Question

The number of (+) ve integral solution of

$\frac{x^2(3x-4)^3(x-2)^4}{(x-5)^5(2x-7)^6}\leq0$ is

Answer 1

3

Answer 2

To solve the inequality $\frac{x^2(3x-4)^3(x-2)^4}{(x-5)^5(2x-7)^6}\leq0$, we follow these steps:

1. Identify Critical Points: Set each factor in the numerator and denominator to zero to find the critical points:

   • $x^2 = 0 \implies x = 0$ (Even power, so no sign change across $x=0$)
   • $(3x-4)^3 = 0 \implies 3x-4 = 0 \implies x = 4/3$ (Odd power, so sign changes across $x=4/3$)
   • $(x-2)^4 = 0 \implies x-2 = 0 \implies x = 2$ (Even power, so no sign change across $x=2$)
   • $(x-5)^5 = 0 \implies x-5 = 0 \implies x = 5$ (Odd power, so sign changes across $x=5$)
   • $(2x-7)^6 = 0 \implies 2x-7 = 0 \implies x = 7/2$ (Even power, so no sign change across $x=7/2$)

   Order the critical points on a number line: $0, 4/3 (\approx 1.33), 2, 7/2 (= 3.5), 5$.

2. Analyze Signs in Intervals (Wavy Curve Method): Pick a test value in the rightmost interval, say $x=6$: $\frac{6^2(3 \cdot 6 - 4)^3(6 - 2)^4}{(6 - 5)^5(2 \cdot 6 - 7)^6} = \frac{36(14)^3(4)^4}{(1)^5(5)^6} = \frac{\text{positive}}{\text{positive}} > 0$. So, for $x > 5$, the expression is positive.

   Now, move left across the critical points, changing the sign only when crossing a critical point with an odd power:

   • For $x > 5$: Positive (+)
   • At $x=5$ (odd power): Sign changes. So, $x \in (7/2, 5)$ is Negative (-).
   • At $x=7/2$ (even power): Sign does not change. So, $x \in (2, 7/2)$ is Negative (-).
   • At $x=2$ (even power): Sign does not change. So, $x \in (4/3, 2)$ is Negative (-).
   • At $x=4/3$ (odd power): Sign changes. So, $x \in (0, 4/3)$ is Positive (+).
   • At $x=0$ (even power): Sign does not change. So, $x \in (-\infty, 0)$ is Positive (+).

   Summary of signs:

   • $x \in (-\infty, 0)$: Positive
   • $x \in (0, 4/3)$: Positive
   • $x \in (4/3, 2)$: Negative
   • $x \in (2, 7/2)$: Negative
   • $x \in (7/2, 5)$: Negative
   • $x \in (5, \infty)$: Positive

3. Determine the Solution Set for $\leq 0$: We need the intervals where the expression is negative ($< 0$) or zero ($= 0$).

   • Where expression is negative: $x \in (4/3, 2) \cup (2, 7/2) \cup (7/2, 5)$. This can be compactly written as $x \in (4/3, 5)$ excluding $x=2$ and $x=7/2$.
   • Where expression is zero: The numerator becomes zero. This occurs at $x=0$, $x=4/3$, and $x=2$.
   • Excluded points: The denominator cannot be zero. So, $x \neq 5$ and $x \neq 7/2$.

   Combining these, the solution set for $\frac{x^2(3x-4)^3(x-2)^4}{(x-5)^5(2x-7)^6}\leq0$ is: $x \in \{0\} \cup [4/3, 7/2) \cup (7/2, 5)$.

4. Find Positive Integral Solutions: We are looking for positive integers ($x > 0$ and $x$ is an integer) that satisfy the inequality.

   • $x=0$ is not a positive integer.
   • Consider the interval $[4/3, 7/2) \cup (7/2, 5)$.
     • $4/3 \approx 1.33$
     • $7/2 = 3.5$
     • $5$
   • Integers in $[1.33, 3.5)$:
     • $x=2$: $2 \in [1.33, 3.5)$, and it makes the expression zero, so it's a solution.
     • $x=3$: $3 \in [1.33, 3.5)$, and it makes the expression negative, so it's a solution.
   • Integers in $(3.5, 5)$:
     • $x=4$: $4 \in (3.5, 5)$, and it makes the expression negative, so it's a solution.

   The positive integral solutions are $2, 3, 4$. The number of positive integral solutions is 3.

The final answer is $\boxed{3}$.

Explanation of the solution:

1. Identify critical points from numerator and denominator: $0, 4/3, 2, 7/2, 5$.
2. Use the wavy curve method. Factors with even powers ($x^2$, $(x-2)^4$, $(2x-7)^6$) do not change the sign across their critical points. Factors with odd powers ($(3x-4)^3$, $(x-5)^5$) change the sign.
3. Test an interval (e.g., $x>5$) to determine the initial sign. For $x>5$, the expression is positive.
4. Move left across critical points, applying sign changes based on power parity.
   • $(5, \infty)$: +
   • $(7/2, 5)$: - (sign change at 5 due to odd power)
   • $(2, 7/2)$: - (no sign change at 7/2 due to even power)
   • $(4/3, 2)$: - (no sign change at 2 due to even power)
   • $(0, 4/3)$: + (sign change at 4/3 due to odd power)
   • $(-\infty, 0)$: + (no sign change at 0 due to even power)
5. Identify regions where the expression is $\leq 0$. This includes intervals where it's negative and points where it's zero.
   • Negative intervals: $(4/3, 2) \cup (2, 7/2) \cup (7/2, 5)$.
   • Zero points (from numerator): $x=0, x=4/3, x=2$.
   • Excluded points (from denominator): $x=5, x=7/2$.
6. Combine: The solution set is $\{0\} \cup [4/3, 7/2) \cup (7/2, 5)$.
7. Find positive integers in this set: $x=2, 3, 4$.
8. Count the solutions: There are 3 positive integral solutions.