1^(1/1^2) = 1
B. a = 2: a^(1/a^2) = 2^(1/2^2) = 2^(1/4)
C. a = 3: a^(1/a^2) = 3^(1/3^2) = 3^(1/9)
D. a = 4: a^(1/a^2) = 4^(1/4^2) = 4^(1/16)
E. a = 5: a^(1/a^2) = 5^(1/5^2) = 5^(1/25)

Option A cannot be the correct answer since the values of each of the other options is greater than 1.

For options B through E, for a^(1/a^2), the values of bases (a) increase, while the values of exponents (1/a^2) decrease, thus, it is not straightforward to choose the option that has the maximum value.

We see that for options B through E, for a^(1/a^2), the rate of increase of the bases (a) is relatively lesser than the rate of decrease of the exponents (1/a^2), thus, though because of the increase of values of bases, the values of a^(1/a^2) increase, their values decrease considerably due to the decrease in the values of exponents. Thus, as we move from option B, C, D, and E, the values decrease. Thus, the maximum possible value of a^(1/a^2) is at a = 2.

The correct answer: B

Hope this helps!

-Jay
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