Given Equation: \([x]^2 - 5[x] + 6 = 0\), where \([x]\) is the greatest integer function.
Step 1: Solve the quadratic equation for \([x]\).
Let \(y = [x]\). The equation becomes:
\(y^2 - 5y + 6 = 0\)
Factorizing: \((y - 2)(y - 3) = 0\)
Solutions: \(y = 2\) or \(y = 3\)
Step 2: Translate back to \([x]\).
Case 1: \([x] = 2\) ⇒ \(x \in [2, 3)\)
Case 2: \([x] = 3\) ⇒ \(x \in [3, 4)\)
Step 3: Combine the intervals.
The solution is \(x \in [2, 3) \cup [3, 4)\) = \([2, 4)\)
Option Analysis:
(A) \(x \in [3, 4]\) - Partial solution (misses [2,3))
(B) \(x \in [2, 4]\) - Includes all solutions (correct)
(C) \(x \in [2, 3]\) - Partial solution (misses [3,4))
(D) \(x \in (2, 3]\) - Incorrect boundaries
Final Answer: \(\boxed{B}\)
Let \( [x] \) denote the greatest integer less than or equal to \( x \). We are given the equation:
\[ [x]^2 - 5[x] + 6 = 0 \]
This is a quadratic equation in \( [x] \). We can factor it as follows:
\[ ([x] - 2)([x] - 3) = 0 \]
This gives two possible values for \( [x] \):
\[ [x] = 2 \quad \text{or} \quad [x] = 3 \]
If \( [x] = 2 \), then \( 2 \leq x < 3 \).
If \( [x] = 3 \), then \( 3 \leq x < 4 \).
Therefore, the solution for \( x \) is the interval \( 2 \leq x < 4 \). In interval notation, this is \( [2, 4) \).
Final Answer: The final answer is \( {[2,4)} \).
You are given a dipole of charge \( +q \) and \( -q \) separated by a distance \( 2l \). A sphere 'A' of radius \( R \) passes through the centre of the dipole as shown below and another sphere 'B' of radius \( 2R \) passes through the charge \( +q \). Then the electric flux through the sphere A is