(P \(\lor\) R) \(\Rightarrow\) Q}
(P \(\Rightarrow\) R) \(\lor\) (Q \(\Rightarrow\) R)}
(P \(\Rightarrow\) R) \(\land\) (Q \(\Rightarrow\) R)
(P \(\land\) R) \(\Rightarrow\) Q
We are given the statement (P \(\Rightarrow\) Q) \(\land\) (R \(\Rightarrow\) Q). We know that P \(\Rightarrow\) Q is equivalent to \(\neg\)P \(\lor\) Q. So, the given statement can be written as: \[ (\neg P \lor Q) \land (\neg R \lor Q). \] Using the distributive law, we can rewrite this as: \[ (\neg P \land \neg R) \lor Q. \] Using De Morgan's law, \(\neg P \land \neg R\) is equivalent to \(\neg(P \lor R)\). So the statement becomes: \[ \neg(P \lor R) \lor Q. \] This is equivalent to (P \(\lor\) R) \(\Rightarrow\) Q.
Let \( y^2 = 12x \) be the parabola and \( S \) its focus. Let \( PQ \) be a focal chord of the parabola such that \( (SP)(SQ) = \frac{147}{4} \). Let \( C \) be the circle described by taking \( PQ \) as a diameter. If the equation of the circle \( C \) is: \[ 64x^2 + 64y^2 - \alpha x - 64\sqrt{3}y = \beta, \] then \( \beta - \alpha \) is equal to:
Let one focus of the hyperbola $ \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 $ be at $ (\sqrt{10}, 0) $, and the corresponding directrix be $ x = \frac{\sqrt{10}}{2} $. If $ e $ and $ l $ are the eccentricity and the latus rectum respectively, then $ 9(e^2 + l) $ is equal to:
Let $ A \in \mathbb{R} $ be a matrix of order 3x3 such that $$ \det(A) = -4 \quad \text{and} \quad A + I = \left[ \begin{array}{ccc} 1 & 1 & 1 \\2 & 0 & 1 \\4 & 1 & 2 \end{array} \right] $$ where $ I $ is the identity matrix of order 3. If $ \det( (A + I) \cdot \text{adj}(A + I)) $ is $ 2^m $, then $ m $ is equal to: