Let α be a solution of the equation 2[x+32]=3[x-64] where [x] is the greatest integer less than or equal to x and let β= prod limitsr=19 sin ((2 r-1/18)) π, then

Question

Let α be a solution of the equation 2[x+32]=3[x-64] where [x] is the greatest integer less than or equal to x and let β= prod limitsr=19 sin ((2 r-1/18)) π, then (A) [α]=[β] (B) α=(2051/8) (C) [α][(1/β)]=1 (D) [(1/α)]+[(1/β)]=28

Tardigrade Admin · Accepted Answer

We have 2[ x ]+64=3[ x ]-192 ⇒[ x ]=256 ⇒ x ∈[256,257) Also, β= sin (π/18) ⋅ sin (3 π/18) ldots ldots sin (9 π/18) ldots ldots sin (17 π/18) = sin 2 (π/18) ⋅ sin 2 (3 π/18) ⋅ sin 2 (5 π/18) ⋅ sin 2 (7 π/18)= sin 2 10° ⋅ sin 2 30° ⋅ sin 2 50° ⋅ sin 2 70° (1/4)( sin 10° ⋅ sin 50° ⋅ sin 70)2=(1/4)(( sin 30°/4))2 ∴ β=(1/4)((1/8))2=(1/256)

Q. Let $α$ be a solution of the equation $2 [x + 32] = 3 [x - 64]$ where $[x]$ is the greatest integer less than or equal to $x$ and let $β = r = 1 \prod 9 sin (\frac{2 r - 1}{18}) π$ , then

$[α] = [β]$

$α = \frac{2051}{8}$

$[α] [\frac{1}{β}] = 1$

$[\frac{1}{α}] + [\frac{1}{β}] = 2^{8}$

Q. Let α be a solution of the equation 2[x+32]=3[x−64] where [x] is the greatest integer less than or equal to x and let β=r=1∏9​sin(182r−1​)π, then

[α]=[β]

α=82051​

[α][β1​]=1

[α1​]+[β1​]=28

Q. Let $α$ be a solution of the equation $2 [x + 32] = 3 [x - 64]$ where $[x]$ is the greatest integer less than or equal to $x$ and let $β = r = 1 \prod 9 sin (\frac{2 r - 1}{18}) π$ , then

$[α] = [β]$

$α = \frac{2051}{8}$

$[α] [\frac{1}{β}] = 1$

$[\frac{1}{α}] + [\frac{1}{β}] = 2^{8}$