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).
Let \( f(x) = \frac{x^2 + 40}{7x} \), \( x \neq 0 \), \( x \in [4,5] \). The value of \( c \) in \( [4,5] \) at which \( f'(c) = -\frac{1}{7} \) is equal to:
The general solution of the differential equation \( \frac{dy}{dx} = xy - 2x - 2y + 4 \) is:
The minimum value of the function \( f(x) = x^4 - 4x - 5 \), where \( x \in \mathbb{R} \), is:
The critical points of the function \( f(x) = (x-3)^3(x+2)^2 \) are:
For the reaction:
\[ 2A + B \rightarrow 2C + D \]
The following kinetic data were obtained for three different experiments performed at the same temperature:
\[ \begin{array}{|c|c|c|c|} \hline \text{Experiment} & [A]_0 \, (\text{M}) & [B]_0 \, (\text{M}) & \text{Initial rate} \, (\text{M/s}) \\ \hline I & 0.10 & 0.10 & 0.10 \\ II & 0.20 & 0.10 & 0.40 \\ III & 0.20 & 0.20 & 0.40 \\ \hline \end{array} \]
The total order and order in [B] for the reaction are respectively: