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Step 1: Understand the Problem
We are given the polynomial
f(x) = 2x^5 + 5x^4 + 10x^3 + 10x^2 + 10x + 10
and an integer a such that all its real roots lie in the interval (a,\,a+1) . We need to find the value of |a| .
Step 2: Find the Derivative of the Polynomial
To analyze how many real roots the polynomial can have, we compute its derivative:
\[
f'(x) = \frac{d}{dx}\bigl(2x^5 + 5x^4 + 10x^3 + 10x^2 + 10x + 10\bigr).
\]
Carrying out the differentiation term by term:
\[
f'(x) = 10x^4 + 20x^3 + 30x^2 + 20x + 10.
\]
Step 3: Show that f(x) is Strictly Increasing
One can factor or carefully re-express f'(x) to check its sign. A known observation (from the solution steps provided) is that it can be represented in a form ensuring it is always positive. Specifically,
\[
f'(x) = 10\bigl(x^4 + 2x^3 + 3x^2 + 2x + 1\bigr).
\]
By appropriate algebraic manipulation or by noticing it stays positive for all real x , we conclude:
\[
f'(x) > 0 \quad \forall\, x \in \mathbb{R}.
\]
Hence, f(x) is strictly increasing on the entire real line.
Step 4: Determine the Number of Real Roots
Since f(x) is a strictly increasing polynomial of odd degree (5th degree), it has exactly one real root.
Step 5: Check Values at Specific Points to Locate the Root
We evaluate the polynomial at x=-1 and x=-2 to find an interval where the sign of f(x) changes:
At x = -1 :
\[
f(-1) = 2(-1)^5 + 5(-1)^4 + 10(-1)^3 + 10(-1)^2 + 10(-1) + 10 \\
= -2 + 5 - 10 + 10 - 10 + 10 = 3 > 0.
\]
At x = -2 :
\[
f(-2) = 2(-2)^5 + 5(-2)^4 + 10(-2)^3 + 10(-2)^2 + 10(-2) + 10.
\]
Calculating step by step:
\[
(-2)^5 = -32 \;,\; 2 \cdot (-32) = -64,
\]
\[
(-2)^4 = 16 \;,\; 5 \cdot 16 = 80,
\]
\[
(-2)^3 = -8 \;,\; 10 \cdot (-8) = -80,
\]
\[
(-2)^2 = 4 \;,\; 10 \cdot 4 = 40,
\]
\[
10(-2) = -20,
\]
so
\[
f(-2) = -64 + 80 - 80 + 40 - 20 + 10 = -34 < 0.
\]
Because f(-2) < 0 and f(-1) > 0 , by the Intermediate Value Theorem, the unique real root lies in the interval (-2,\, -1) .
Step 6: Identify the Required Interval and Conclude |a|
We see that the interval for the real root is (-2, -1) . This matches (a, a+1) with a = -2 . Therefore:
\[
|a| = |-2| = 2.
\]
