If \( 2 \) is a solution of the inequality \( \frac{x-a}{a-2x}<-3 \), then \( a \) must lie in the interval:
First, substitute \( x = 2 \) into the inequality and simplify: \[ \frac{2-a}{a-4}<-3. \] Multiply both sides by \( a-4 \) (assuming \( a \neq 4 \)) to avoid reversing the inequality: \[ 2 - a<-3(a - 4). \] Expanding and simplifying yields: \[ 2 - a<-3a + 12 \quad \Rightarrow \quad 2a<10 \quad \Rightarrow \quad a<5. \]
Now, because the inequality assumes \( a - 4 \) is positive (so we do not reverse the inequality sign when multiplying), it implies \( a>4 \).
Conclusion: Combining \( a>4 \) and \( a<5 \), we find that \( a \) must lie in the interval \( (4,5) \), matching option (A).
The focus of the parabola \(y^2 + 4y - 8x + 20 = 0\) is at the point:
Let \( S \) denote the set of all subsets of integers containing more than two numbers. A relation \( R \) on \( S \) is defined by:
\[ R = \{ (A, B) : \text{the sets } A \text{ and } B \text{ have at least two numbers in common} \}. \]
Then the relation \( R \) is:
The centre of the hyperbola \(16x^2 - 4y^2 + 64x - 24y - 36 = 0\) is at the point: