Question

f(10) = 5, then value of '$\alpha$' such that f($\alpha$) = 0 (where $\alpha$ > 1). Let all the roots of the equation $x^{1008} + a_1x^{1007} + a_2x^{1006} + ... + a_{1006}x^2 - 1008x + 1 = 0$ are real and positive

then $\frac{a_2}{|a_1|}$ is

Answer 1

503.5

Answer 2

Let the given polynomial be $P(x) = x^{1008} + a_1x^{1007} + a_2x^{1006} + ... + a_{1006}x^2 - 1008x + 1 = 0$. The degree of the polynomial is $n=1008$. Let the roots of the equation be $r_1, r_2, ..., r_{1008}$. We are given that all roots are real and positive.

From Vieta's formulas, for a monic polynomial $x^n + c_1x^{n-1} + c_2x^{n-2} + ... + c_{n-1}x + c_n = 0$:

1. Sum of roots: $\sum_{i=1}^n r_i = -c_1$
2. Sum of products of roots taken two at a time: $\sum_{1 \le i < j \le n} r_i r_j = c_2$
3. Product of roots: $\prod_{i=1}^n r_i = (-1)^n c_n$
4. Sum of products of roots taken $n-1$ at a time: $\sum_{k=1}^n \frac{\prod r_i}{r_k} = (-1)^{n-1} c_{n-1}$

Comparing the given polynomial with the general form: The coefficient of $x^{1007}$ is $a_1$. So, $c_1 = a_1$. The coefficient of $x^{1006}$ is $a_2$. So, $c_2 = a_2$. The coefficient of $x^1$ is $-1008$. So, $c_{1007} = -1008$. The constant term (coefficient of $x^0$) is $1$. So, $c_{1008} = 1$.

Applying Vieta's formulas:

1. Sum of roots: $\sum_{i=1}^{1008} r_i = -a_1$
2. Sum of products of roots taken two at a time: $\sum_{1 \le i < j \le 1008} r_i r_j = a_2$
3. Product of roots: $\prod_{i=1}^{1008} r_i = (-1)^{1008} \times 1 = 1$
4. Sum of products of roots taken $1007$ at a time: $\sum_{k=1}^{1008} \frac{\prod r_i}{r_k} = (-1)^{1007} (-1008)$ Since $\prod r_i = 1$, this simplifies to $\sum_{k=1}^{1008} \frac{1}{r_k} = 1008$.

Now, we use the AM-GM inequality. For $n$ positive real numbers $x_1, x_2, ..., x_n$, the AM-GM inequality states: $\frac{x_1 + x_2 + ... + x_n}{n} \ge (x_1 x_2 ... x_n)^{1/n}$ Equality holds if and only if $x_1 = x_2 = ... = x_n$.

Apply AM-GM to the reciprocals of the roots, $1/r_1, 1/r_2, ..., 1/r_{1008}$: $\frac{\frac{1}{r_1} + \frac{1}{r_2} + ... + \frac{1}{r_{1008}}}{1008} \ge \left(\frac{1}{r_1} \frac{1}{r_2} ... \frac{1}{r_{1008}}\right)^{1/1008}$ Substitute the sum of reciprocals and product of roots: $\frac{1008}{1008} \ge \left(\frac{1}{1}\right)^{1/1008}$ $1 \ge 1$

Since the equality holds in the AM-GM inequality, it implies that all the terms must be equal: $\frac{1}{r_1} = \frac{1}{r_2} = ... = \frac{1}{r_{1008}}$ This means $r_1 = r_2 = ... = r_{1008}$. Since their product is $\prod r_i = 1$, we must have $r_i^{1008} = 1$. As the roots are positive, this implies $r_i = 1$ for all $i=1, ..., 1008$.

So, all roots of the polynomial are 1. This means the polynomial must be $(x-1)^{1008}$. We can expand $(x-1)^{1008}$ using the binomial theorem: $(x-1)^{1008} = \binom{1008}{0}x^{1008}(-1)^0 + \binom{1008}{1}x^{1007}(-1)^1 + \binom{1008}{2}x^{1006}(-1)^2 + ... + \binom{1008}{1007}x^1(-1)^{1007} + \binom{1008}{1008}x^0(-1)^{1008}$ $(x-1)^{1008} = x^{1008} - \binom{1008}{1}x^{1007} + \binom{1008}{2}x^{1006} - ... - \binom{1008}{1}x + 1$

Comparing the coefficients with the given polynomial $x^{1008} + a_1x^{1007} + a_2x^{1006} + ... - 1008x + 1 = 0$:

1. Coefficient of $x^{1007}$: $a_1 = -\binom{1008}{1} = -1008$.
2. Coefficient of $x^{1006}$: $a_2 = \binom{1008}{2} = \frac{1008 \times (1008-1)}{2} = \frac{1008 \times 1007}{2}$. (We can also verify the coefficient of $x$: $-\binom{1008}{1007} = -\binom{1008}{1} = -1008$, which matches the given polynomial).

We need to find the value of $\frac{a_2}{|a_1|}$. $|a_1| = |-1008| = 1008$. $\frac{a_2}{|a_1|} = \frac{\frac{1008 \times 1007}{2}}{1008} = \frac{1007}{2} = 503.5$.

The first part of the question "f(10) = 5, then value of '$\alpha$' such that f($\alpha$) = 0 (where $\alpha$ > 1)." seems unrelated and incomplete. The number '3' at the end is likely the intended answer for the question asked, but based on the provided polynomial and standard mathematical principles, the calculated answer is $503.5$. Assuming the problem statement for the polynomial is exactly as written and standard mathematical rules apply, $503.5$ is the correct result.