[[Christmas]] New year Streak 84/88: Calculus Rush #2

Let us define:

$\large \displaystyle f(x) = ax^2 + bx + c$

Then:

$\large \displaystyle g(x) = a[\ln(x)]^2 + b\ln(x) + c$

Also:

$\large \displaystyle g'(x) = 2a\frac{\ln(x)}{x} + \frac{b}{x}$

$\large \displaystyle g''(x) = 2a\frac{[1 - \ln(x)] }{x^2} - \frac{b}{x^2}$

Condition $1$ says that if $g'(p) = 0$ , also $g''(p^2) = 0$ . Thus:

$\large \displaystyle g'(p) = 2a\frac{\ln(p)}{p} + \frac{b}{p} = 0$

$\color{#20A900} \boxed{ \large \displaystyle p = e^{-\frac{b}{2a}} }$

Putting this in $g''(x)$ :

$\large \displaystyle g''(p^2) = 2a\frac{[1 - \ln(p^2)] }{p^2} - \frac{b}{p^2} = 0$

$\large \displaystyle \frac{2a ( 1 + b/a) - b}{e^{-\frac{b}{a}}} = 0$

$\color{#20A900} \boxed{ \large \displaystyle b = -2a }$

Looking at condition $2$ , since the line passes through point $(k, 0)$ , its equation is:

$\large \displaystyle r(x) = m(x-k)$

Where $m$ is the inclination. If we consider the tangent point with $g(x)$ as $(q, g(q))$ , the inclination $m$ will be equal to $g'(q)$ . That is:

$\color{#20A900} \boxed{ \large \displaystyle m = \frac{2a\ln(q) + b}{q} }$

But $(q, g(q))$ is a solution for both the line $r(x)$ and $g(x)$ . So:

$\large \displaystyle r(q) = g(q)$

$\large \displaystyle \frac{2a\ln(q) + b}{q} (q - k) = a [ \ln(q) ]^2 + b \ln(q) + c$

Recalling that $b = -2a$ :

$\large \displaystyle \frac{2a\ln(q) -2a}{q} (q - k) = a [ \ln(q) ]^2 - 2a\ln(q) + c$

This equation gives only one solution $q$ $\forall k \geq 0$ . This means that $k = 0$ also gives only one solution:

$\large \displaystyle \frac{2a\ln(q) -2a}{q} q = a [ \ln(q) ]^2 - 2a\ln(q) + c$

Since $g(x)$ is only defined for $x > 0$ , then $q \neq 0$ :

$\large \displaystyle 2a\ln(q) -2a = a [ \ln(q) ]^2 - 2a\ln(q) + c$

$\large \displaystyle a [ \ln(q) ]^2 - 4a\ln(q) + c + 2a = 0$

Since it only has one solution $q$ , the discriminant must be $0$ (note this equation is quadratic in $\ln(q)$ ). Thus:

$\large \displaystyle (4a)^2 = 4a(c+2a)$

$\color{#20A900} \boxed{ \large \displaystyle c = 2a }$

So:

$\color{#20A900} \boxed{ \large \displaystyle f(x) = a(x^2 - 2x + 2) }$

Thus:

$\color{#3D99F6} \boxed{ \large \displaystyle \frac{f(10)}{f(2)} = \frac{82a}{2a} = 41 }$