Given $${f_1},$$ $${f_3},$$ and $$f$$ in canonical sum of products form (in decimal) for the circuit. $$${f_1} = \sum {m\left( {4,5,6,7,8} \right)} $$$ $$${f_3} = \sum {m\left( {1,6,15} \right)} $$$ $$$f = \sum {m\left( {1,6,8,15} \right)} $$$
Then $${f_2}$$ is

$$\mathop {\lim }\limits_{x \to \infty } {{x - \sin x} \over {x + \cos \,x}}\,\,Equals$$

A binary tree with $$n>1$$ nodes has $${n_1}$$, $${n_2}$$ and $${n_3}$$ nodes of degree one, two and three respectively. The degree of a node is defined as the number of its neighbours.

Starting with the above tree, while there remains a node $$v$$ of degree two in the tree, add an edge between the two neighbours of $$v$$ and then remove $$v$$ from the tree. How many edges will remain at the end of the process?

A binary tree with $$n>1$$ nodes has $${n_1}$$, $${n_2}$$ and $${n_3}$$ nodes of degree one, two and three respectively. The degree of a node is defined as the number of its neighbours.

$${n_3}$$ can be expressed as: