Number of Real Roots of Quintic Polynomial

To determine the number of real roots of the quintic polynomial function $f(x) = x^5 - 5x + a$ for a given $a \in \mathbb{R}$, we analyze the local extrema and monotonicity of $f(x)$.

First, we find the first derivative of $f(x)$: $f'(x) = 5x^4 - 5 = 5(x^4 - 1) = 5(x^2 - 1)(x^2 + 1) = 5(x - 1)(x + 1)(x^2 + 1)$

Setting $f'(x) = 0$ gives the critical points: $x = -1 \quad \text{and} \quad x = 1$

We analyze the sign of $f'(x)$ to determine the intervals of monotonicity:

• For $x \in (-\infty, -1)$, $f'(x) > 0$, so $f(x)$ is strictly increasing.
• For $x \in (-1, 1)$, $f'(x) < 0$, so $f(x)$ is strictly decreasing.
• For $x \in (1, \infty)$, $f'(x) > 0$, so $f(x)$ is strictly increasing.

Therefore:

• $x = -1$ is a point of local maximum with local maximum value: $f(-1) = (-1)^5 - 5(-1) + a = a + 4$
• $x = 1$ is a point of local minimum with local minimum value: $f(1) = 1^5 - 5(1) + a = a - 4$

We also observe the asymptotic behavior of $f(x)$: $\lim_{x \to -\infty} f(x) = -\infty \quad \text{and} \quad \lim_{x \to \infty} f(x) = \infty$

Now, we evaluate the number of real roots in different cases for $a$:

1. Case 1: $a > 4$ Here, $f(1) = a - 4 > 0$ and $f(-1) = a + 4 > 8 > 0$.

   • On the interval $(-\infty, -1]$, $f(x)$ increases continuously from $-\infty$ to $a + 4 > 0$. By the Intermediate Value Theorem, there is exactly one real root in $(-\infty, -1)$.
   • On the interval $[-1, 1]$, $f(x)$ decreases from $a + 4 > 0$ to $a - 4 > 0$. Since $f(x) > 0$ throughout this interval, there are no real roots in $[-1, 1]$.
   • On the interval $[1, \infty)$, $f(x)$ increases from $a - 4 > 0$ to $\infty$. Since $f(x) > 0$ throughout this interval, there are no real roots in $[1, \infty)$.

   Thus, for $a > 4$, $f(x)$ has only one real root. This makes option (B) correct and option (A) false.

2. Case 2: $a < -4$ Here, $f(-1) = a + 4 < 0$ and $f(1) = a - 4 < -8 < 0$.

   • On $(-\infty, -1]$, $f(x)$ increases from $-\infty$ to $a + 4 < 0$, so there are no real roots.
   • On $[-1, 1]$, $f(x)$ decreases from $a + 4 < 0$ to $a - 4 < 0$, so there are no real roots.
   • On $[1, \infty)$, $f(x)$ increases from $a - 4 < 0$ to $\infty$, giving exactly one real root.

   Thus, for $a < -4$, $f(x)$ has only one real root. This makes option (C) false.

3. Case 3: $-4 < a < 4$ Here, $f(-1) = a + 4 > 0$ and $f(1) = a - 4 < 0$.

   • On $(-\infty, -1]$, $f(x)$ increases from $-\infty$ to $f(-1) > 0$, yielding one real root.
   • On $[-1, 1]$, $f(x)$ decreases from $f(-1) > 0$ to $f(1) < 0$, yielding one real root.
   • On $[1, \infty)$, $f(x)$ increases from $f(1) < 0$ to $\infty$, yielding one real root.

   Thus, for $-4 < a < 4$, $f(x)$ has three real roots. This makes option (D) correct.

Hence, the correct options are (B) and (D).