Animated Solution for Mathematics - Basic Mathematics: Let S={x∈[−6,3]−{−2,2}:∣x∣−2∣x+3∣−1≥0} and T={x∈Z:x2−7∣x∣+9≤0}. Then the number of elements in S∩T is
Select Answer:
Visualized Solution
Defining the Sets S and T
We need to find the number of elements in S∩T.
Set S: x∈[−6,3]∖{−2,2} satisfying ∣x∣−2∣x+3∣−1≥0.
Set T: Integers x satisfying x2−7∣x∣+9≤0.
Analyzing Set S: The Numerator
Consider the inequality: ∣x∣−2∣x+3∣−1≥0
Critical points of the numerator: ∣x+3∣−1=0
∣x+3∣=1⟹x+3=1 or x+3=−1
x=−2 or x=−4
Analyzing Set S: The Denominator
Critical points of the denominator: ∣x∣−2=0
∣x∣=2⟹x=2 or x=−2
Note: The denominator cannot be zero, so x=2 and x=−2.
Solving the Inequality for S
Critical points: −4,−2,2.
The expression ∣x∣−2∣x+3∣−1≥0 is positive when numerator and denominator have the same sign.
Solving this yields: x∈(−∞,−4]∪(2,∞).
Applying Domain Constraints to S
Given domain for S: x∈[−6,3]∖{−2,2}
Intersecting with our solution: S=((−∞,−4]∪(2,∞))∩[−6,3]
S=[−6,−4]∪(2,3]
Analyzing Set T: Quadratic Inequality
Set T condition: x2−7∣x∣+9≤0 for x∈Z
Notice that x2=∣x∣2.
Let t=∣x∣. The inequality becomes: t2−7t+9≤0
Finding Roots for Set T
Solve t2−7t+9=0 using the quadratic formula:
t=2(1)−(−7)±(−7)2−4(1)(9)
t=27±49−36=27±13
Determining Integer Values for ∣x∣
13≈3.6
Roots are approximately 27−3.6=1.7 and 27+3.6=5.3
So, 1.7≤∣x∣≤5.3
Since x∈Z, ∣x∣ must be an integer: ∣x∣∈{2,3,4,5}
Listing the Elements of Set T
If ∣x∣∈{2,3,4,5}, then x can be positive or negative.
T={−5,−4,−3,−2,2,3,4,5}
Finding the Intersection S∩T
Integers in S: {−6,−5,−4,3}
Elements of T: {−5,−4,−3,−2,2,3,4,5}
Common elements (S∩T): {−5,−4,3}
Final Conclusion
The elements in S∩T are {−5,−4,3}.
The number of elements is 3.
00:00 / 00:00
The Sigma Insight: Properties of Modulus Function
Solution Diagram
Analyzing the Setup
Welcome, fellow traveler of the mathematical landscape! Today, we are embarking on a journey to solve a problem that tests your precision, logic, and ability to navigate the traps of absolute values and rational inequalities.
We are tasked with finding the intersection S∩T for two sets defined under specific constraints. Let us begin our systematic approach.
Phase 1
The Rational Inequality (Set S)
Our first challenge is to define Set S using the condition:
∣x∣−2∣x+3∣−1≥0
The domain is constrained to x∈[−6,3]∖{−2,2}.
To solve this, we identify the critical points. The numerator ∣x+3∣−1 vanishes when ∣x+3∣=1, which yields x=−2 or x=−4.
The denominator ∣x∣−2 vanishes when ∣x∣=2, giving x=2 or x=−2. Since the denominator cannot be zero, x=2 and x=−2 are strictly excluded from the solution set.
Using the Wavy Curve Method across the critical points {−4,−2,2}, we determine the expression is non-negative when x∈(−∞,−4]∪(2,∞).
Intersecting this result with the domain x∈[−6,3]∖{−2,2}, we obtain:
S=[−6,−4]∪(2,3]
Phase 2
The Quadratic Elegance (Set T)
Now, we define Set T by the condition x2−7∣x∣+9≤0, where x∈Z. Noting that x2=∣x∣2, we substitute t=∣x∣ to get the quadratic inequality:
t2−7t+9≤0
Applying the quadratic formula, the roots are:
t=27±49−36=27±13
Given 13≈3.6, the roots are approximately 1.7 and 5.3. Thus, the inequality holds for 1.7≤∣x∣≤5.3.
Since x must be an integer, ∣x∣ can only take the values 2,3,4, or 5. Consequently, the set T is:
T={−5,−4,−3,−2,2,3,4,5}
Phase 3
The Final Convergence
We now find the intersection S∩T by checking which elements of T reside within the intervals of S=[−6,−4]∪(2,3].
Evaluating the elements of T:
−5 is in [−6,−4].
−4 is in [−6,−4].
3 is in (2,3].
The values −3,−2,2,4, and 5 do not satisfy the conditions of S.
Thus, the intersection is:
S∩T={−5,−4,3}
The number of elements in the intersection is 3. We have successfully conquered the problem through rigorous logical steps.