To find the number of real roots for the equation $x^7 + 4x^5 + 2x^3 + x + 1 = 0$, we can analyze the function using calculus and the properties of polynomials.
1. Identify the Function
Let $f(x) = x^7 + 4x^5 + 2x^3 + x + 1$. We want to find the number of values for $x$ where $f(x) = 0$.
2. Analyze the Derivative
The first derivative tells us about the slope and whether the function is increasing or decreasing:
$$f'(x) = 7x^6 + 20x^4 + 6x^2 + 1$$
Notice the following about $f'(x)$:
All the powers of $x$ are even ($x^6, x^4, x^2$).
All the coefficients are positive.
The constant term is $+1$.
Since $x^{2n} \ge 0$ for all real $x$, then $f'(x) \ge 1$ for all real $x$. Because the derivative is strictly positive everywhere, the function $f(x)$ is strictly increasing over its entire domain.
3. Determine the Number of Roots
Existence: $f(x)$ is a polynomial of odd degree (degree 7). As $x \to \infty$, $f(x) \to \infty$. As $x \to -\infty$, $f(x) \to -\infty$. By the Intermediate Value Theorem, the graph must cross the x-axis at least once.
Uniqueness: Since the function is strictly increasing, it can cross the x-axis exactly once. If it crossed twice, the function would have to decrease at some point to return to the axis, which would require $f'(x)$ to be zero or negative.
Conclusion
The equation has exactly 1 real root.
Correct Option: A. 1
