JEE Advance - Mathematics (1996 - No. 2)
Let $${A_n}$$ be the area bounded by the curve $$y = {\left( {\tan x} \right)^n}$$ and the
lines $$x=0,$$ $$y=0,$$ and $$x = {\pi \over 4}.$$ Prove that for $$n > 2,$$
$${A_n} + {A_{n - 2}} = {1 \over {n - 1}}$$ and deduce $${1 \over {2n + 2}} < {A_n} < {1 \over {2n - 2}}.$$
lines $$x=0,$$ $$y=0,$$ and $$x = {\pi \over 4}.$$ Prove that for $$n > 2,$$
$${A_n} + {A_{n - 2}} = {1 \over {n - 1}}$$ and deduce $${1 \over {2n + 2}} < {A_n} < {1 \over {2n - 2}}.$$
The question requires proving a recursive relationship for the area A_n.
The question only requires numerical calculation of the area A_n for specific values of n.
The question necessitates using integration by parts to establish the relationship.
The question can be solved by direct substitution without any integration.
The question involves comparing A_n with bounds derived from the established recurrence.