The volume of 0.1N dibasic acid sufficient to neutralize 1 g of a base that furnishes 0.04 mole of OH⁻ in aqueous solution is :

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

If    $${{{}^{n + 2}C{}_6} \over {{}^{n - 2}{P_2}}}$$ = 11, then n satisfies the equation :

The sum $$\sum\limits_{r = 1}^{10} {\left( {{r^2} + 1} \right) \times \left( {r!} \right)} $$ is equal to :