If sin - 1 x + sin - 1 y + sin - 1 z = 3 π 2 and f ( 1 ) = 2 , f ( p + q ) = f ( p ) · f ( q ) , ∀ p , q ∈ R , then x f ( 1 ) + y f ( 2 ) + z f ( 3 ) - ( x + y + z ) x f ( 1 ) + y f ( 2 ) + z f ( 3 ) is equal to

Step-by-step solution

∵   - π 2 ≤ sin - 1 x ≤ π 2 , - π 2 ≤ sin - 1 y ≤ π 2 and - π 2 ≤ sin - 1 z ≤ π 2 Given that, sin - 1 x + sin - 1 y + sin - 1 z = 3 π 2 Which is possible only when sin - 1 x = sin - 1 y = sin - 1 z = π 2 ⇒   x = y = z = 1 Put p = q = 1 Then, f ( 2 ) = f ( 1 ) f ( 1 ) = 2 × 2 = 4 and put p = 1 , q = 2 then, f ( 3 ) = f ( 1 )   f ( 2 ) = 2 · 2 2 = 8 ∴   x f ( 1 ) + y f ( 2 ) + z f ( 3 ) - x + y