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Question: Let ƒ(x) = x – [x], where [x] denotes the greatest integer ≤ x and g(x) =\(f(x) = \frac{a^{x} - a^{-...

Let ƒ(x) = x – [x], where [x] denotes the greatest integer ≤ x and g(x) =f(x)=ax−a−xax+a−xf(x)=sin⁡xf(x) = \frac{a^{x} - a^{- x}}{a^{x} + a^{- x}}f(x) = \sin x , then g(x) is equal to –

A

0

B

1

C

–1

D

None of these

Answer

–1

Explanation

Solution

As 0 ≤ x – [x] < 1 ∀ x ∈ R, 0 ≤ ƒ(x) < 1.

∴{ƒ(x)}2n = 0.

Thus, for x ∈ R, g(x) ={f(x)}2n−1{f(x)}2n+1\frac { \{ f ( \mathrm { x } ) \} ^ { 2 \mathrm { n } } - 1 } { \{ f ( \mathrm { x } ) \} ^ { 2 \mathrm { n } } + 1 }= 0−10+1\frac { 0 - 1 } { 0 + 1 } = –1.