Powers and Common Tangents.

Calculus Level 3

The problem below is a generalization of a brillant problem of the week.

Let $n$ be a positive integer and

$\large \begin{cases} f(x) = x^{2n} \\ g(x) = x^{2n + 1} \ \end{cases}$

The two curves above have a common tangent.

(1) Find four linear functions $\alpha(n), \beta(n), \lambda(n)$ and $\gamma(n)$ for which the slope $m$ of the common tangent can be expressed as $m = \dfrac{\alpha(n)^{\alpha(n)\beta(n)}}{\lambda(n)^{\lambda(n)\gamma(n)} \:\ \gamma(n)^{\alpha(n)\gamma(n)}}$ .

(2) Find the value of $n$ for which $\alpha(n) + \beta(n) + \lambda(n) + \gamma(n) = 59$ .

The answer is 6.

1 solution

Rocco Dalto
Dec 8, 2018

$f'(a) = 2na^{2n - 1} = g'(b) = (2n + 1)b^{2n} \implies b = w_{n}a^{\frac{2n - 1}{2n}}$ , where $w_{n} = (\dfrac{2n}{2n + 1})^{\frac{1}{2n}}$ .

$A:(a, a^{2n})$ and $B:(w_{n}a^{\frac{2n - 1}{2n}}, w_{n}^{2n + 1}a^{\frac{4n^2 - 1}{2n}})$ $\implies m_{AB} = a^{2n - 1}(\dfrac{w_{n}^{2n + 1} - a^{\frac{1}{2n}}}{w_{n} - a^{\frac{1}{2n}}}) = 2na^{2n - 1} \implies$

$(2n - 1)a^{\frac{1}{2n}} = w_{n}(2n - w^{2n}) \implies$ $a = (w_{n}(\dfrac{2n - w^{2n}}{2n - 1}))^{2n}$

$w_{n} = (\dfrac{2n}{2n + 1})^{\frac{1}{2n}} \implies$

$a = (\dfrac{2n}{2n + 1})(\dfrac{4n^2}{4n^2 - 1})^{2n} \implies m_{AB} = f'(a) = 2n((\dfrac{2n}{2n + 1})(\dfrac{4n^2}{4n^2 - 1})^{2n})^{2n - 1} =$

$\dfrac{(2n)^{2n(4n - 1)}}{(2n + 1)^{(2n + 1)(2n - 1)} (2n - 1)^{2n(2n - 1)}}$ $= \dfrac{\alpha(n)^{\alpha(n)\beta(n)}}{\lambda(n)^{\lambda(n)\gamma(n)} \:\ \gamma(n)^{\alpha(n)\gamma(n)}}$

$\implies \alpha(n) + \beta(n) + \lambda(n) + \gamma(n) = 10n - 1 = 59 \implies n = \boxed{6}$ .

Below is a graph for case $\bf (n = 2)$ :

Powers and Common Tangents.

The answer is 6.

1 solution

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