Minimum Supremum

Algebra Level 5

$\large f_\alpha(\alpha x ) = f_\alpha(x + \alpha)$

The function $f_{\alpha} \colon \mathbb{R} \to \mathbb{R}$ satisfies the above functional equation for all $x$ where $\alpha$ is real number $\neq 0, 1$ .

It is given that $f_{12}(2) = 3$ . Let $m$ be the minimum value of the least upper bound of $k$ such that $f_{12}(k) = 3$ .

If $m$ can be written as $\dfrac{A}{B}$ , where $A$ and $B$ are coprime positive integers, find $A+B$ .

Bonus: Find all functions $f_{\alpha}(x)$ that satisfy the given functional equation.

The answer is 155.

1 solution

Pranshu Gaba
Mar 20, 2016

[I will add complete solution soon. ]

The function $f_{\alpha}$ behaves differently depending on whether $\alpha$ is positive or negative. Since we are dealing with positive $\alpha$ in the problem, ( $\alpha = 12$ ), we will analyze $f_{\alpha}$ for $\alpha > 0$ .

We can define $f_{\alpha} (x)$ arbitrarily in the range $0 \leq x\leq \alpha$ .

Let $x = y + \frac{\alpha}{\alpha - 1}$ . We get

$f_{\alpha} \left(\alpha y + \frac{\alpha^{2}}{\alpha - 1} \right) = f_{\alpha} \left( y + \frac{\alpha^{2}}{\alpha - 1} \right)$

$m = \frac{\alpha^{2}}{\alpha - 1} = \frac{12^{2}}{ 12 - 1} = \frac{144}{11}$

The answer is $144 + 11 = \boxed{155}$ $_\square$

Minimum Supremum

The answer is 155.

1 solution

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