A calculus problem by Parth Thakkar

Calculus Level 3

Let $f(x) = (x-1)(x-2)\ldots(x-n)$ where $n \in \mathbb{N}$ be a function from $\mathbb{R}$ to $\mathbb{R}$ .

Let $\int \dfrac{ f(x)f''(x) - (f'(x))^2 } { (f(x))^2 } dx = u(x) + C$ where $C$ is some arbitrary constant.

If the number of (real) solutions to the equation $u(x) = 5$ is $\alpha_n$ where $\alpha_n \in \mathbb{N}$ , then the minimum possible value of $n$ (that is, for the equation to have $\alpha_n$ roots) can be written as $k \alpha_n$ , where $k \in \mathbb{N}$ and is independent of $n$ or $\alpha_n$ . What is the value of $k$ ?

The answer is 1.00.

1 solution

Parth Thakkar
Mar 9, 2014

It's easy to see that $u(x) = \dfrac{ f'(x) } { f(x) }$ . You may add a constant to it, but that doesn't matter here. The equation $u(x) = 5$ is then equivalent to $f'(x) = 5 f(x)$ . Note that $f(x)$ is an $n^{\text{th}}$ degree polynomial, and $f'(x)$ is $(n-1)^{\text{th}}$ degree polynomial. So, the equation is of degree $n$ . Hence, it will have at most $n$ (real) roots.

If the last part is not clear, let's see it this way. The solutions to that equation will be the roots of the polynomial $5f(x) - f'(x)$ which has at most $n$ (real) roots.

In the question, the number of (real) roots is given to be $\alpha_n$ . So $n$ must be at least $\alpha_n$ .

This was my first question, so please excuse me for not using the phrase "real roots". I will be more careful from next time! The question has been edited now anyways!

A calculus problem by Parth Thakkar

The answer is 1.00.

1 solution

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