Let `n` be a positive integer such that `sinpi/(2n)+cospi/(2n)=(sqrt(n))/2dot` Then `6lt=nlt=8` (b) `4

Options

A. `n in [6, 8]`
B. `n in (4, 8]`
C. `n in [4, 8)`
D. `n in (4, 8)`

Answer

D

Explanation

We have,
`"sin" (pi)/(2n) + "cos" (pi)/(2n) = (sqrt(n))/(2)`
`rArr "sin" ((pi)/(2n) + (pi)/(4)) = (sqrt(n))/(2sqrt(2)) " and " 1 + "sin" (pi)/(n) = (n)/(4)`
`rArr "sin" ((pi)/(2n) + (pi)/(4)) = (sqrt(n))/(2 sqrt(2)) " and sin" (pi)/(n) = (n)/(4) -1`
`rArr (sqrt(n))/(2sqrt(2)) lt 1 " for all" n gt 2 " and " (n)/(4) - 1 gt0 "for all "n gt 1`
`rArr n lt 8 " for all" n gt 2 "and" n gt 4" for all" n gt 1`
`rArr n in (2, 8) " and " n in (4, oo)`
`rArr n in (4, 8)`