Question

If the expression $x(e^x-1)(x-1)(x-2)^3(x-3)^5 \ge 0$ then the true solution set is

• A. [0,1][2,3]
• B. (-∞,0][1,2][3,∞]
• C. [1,2][3,∞)∪{0}
• D. (-∞,0][2,∞)

Answer 1

Answer: (C)

[1,2][3,∞)∪{0}

Answer 2

The inequality to solve is $x(e^x-1)(x-1)(x-2)^3(x-3)^5 \ge 0$.

1. Identify Critical Points: The critical points are the values of $x$ where any of the factors become zero.

• $x = 0$
• $e^x - 1 = 0 \implies e^x = 1 \implies x = 0$
• $x - 1 = 0 \implies x = 1$
• $x - 2 = 0 \implies x = 2$
• $x - 3 = 0 \implies x = 3$

The distinct critical points are $0, 1, 2, 3$. These points divide the number line into intervals: $(-\infty, 0)$, $(0, 1)$, $(1, 2)$, $(2, 3)$, $(3, \infty)$.

2. Analyze the Sign of Each Factor: Let $f(x) = x(e^x-1)(x-1)(x-2)^3(x-3)^5$. We need to determine the sign of $f(x)$ in each interval.

• Factor $x$: Positive for $x>0$, negative for $x<0$.
• Factor $e^x-1$: Positive for $x>0$, negative for $x<0$. At $x=0$, $e^x-1=0$.
• Factor $x-1$: Positive for $x>1$, negative for $x<1$.
• Factor $(x-2)^3$: The sign is determined by $(x-2)$. Positive for $x>2$, negative for $x<2$.
• Factor $(x-3)^5$: The sign is determined by $(x-3)$. Positive for $x>3$, negative for $x<3$.

3. Sign Analysis in Intervals:

• Interval $(-\infty, 0)$:

  • $x$: -
  • $e^x-1$: - (since $x<0 \implies e^x < 1$)
  • $x-1$: -
  • $(x-2)^3$: -
  • $(x-3)^5$: -
  • $f(x) = (-) \times (-) \times (-) \times (-) \times (-) = (-)$. So $f(x) < 0$.

• Interval $(0, 1)$:

  • $x$: +
  • $e^x-1$: + (since $x>0 \implies e^x > 1$)
  • $x-1$: -
  • $(x-2)^3$: -
  • $(x-3)^5$: -
  • $f(x) = (+) \times (+) \times (-) \times (-) \times (-) = (-)$. So $f(x) < 0$.

• Interval $(1, 2)$:

  • $x$: +
  • $e^x-1$: +
  • $x-1$: +
  • $(x-2)^3$: -
  • $(x-3)^5$: -
  • $f(x) = (+) \times (+) \times (+) \times (-) \times (-) = (+)$. So $f(x) > 0$.

• Interval $(2, 3)$:

  • $x$: +
  • $e^x-1$: +
  • $x-1$: +
  • $(x-2)^3$: + (since $x>2$)
  • $(x-3)^5$: -
  • $f(x) = (+) \times (+) \times (+) \times (+) \times (-) = (-)$. So $f(x) < 0$.

• Interval $(3, \infty)$:

  • $x$: +
  • $e^x-1$: +
  • $x-1$: +
  • $(x-2)^3$: +
  • $(x-3)^5$: + (since $x>3$)
  • $f(x) = (+) \times (+) \times (+) \times (+) \times (+) = (+)$. So $f(x) > 0$.

4. Include Critical Points: The inequality is $f(x) \ge 0$. This means we include the intervals where $f(x) > 0$ and the points where $f(x) = 0$. The points where $f(x) = 0$ are $0, 1, 2, 3$.

5. Combine Results: From the sign analysis, $f(x) > 0$ for $x \in (1, 2)$ and $x \in (3, \infty)$. Including the roots where $f(x)=0$: The solution set is $\{0\} \cup [1, 2] \cup [3, \infty)$.

6. Match with Options: The solution set $\{0\} \cup [1, 2] \cup [3, \infty)$ matches option (C).