# Difference between revisions of "2020 AMC 12B Problems/Problem 22"

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<math>\textbf{(A)}\ \frac{1}{16} \qquad\textbf{(B)}\ \frac{1}{15} \qquad\textbf{(C)}\ \frac{1}{12} \qquad\textbf{(D)}\ \frac{1}{10} \qquad\textbf{(E)}\ \frac{1}{9}</math> | <math>\textbf{(A)}\ \frac{1}{16} \qquad\textbf{(B)}\ \frac{1}{15} \qquad\textbf{(C)}\ \frac{1}{12} \qquad\textbf{(D)}\ \frac{1}{10} \qquad\textbf{(E)}\ \frac{1}{9}</math> | ||

− | == | + | ==Solution1== |

+ | |||

+ | Set <math> u = t2^{-t}</math>. Then the expression in the problem can be written as <cmath>t2^{-t} - 3t4^{-t} = u - 3u^2 = \frac{1}{12}- 3 (\frac{1}{6} - u)^2.</cmath> The maximal value is thus <math>\frac{1}{12}</math>. | ||

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+ | ==Solution2== | ||

First, substitute <math>2^t = x (log_2{x} = t)</math> so that | First, substitute <math>2^t = x (log_2{x} = t)</math> so that |

## Revision as of 00:57, 8 February 2020

## Problem 22

What is the maximum value of for real values of

## Solution1

Set . Then the expression in the problem can be written as The maximal value is thus .

## Solution2

First, substitute so that

Notice that

When seen as a function, is a synthesis function that has as its inner function.

If we substitute , the given function becomes a quadratic function that has a maximum value of when .

Now we need to check that can have the value of in the range of real numbers.

In the range of (positive) real numbers, function is a continuous function whose value gets infinitely smaller as gets closer to 0 (as also diverges toward negative infinity in the same condition). When , , which is larger than .

Therefore, we can assume that equals to when is somewhere between 1 and 2 (at least), which means that the maximum value of is .