Consider the following two propositions:

$$P1: \sim (p \to \sim q)$$

$$P2:(p \wedge \sim q) \wedge (( \sim p) \vee q)$$

If the proposition $$p \to (( \sim p) \vee q)$$ is evaluated as FALSE, then :

If $${1 \over {2\,.\,{3^{10}}}} + {1 \over {{2^2}\,.\,{3^9}}} + \,\,.....\,\, + \,\,{1 \over {{2^{10}}\,.\,3}} = {K \over {{2^{10}}\,.\,{3^{10}}}}$$, then the remainder when K is divided by 6 is :

Let f(x) be a polynomial function such that $$f(x) + f'(x) + f''(x) = {x^5} + 64$$. Then, the value of $$\mathop {\lim }\limits_{x \to 1} {{f(x)} \over {x - 1}}$$ is equal to:

Let E₁ and E₂ be two events such that the conditional probabilities $$P({E_1}|{E_2}) = {1 \over 2}$$, $$P({E_2}|{E_1}) = {3 \over 4}$$ and $$P({E_1} \cap {E_2}) = {1 \over 8}$$. Then :