The negation of \( \sim S \vee ( \sim R \wedge S) \) \(\text{ is equivalent to}\)
Let A and B be sets. \[A \cap X = B \cap X = \varnothing \quad \text{and} \quad A \cup X = B \cup X \quad \text{for some set } X,\ \text{find the relation between } A \text{ and } B.\]
If \( f(x) = \lim_{x \to 0} \frac{6^x - 3^x - 2^x + 1}{\log_e 9 (1 - \cos x)} \) \(\text{ is a real number, then }\) \( \lim_{x \to 0} f(x) = \)
Consider the following frequency distribution table. \[ \begin{array}{|c|c|} \hline \textbf{Class Interval} & \textbf{Frequency} \\ \hline 10-20 & 180 \\ \hline 20-30 & f_1 \\ \hline 30-40 & 34 \\ \hline 40-50 & 180 \\ \hline 50-60 & 136 \\ \hline 60-70 & 50 \\ \hline 70-80 & f_2 \\ \hline \end{array} \] If the total frequency is 685 and the median is 42.6, then the values of \( f_1 \) and \( f_2 \) are
Let \[ f(x) = \frac{x^2 - 1}{|x| - 1}. \] \(\text{Then the value of}\) \[ \lim_{x \to 1} f(x) \text{ is:} \]
If \( |x - 6| = |x^2 - 4x| - |x^2 - 5x + 6| \), \(\text{ where \( x \) is a real variable.}\)
If \[ \mathbf{a} = \hat{i} - \hat{k}, \mathbf{b} = x\hat{i} + \hat{j} + (1 - x)\hat{k}, \mathbf{c} = y\hat{i} + x\hat{j} + (1 + x - y)\hat{k}, \] \(\text{then }\) [\(\mathbf{a}\) \(\mathbf{b}\) \(\mathbf{c}\)] \(\text{ depends on:}\)