Question

Given that

$a, b, c, d, e, k \in \mathbb{R}, ad = 25e$,

$f(x) = x^5 + ax^4 + bx^3 + cx^2 + dx + e$,

$g(x) = 10x^6 + 12ax^5 + 15bx^4 + 20cx^3 + 30dx^2 + 60ex + k$.

If $f(3) = 1$, and polynomial $g(x) = 0$ has six positive real roots, then

• A. 2a + b = 20
• B. 2b + c = 40
• C. 2c + d = 80
• D. 2e + a = -54

Answer 1

Answer: (A)

2a + b = 20

Answer 2

1. Relate $g(x)$ and $f(x)$ through differentiation: Calculate the derivative of $g(x)$: $g'(x) = \frac{d}{dx}(10x^6 + 12ax^5 + 15bx^4 + 20cx^3 + 30dx^2 + 60ex + k)$ $g'(x) = 60x^5 + 60ax^4 + 60bx^3 + 60cx^2 + 60dx + 60e$ $g'(x) = 60(x^5 + ax^4 + bx^3 + cx^2 + dx + e) = 60f(x)$.

2. Apply Rolle's Theorem: Since $g(x)$ has six positive real roots, by Rolle's Theorem, its derivative $g'(x)$ must have five positive real roots. Therefore, $f(x) = 0$ must have five positive real roots.

3. Deduce coefficient signs: For a polynomial to have all positive real roots, its coefficients must alternate in sign, starting with a positive leading coefficient. For $f(x) = x^5 + ax^4 + bx^3 + cx^2 + dx + e$:

   • Coefficient of $x^5$ is $1 > 0$.
   • $a < 0$.
   • $b > 0$.
   • $c < 0$.
   • $d > 0$.
   • $e < 0$.

4. Utilize the condition $ad = 25e$: Given $a<0$, $d>0$, and $e<0$, the condition $ad = 25e$ is consistent with these sign requirements ($ad < 0$ and $25e < 0$).

5. Assume a specific form for $f(x)$: The condition $ad = 25e$ and the requirement of having five positive real roots strongly suggest that $f(x)$ might be a power of a linear term with a single repeated root. Let's assume $f(x) = (x-r)^5$ for some positive real number $r$ (since all roots are positive). Expanding $(x-r)^5$: $x^5 - 5rx^4 + 10r^2x^3 - 10r^3x^2 + 5r^4x - r^5$. Comparing coefficients with $f(x) = x^5 + ax^4 + bx^3 + cx^2 + dx + e$: $a = -5r$ $b = 10r^2$ $c = -10r^3$ $d = 5r^4$ $e = -r^5$ Now, check $ad = 25e$: $ad = (-5r)(5r^4) = -25r^5$. $25e = 25(-r^5) = -25r^5$. The condition $ad=25e$ is satisfied.

6. Use the condition $f(3) = 1$: Substitute $x=3$ into $f(x) = (x-r)^5$: $f(3) = (3-r)^5 = 1$. Since $r$ is a real number, $3-r = 1$, which implies $r = 2$.

7. Calculate the coefficients: Substitute $r=2$ into the expressions for $a, b, c, d, e$: $a = -5(2) = -10$ $b = 10(2^2) = 10(4) = 40$ $c = -10(2^3) = -10(8) = -80$ $d = 5(2^4) = 5(16) = 80$ $e = -(2^5) = -32$

8. Test the options: (A) $2a + b = 2(-10) + 40 = -20 + 40 = 20$. This is TRUE. (B) $2b + c = 2(40) + (-80) = 80 - 80 = 0$. This is FALSE. (C) $2c + d = 2(-80) + 80 = -160 + 80 = -80$. This is FALSE. (D) $2e + a = 2(-32) + (-10) = -64 - 10 = -74$. This is FALSE.

Therefore, the correct option is (A).