The number of sp² hybrid orbitals in a molecule of benzene is :

Let a, b $$ \in $$ R, a $$ \ne $$ 0 be such that the equation, ax² – 2bx + 5 = 0 has a repeated root $$\alpha $$, which is also a root of the equation, x² – 2bx – 10 = 0. If $$\beta $$ is the other root of this equation, then $$\alpha $$² + $$\beta $$² is equal to :

Let a function ƒ : [0, 5] $$ \to $$ R be continuous, ƒ(1) = 3 and F be defined as :

$$F(x) = \int\limits_1^x {{t^2}g(t)dt} $$ , where $$g(t) = \int\limits_1^t {f(u)du} $$

Then for the function F, the point x = 1 is :

In the expansion of $${\left( {{x \over {\cos \theta }} + {1 \over {x\sin \theta }}} \right)^{16}}$$, if $${\ell _1}$$ is the least value of the term independent of x when $${\pi \over 8} \le \theta \le {\pi \over 4}$$ and $${\ell _2}$$ is the least value of the term independent of x when $${\pi \over {16}} \le \theta \le {\pi \over 8}$$, then the ratio $${\ell _2}$$ : $${\ell _1}$$ is equal to :