Question

$(x^2 + 1) (x-2)^2 (x-3) < 0$, then x belongs to

• A. (-\infty, 2) \cup (2, 3)
• B. (-\infty, 3)
• C. (2, 3)
• D. none of these

Answer 1

Answer: (A)

(-\infty, 2) \cup (2, 3)

Answer 2

The given inequality is $(x^2 + 1) (x-2)^2 (x-3) < 0$.

1. Analyze the factors:

   • $(x^2 + 1)$: This term is always positive for all real values of $x$ since $x^2 \ge 0$.
   • $(x-2)^2$: This term is always non-negative. It is zero when $x=2$ and positive for all other real values of $x$.
   • $(x-3)$: This term is negative for $x < 3$, zero at $x=3$, and positive for $x > 3$.

2. Simplify the inequality: Since $(x^2 + 1)$ is always positive, it does not affect the sign of the inequality. We can divide both sides by $(x^2+1)$ without changing the inequality sign. The inequality becomes $(x-2)^2 (x-3) < 0$.

3. Determine conditions for the simplified inequality: For the product $(x-2)^2 (x-3)$ to be negative, we must have:

   • $(x-2)^2 > 0$ (since it cannot be negative, it must be positive)
   • $(x-3) < 0$ (to make the product negative)

4. Solve the conditions:

   • $(x-2)^2 > 0$ implies that $x-2 \neq 0$, so $x \neq 2$.
   • $(x-3) < 0$ implies that $x < 3$.

5. Combine the conditions: We need $x < 3$ and $x \neq 2$. This means $x$ can be any real number less than 3, except for 2. This set of values can be represented as the union of two intervals: $(-\infty, 2)$ and $(2, 3)$.

6. Final Solution: The solution set is $(-\infty, 2) \cup (2, 3)$.

Visualizing the Sign Analysis (Wave Curve Method)

The critical points are the roots of the factors: $x=2$ (from $(x-2)^2$) and $x=3$ (from $x-3$). The term $(x-2)^2$ has a root at $x=2$ with multiplicity 2 (even). The term $(x-3)$ has a root at $x=3$ with multiplicity 1 (odd). The leading term of the polynomial $(x^2+1)(x-2)^2(x-3)$ is $x^2 \cdot x^2 \cdot x = x^5$, which is positive. Thus, for very large positive $x$, the expression is positive.

The inequality $(x^2 + 1) (x-2)^2 (x-3) < 0$ holds when the expression is negative. This occurs in the intervals $(-\infty, 2)$ and $(2, 3)$. Since the inequality is strict, the roots $x=2$ and $x=3$ are not included.