Question

If sin^–1 x + sin^–1y + sin^–1z = $\frac { 3 \pi } { 2 }$and f(1) = 2, f(p +q) = f(p). f(q) " p, q Ī R, then

$\frac { x + y + z } { x ^ { f ( 1 ) } + y ^ { f ( 2 ) } + z ^ { f ( 3 ) } }$=

• A. 0
• B. 1
• C. 2
• D. 3

Answer 1

Answer: (C)

2

Answer 2

– $\frac { \pi } { 2 }$ £ sin^–1 x £ $\frac { \pi } { 2 }$

\ sin^–1 x + sin^–1 y + sin^{– 1}z = $\frac { 3 \pi } { 2 }$

Ū sin^–1 x = sin^–1 y = sin^{– 1}z = $\frac { \pi } { 2 }$

Ū x = y = z = 1

Also f(p + q) = f(p). f(q) " p, q Ī R …(1)

Given f(1) = 1

from (1),

F(1 + 1) = f(1). F(1) Ž f(2) =1² = 1

from (2), f(2 + 1) = f(2) . f(1)

Ž f(3) = 1². 1 = 1³ = 1

Now given expression = 3 – $\frac { 3 } { 3 }$ = 2