JEE Main · 2020 · 08 Jan 2020 · Shift 1 · Sets Relations and Functions

The inverse function of

f (x) = \frac{8 ^{2 x} - 8 ^{- 2 x}}{8 ^{2 x} + 8 ^{- 2 x}}, x \in (- 1, 1)

, is

Answer: (D)

Solution

Set

y = \frac{8 ^{2 x} - 8 ^{- 2 x}}{8 ^{2 x} + 8 ^{- 2 x}}

and solve for

x

. Cross-multiplying gives

8^{2 x} (1 - y) = 8^{- 2 x} (1 + y)

, so

8^{4 x} = \frac{1 + y}{1 - y}

and

4 x = lo g_{8} (\frac{1 + y}{1 - y})

. Hence

x = \frac{1}{4} lo g_{8} (\frac{1 + y}{1 - y}) = \frac{1}{4} (lo g_{8} e) ln (\frac{1 + y}{1 - y})

, and writing the inverse in its own variable,

f^{- 1} (x) = \frac{1}{4} (lo g_{8} e) lo g_{e} (\frac{1 + x}{1 - x})

JEE Main · 2020 · 08 Jan 2020 · Shift 1 · Single correct option