Question

Let 

\(S=\left\{x∈[-6,3]-\left\{-2,2 \right\} :\frac{|x+3|-1}{|x|-2}>=0 \right\} \space and \space \left\{T={x∈Z:x^2-7|x|+9<=0}\right\}\)

Then the number of elements in S ⋂ T is

• A. 7
• B. 5
• C. 4
• D. 3

Answer 1

The Correct Option is D

 To solve this problem, we need to find the number of elements in the intersection of the sets \( S \) and \( T \). Let's determine each set step-by-step.

Finding set \( S \):

Given:

\(S = \{ x \in [-6, 3] - \{-2, 2\} : \frac{|x+3|-1}{|x|-2} \geq 0 \}\)

1. Determine the domain: \( x \) must be within the interval \([-6, 3]\) except at \( x = -2 \) and \( x = 2 \).
2. Solve the inequality: \(\frac{|x+3|-1}{|x|-2} \geq 0\).
   1. The expression is undefined for \( x = 2 \), so we exclude 2.
   2. Numerator: \( |x+3| - 1 \) can be non-negative for \( |x+3| \geq 1 \):
      
      • \( x + 3 \geq 1 \rightarrow x \geq -2 \)
      • \( -(x + 3) \geq 1 \rightarrow x \leq -4 \)
   3. Denominator: \( |x| - 2 \geq 0 \) implies:
      • \( x \geq 2 \)
      • -\(x \geq 2 \rightarrow x \leq -2 \)
   4. Combining:\( x \leq -4 \) or \( x \geq 2 \)
3. Elements of \( S \): Intersection of inequalities within domain leads to \( S = \{-6, -5, -4, 2, 3\} \).

Finding set \( T \):

Given:

\(T = \{ x \in \mathbb{Z} : x^2 - 7|x| + 9 \leq 0 \}\)

1. Consider separation into cases based on absolute value:
   1. For \( x \geq 0 \): \( x^2 - 7x + 9 \leq 0 \).
   2. Find roots by solving \( x^2 - 7x + 9 = 0 \).
      • Roots: \( x = \frac{7 \pm \sqrt{49 - 36}}{2} = \frac{7 \pm \sqrt{13}}{2} \), not exact integers.
2. For \( x \leq 0 \), \( y = -x \) implies \( y^2 - 7y + 9 \leq 0 \). Similar derivation leads to no integer values.

Intersection of \( S \cap T \): Consider elements common to both sets.

• Elements of \( S \): \{-6, -5, -4, 3\}
• Common elements in \( \{3\} \) since others do not satisfy both sets.

The number of elements in \( S \cap T \) is 3, hence the correct answer is 3.