Let    A_n = $$\left( {{3 \over 4}} \right) - {\left( {{3 \over 4}} \right)^2} + {\left( {{3 \over 4}} \right)^3}$$ $$-$$. . . . . + ($$-$$1)^n-1 $${\left( {{3 \over 4}} \right)^n}$$    and    B_n = 1 $$-$$ A_n.
Then, the least dd natural numbr p, so that B_n > A_n , for all n$$ \ge $$ p, is :

$$\mathop {\lim }\limits_{x \to 0} {{x\tan 2x - 2x\tan x} \over {{{\left( {1 - \cos 2x} \right)}^2}}}$$ equals :

Let f(x) = $$\left\{ {\matrix{ {{{\left( {x - 1} \right)}^{{1 \over {2 - x}}}},} & {x > 1,x \ne 2} \cr {k\,\,\,\,\,\,\,\,\,\,\,\,\,\,} & {,x = 2} \cr } } \right.$$

Thevaue of k for which f s continuous at x = 2 is :

If    f(x) = sin^-1 $$\left( {{{2 \times {3^x}} \over {1 + {9^x}}}} \right),$$ then f'$$\left( { - {1 \over 2}} \right)$$ equals :